Singapore Mthematical oiety . Singapore Mathematical Oymiad (SMO)
201
(Junior Section)
Tuesday, 1 June 20 10
0930- 1200
Important:
Answer ALL 35 questions. Enter yur answers on the answer sheet provided. For the multiple choice questions, enter ur nswer on the answer sheet b shding th bubble containing the letter (A, B C D or E) corresponding to the correct answer. or the other short questions, write our nswer on the answer sheet and shade the appropriate bubbles below our answe No steps are needed to just our answers. Each question carries1 mark. No calculators are allowed.
PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO D S.
1
Multipe Choie Questions 1 mong the ve eal numbes below, which one is the smallest?
()
(A)
(C) 2010;
2010
D
;
2009
2009
.
2010
2 mong the ve integes below, which one is the lgest?
3
mong the fou statements on eal numbes below, how my of them e coect? "If a
"If a
(A) 0;
() 1;
5.
() 2010;
"If a2
D3 4.
(C) 2;
4. What is e lgest intege less than or equal to (A) 2009;
"If a
(C) 2011;
3 x (2010)+4 x 2010+1?
+
D 2012; None of he above.
The conditions of te oad beteen Town A and Town B can be classied as up slope, oizontal o down slope nd total lengt of each type of oad is the same. A cyclist tavels om Town A to Town B with unifom speeds 8 k, 12 k and 24 k on the up slope, oizont and down slope oad espectively. What is the aveage speed of his jouey?
12kf -
Town A
(A) 12 k;
(B)� k;
Town B
(C 16 k;
2
17 k E 18 k
6. In te diagra, M and E ae eqilatera tinges Given that LE LAE = X0, wht is the vae of x?
62° and
A
(A) 100;
(B)
118; (C) 120;
(D) 122; (E) 135.
7 A caener wishes to ct a wooden 3 x 3 x 3 cbe into twenty seven 1 x 1 x 1 cbes He can do hs easiy by mng 6 cuts thrgh te cube, keeping the pieces together n e cbe shape as shown:
. . . . :· .
·····················
What is te miimm nmber of cuts needed if he is alowed to reaange the pieces after each cut? (A) 2;
(B) 3;
(C) 4; (D) 5; (E) 6.
8. What is he ast digit of 7 ? (A) 1;
(B) 3; (C) 5; (D) 7; (E) 9 3
9 Give that n is a odd iteger less tha 1000 ad the product of a its digits is 252 How ay such itegers e there ? (A) 3;
(B) 4;
(C) 5;
D 6;
E 7.
10 What is e valu of
(A) 451856;
(B) 691962;
(C) 903712;
D 1276392;
E 1576392
Short Questions
11. Let x ad y be ea umbers satisfyig y=
2008x2009 + 2010x- 2011
2008x2009 2010· 201-
Fid te value of .
12 For itegers ah . . , an E {1, 2, 3, . , 9}, we use the otatio aa •••an to deote the umber n n 10 a10 a · · · 10an an. or example, whe a = 2 d b = 0, deotes te umber 20 Give tat ab = b ad ace = (ba)• id he value of ab. 13 Gie that (m- 2) is a positive ieger ad it is also a factor of 3m- 2m10 Fid te sum of all such values of m 14. In tiagle ABC AB = 32 m, AC = 36 ad BC = 44 m. If M is the midpoit of BC d the lg of AM i m. A
B
M
4
c
15. Evaluate
678+ 690+72+ 74+...+ 1998+200 3+9+15+21+... 2+ 333
16 Esher and Frida ae supposed to l a rectangul ray of 16 columns and 10 rws, with he numbers 1 t 160 Esther chose to do it row-wise so that he rst row i ubered 1, 2, , 16 and the secnd row is 7, 18, . .., 32 and so on. Frida chose to do i column wse, so that her rst colmn has , 2, . ., 0, and he second comn has 1, 12, , 20 nd s on. Comping Esther's ray with Fria's aay, we notice hat some numbers ocupy the same position. Find the sum of the nubers n these positions. 17
... ... 45
.. 3 19 . . ... .. ... ... 146 47 ... 2 18 ... .
... 16 . . . 32 . . ... . . ... .. . 16 .
.
1 2 .. . ..
11 12 ... ...
21 22 .. ...
10
2
30
Esther
.. ... ... .. . ..
... ... .. . ... ...
151 152 .. ..
160
Fida
17. The sum of two integers A and B is 21. If the owes common multiple of A and B is 14807, write down the ger of the wo ntegers A or B
18. A sequence f olynoils an(x) e dene recursivey by a0x a(x) an(X)
=
=
=
1, x+x+1, (�+1a-IX) a_z(X), for all n
2
2.
For
az(x) x)
=
=
=
Evaluate
(�+1)( +x+ 1) 1 x4 + x + 2+x, (� +1)(x4+x+2�+ x (�+x+1) =
-
x7 +x6+2 +2x4 x +� 1.
a 010(). 2
19 A triangl AB is nsribed i a secrcle of rais 5. f AB of s where s A+B
5
=
0, nd the maxmm value
20. Fin the ast to igits o 2011 <>. 21. Your nationa footba coach brought a squa of 18 ayers to the 2010 Wor Cu, onsisting of 3 goakeepes, 5 efeners, 5 ieders an 5 stikers. ieers ae versatie enough to pay as both efeners an mideders, whie the oer payers can ony pay in their esignate positions. How many possbe teams of 1 goakeeper, 4 efeners, 4 mieers an 2 srikers can the coach el
22. Given that 169(157-77x +10020-100x = 2677x-1571000x-2010, n the vaue of x. 23. Evauate
2020-20100)20100-1002000 +20100 2010-10
24. hen 15 is ae to a number x, it becomes a squae number. When 74 is subracte om x, e resut is again a squae number. Fin te number x. 25. Given that x an e positve integers suc h tht 56: x+: 59 an 0.9 e vaue of f-x.
<
:
<
0.91, n
26. Let A' nd BB' be two ine segmnts whih e peencua to A' B'. The engths of A', BB' an A'B' ae 680, 2000 an 2010 respecvy. Fin the minima ength of +XB where X is a point between A' n B.
B
I
0 0 0 C
0 1
lA'
2010
1
B'
x n is enote b n! For exampe 4! = 1 x 2 x 3 x 4 = 24. Let 2. The prouct 1 x 2 x 3 x M= 1! x 2! x 3! x 4! x 5! x 6! x 7! x 8! X 9!. How many fators of Me perfect sques? ·
·
·
6
28. Stting om any of the L's, the word LEVE c be speled by moving either up, down, le or right to an adjacent lette. If the same ltter may be used twice in each spell, how many dierent ways e there to spell te word LEVEL?
L
L
L
E
L
E
v
E
L
E
L
L
29. et ABC be a recgle wit AB
=
10. Draw circles C1 nd C2 with diaeters AB and C
C1 shaded respectiely. tand P, QC, be th intersection ofthe nd C2.region? If te circle with diameter PQ is tangent o AB then what is tepoints area of
30. Find the least pime factor of 1000 .. ·00 1. " 201-may
7
31 nsider the identity 1 + 2 +· = � ( 1. If we set Px) = � xx + ), then it is the unique poynomia such that for a positive nteger , P1 ()=1 + 2 +· · · + . In genera, for each positive integer k, there is a uniue ponomia P (x) such that Pk(n) =1k + 2k + 3k + ·· + nk
Find the vaue of 2o10C � ) -
for each =1, 2, . . . .
.
32. Given that ABCD is a sque. oints E and lie on the side BC ad CD respectivey. such that BE=C = AB. G is the intersection of B and DE. If rea of ABD rea of ABCD
m
is in its owet term, nd the vaue f m + .
E
C
33. It is kown that there is ony one pair of positive integers and such that : and + b2 + 8=2010. Fid the vaue of + . 3. The digits of the number 123456789 can e reanged o form a nuer that is divisibe by 11. For eampe, 12347569, 592681 ad 98453126. Ho many such nmbes e there? 35. Suppose the three sides of a tria ng ed e al integers, and its ea equas th per iter (in numbers). What is the agest possibe aea f the ed?
8
Singapore Mthemticl iey Singapore Mathemtcal Olympiad (SMO)2010 (Junior Section Solutions)
. Ans: (E) It is te only number less tan . 2. As: (D) Oter tan (D), all numbers e less tan 20. Now 37 39, te reslt follows.
>
200. Tus 200
<
(37
<
3. Ans: (B) Only te tird statement is coect: a < b implies a+ c < b + c. For other statements, counterexaples can be ten as a= , b= ; c= 0 and a= 0, b= - respectively. 4.
Ans: (C) Since (2010+ )= 200 + 3 x 2010+ 3 x 2010+ . Te result follows.
5
Ans: (A) Let te distance between TownA and own B be 3s. Te total time ten for up slpe, orizontal and down slop road e �, and respectively. His average speed for te
3s
W 0 1e JUey IS
8+ s
s
+
s
4
=
12 k.
6. Ans: (D) Observe tat !B is congruent to AE (using SAS, 60° LEB= LB). Tus LAE= LB. We get
B= A, = E
o = =
LBE 360° LAE360° L- LE
LE+ LE = =
62° + 60°= 22°
22.
9
and LAE=
A
D
7. As: (E) There is no way to reduce the cuts to fewer tha 6: Just cosider the middle cube (the oe which has o expose urfaces i e begig), each of the its sides requires at least oe ct. 8. As: (B)
The last digit of 7k is 1, 7, 9, 3 respctively for k=< 0, 1, 2, 3 (mo 4). Sice 7 = (-1? = 3 (m�d 4), e last digit of 7 is 3. 9. As: (C) 252 = 2 x 2 x 3 x 3 x 7. We c have: 667, 497, 479, 947, 749.
10. As: (C) e oly eed a rough esate to rule out the wrog aswers. - 33 ad 2.2, so te sum is 5 . 58 + (1.1)8 5. 58 = 30.254 304 80000. Thus (C) Of course the exact swer c be otained by calculatios, for example,
t (�)< «)l( � t (�)< «)'(- � )8-1 +
=2
.I (�) m)i( v)8-i
zen
1
)8-1
11. ns: 201. Let
a=
2008x+ 2009 2010 x- 201 1·
Then a � 0 and- a � 0 snce they e unde e sque oot. Hence a= 0. Thus y= 2010.
12. ns: 369. It s easy to see hat b= 5, ab= 25 or b= 6, ab= 36. Upon checkng, b 6, a= 3 nd so 2 (b a?= 63 = 3969. Theefoe c 9. Hence abc= 369.
13. Ans: 5 1. Snce
2 3m - 2m+ 10 18 3m 4+ m- 2 m- 2 an ntege. Thus m- 2 s a faco of 18. m- 2= 1, 2, 3, 6, 9 , 18, us m 3, 4, 5, 8, 1 1, 20. Th equed sum s 51. -
14. Ans: 26. Usng the Medan Fomula,A_
=
2 2 2 AB + AC - BC ·
. ThusA M= 26 m.
4 15. Ans: 16. Note that boh numeato and denomnato e thmec pogessions. Removng common facos gives
2X
=
113+ 115 + 117 ... + 333 335 1+ 3+ 5 + ...+ 109+ 111
2 2 X 12 (335 113) + 1
:
16.
16. As: 322. The numbe n he -th ow and -th column has value16(- 1) + c n Esthe's ay and value 10(c- 1) + in Fda's ray. o we need to solve
16(- 1) c= 10(c- 1)+
�
5r= 3c 2.
Thee e exctly fou solutons
(, c)= {(1, 1), (4, 6), (7, 11), (10, 16) }.
1
17. As: 1139. A direct way to solve the problem is to factor 14807 irectly. Alteatiely, one ay hope for and B to have common factors to siplify the problem. This is a good stategy because of the following fact: "The common + Bgreatest nd lcm(, B)." divisor of and B, equals the greatest commo divisor of
2010 is easily factored as 2 X 3 5 67. Checking that 67 is also a factor of 1807, we ca conclude that 67 is also a factor of and B. The problem is reduced to nding a nd such at
2010 =30 and ab= =221. 67 Since 221 can be factored easily, a and ust be 13 and 17. So the aswer is 1767=1139. a + =
18. Ans: 4021. a(1) is the simple recuence relation f = 2f - f fo = 1 and ! =3. Using standd technique or siply guess nd ve at f=2n + 1. So aoio(1)= =2(2010) + 1. 19. Ans: 200. ABC must be a right-angled iangle. Let x = AC nd y = BC, by Pythaoras theorem r+ =10• i =(x + y) =r + y + 2=100 + 2 ea of ABC.
Maximum ea occurs when x=y, i.e. LC B=45°. So x=y= -.
20. Ans: 01. Note that 2011 = 11 (mod 100) and 1 = 21,11 (mo 100). Since 010 is divisible by 10,
2011
=
11
·
·
11
=
=
31 (mod 100) etc. So 11
_
1
1 (od 100).
21. ns: 2250.
i ;) (� chies r goalkeepers, kers nd midelders espectively. The remainin idelder and deenders can all play s defenders, ence total number of ossibilites ae 3X
lOX 5
12
X
(�)=2250.
2. ns: 31. Let a= 1001x ' 041 and b= 000x-2010. Then the equaton becomes +b= 2ab. Thus (a- b?= 0. The esut folows. 3. Ans 00. Let x= 010 and y= 10. The numeto becomes [(x + y)- x ( �i ii ) (xy) + x]
-
= (� + x + i ) Cx-y)(x+ y) (� - x + i ) = Y(� -)(x +) = y(x6 -l). ·
Hece the swe s 100. 4. Ans: 010. Let 15+x= and x74= • We hav = 89= 189. (-)(+)= 189. Let -= 1 ad += 89. Soving = 45 and = 4. Thus the numbe x s 4515= 010. 5. Ans: 177. Fom 0.9y < x < 0.91y, we ge 0.9y + y < x + y < 0.91y + y. Thus 0.9y + y < 59 d 0.91y + y> 5. I foows tht y < 3.05 and y> 9.3. Ths y= 30 o 31. f y= 0, then 7 < x < 7.3, no ntege vaue of x. f y= 31, then 7.9 < x < 8.1, tus x= 28. Thus y x= (31 8)(31-8) = 177. . Ans: 3350. Te the eecion wth espect to A' B'. L A" be the mage of A. The the minma length (2000 + 80)= 3350. is equa to the ength of A" B =
2. Ans: 7. M
= !!3!4!5!!7!8!9! = X 3 X 46 X 5 X X 7 X 8 X 9 = 2 X 3 X 5 X 7
A perfect sque facto o M must be of te fom 23Y5z7, where x, y, and e woe numbes such that x � 30, y � 13, z � 5, � 3. Hence, the me of peect sque factos of M s 173= 7. 1
28. Ans: 144. There e total 12 was starting fro any of the 's to reach te middle V. Hence te total number of ays to spell the wod EVE is 12 = 144.
29. Ans 25. Le N be the midpoint of CD. Ten LPNQ = 90°. So Q = 5 V. B
A
D
N
c
Ten the area f the shade region is
3. Ans 11. Clearly 11 is a fator and 2, 3, 5 e not. We only need to rle out 7. 1+1 becaus 1 -1 mod 7). =
31. ns . Let k be a poive ven numb�r.
14
=
4 (mod 7)
Dene j(x) Pk(x)- Pk(x - 1). Ten j(n) nk for ll integer n � . polynomia We must ave j() x. In picul, for integers n � 2,
2
Not hatj s a
Pk(-n+ 1)- Pk(n) (-n+ 1 (n- 1)\ Pk-n+ 2)- Pk-n+ 1) -n+ 2) (n- 2i,
Pk(O- k(-1) (O Ok, Pk(1)- Pk(O) 1) 1k Summing these equaities Pk(l)- Pk-n) 1k+ Ok+ 1k+
Dene g(x) Pk(-x) + Pk(x - 1). Ten gn) polyoma g(x) 0 n ptcul Pk -�)+ Pk( -�) 0 , i Pk( -�)
0
·
·
+
·
(n- 1)k Tat is
for all integers n
�
2.
Since g is a
32
Ans: 23 Jon B and CG an note tat
0.
2
E
F Assum te lngt o B is . Let te ea of BGE n FGC b x esctiely en as of EC an F ad y. Sice a o BFC i · w a
3y+ 2x= a o VE=13. o x n= � . s 3+ =16 Sl 4 ra G =1 3( )=l _ 5 � a 4 4 =9 = 4. t w
_
m
n
1
+
=
·
33. Ans: 42. Since � 1,20 10 =++8 � 1++8. +8-2009 0. Hwever +8-209=0 as an integer sutin 41. S = 1 and =41. The resut fws. 34. Ans: 31 680. Let X and Y be the sum f te digits at even and dd psitins respectivey. Nte at 1 + 2 + 3 + + 9=45. We have X+ Y=45 and 11 divides IX - Yl. It's esy t see X= 17 and Y = 28; r X = 28 and Y = 17. Hence we spit the di gits int 2 sets whse sum is 17 and 28 respectivey.
There e 9 ways fr 4 digits t sum t 17: {9,5,2,1}, 9,4,3,1}, 8,6,2,1}, {8,5,3,1}, {8,4,3,2}, {7,6,3,1},{7,5,4,1},7,5,3,2}, 6,5,4,2}. Thre e 2 ways fr 4 dgts sum to 28: {9,8,7,4}, 9,8,6,5}. Tus the ttal numer f ways is 11 x 4! X 5! = 3168. 35. Ans. 60. Let te three sides f the triange be ,, respectivey. Ten )( - ( - )=+ + =2, where =
. Nte tat is an integer; therwise (-)(-)(- ) is a nnnteger.
Let x= - , = s - and z = - . hen x,,z ae psitive integers satsfying z=x + + z). Assume tat x � � z. Then z 12x,i.e. , z 12,and thus z 3. If z = 1, = 4(x + + 1) impies (x - 4)( - 4 = 20 = 20 1 = 1 2 = 5 4. S (x,)=(4,5),(14,6),(9,8).
·
If z=2,2=4(x+ + 2) impies (x- 2) 2)=8=8· 1=4 2. S (x,=(10 ,3),(6,4).
If z=3,3=(x + + 3) impies (3x- 4)3-4 )=52, wc s n sutin x � � 3. Te aea is 6 ,2,36,30,24, respectiey; and te agst pssibe vaue is 60.
16
Singapore Mathematcal Society Od (O)
2010
(Junior Section, Round 2) Saturday, 25 June 2010
0930-1230
INSTRUCTIONS TO CONTETATS 1 Answer ALL 5 questions. 2 Show all the steps in your working. 3 Each question carries 10 mark. 4. No calculators are allowd.
1. Let the diagonals of te square ABCD intersect at S and let P be the mipoint of A et M be the intersection of AC and PD and N the intersection of D an PC. A circle is incribed in the quadrilateral PMSN. Prov that the radius of the circle is MP- MS.
2. Fin t sum of all the 5-diit integers which are not multiples f a whose digits are , 3, 4, 7, 9. 3. Let a, a2, . . . , an be positive integers not necessarily distint but with at least ve distinct value. Supose that for any s i < j s n, there exist k £ oth dierent from i an j such that ai + aj � a + a. Wht is the smalest possible vaue of ? 4. A stuent divides an integer m by a positive integer n, where n s 10, an claims
tat
m _ n
=
167al2 ·
· · · .
Show the stuent mus be wrong.
5. The numbers f, �, . . , are ritten on a blackboar. A student chooses any two of the numbers say x, y, erases them and then writes don x + y + xy He continues o o this unti ony one number is left on the backboard. What is this number?
7
Singapore Mathematical Olypiad (SMO) Junior Section, Round
2
00
solutions)
1. et 0 be the centre and the radius of the circle et X, Y be its points of contact with the sides PM, MS, respectively
Since OY j MS and LSO= LASP=45°, SY=YO=. lso LPX=LPDA ( since OP II DA) and LOXP=LPAD=9°. herefore 60XP 6PAD. Hence X/XP =PA/AD= 1/2. Hence PX = 2. herefore PM-MS = 2 + MX MY-=r. D
A
c
p
B
2 First note tat an integer is divisible by 11 if and only i the alternating sum of the digits is divisible by 11. In our case, these are the integers where 1,4 and 7 are at the odd positions et S be the sum of all the 5igit integers formed by 1 3, 7, 9 and let T be the sum of those which are multiples of 11. hen s=4!(1 + 3 + 4 + 7 + 9)(1
=39993 T=2!2!(1 + 4 + 7)(1
+ 1 + 1 + 1 + 1)
+ 1 + 1) + 3!(3 + 9)(1 + 1)=55758.
hus the sum is 39993-55758=58423.
·
·
3 : a2 : an. Suppose x < y are the two smallet values hen a1 = x and s be the smallest index such thata8 = y No there are two oter terms whose let a1 sum is x + y hus we have a2 = x and as+I = Since a1 + a2 = 2x, we must have a3 = a4 = x. Similarly, by considering the largest two values < z, we have an =an-I = an-2 = an-3 =z and another two terms equal to Since there is one other value, tee are at least 4 + 2 + 4 + 2 + 1=13 terms he following 13 numbers
8
satisf the require property: 1, 1, , 1, 2, 2 3, 4 4, 5, , 5, 5 Thus te smallest ossible value of n is 13.
. We hav
m < 0 167 < 0 168 n ·
·
167n: 000m < 168n.
=
Multiply by 6, we get 002n 6000m < 008n
6000- 100n < 8n: 800.
But 6000m -lOOOn 2 > 0. Thus 6000m - 000n 000 since it is a ultiple of 1000. We thus get a contradiction.
5. We shall prove by inuction that if the srcinal numbers are a1, ..., an, n 2, then the last number is ( + a1) ( + an) - . ·
·
·
2, the base case. Now suppose it is rue for The assertion is certainly true for n n k 2. Consier k + numbers a1, ..., akl writen on the board. Aer one operation, we are le with k numbers. Without loss of generality, we can assume that the stuent erases ak an akl an writes bk ak+akl+ akakl (+ak)(+al)- . Aer a rther k operations, we are le with the number =
=
=
=
This completes the proof of the inuctive step. Thus the last number is
( �) ( ... ( 1+
1+
1+
19
1 20 0
)
-1
=
20 0
Singapore Mathemaical Sociey Sngapore Mathematical Oly mpad (SM) (Senor Seton)
Tuesa 1 June 2010
090 - 120 hs
Imortant: Answer A 35 questions. Enter your answers on the anser heet provided. For the multiple choice uestions, ener you answer on the answer sheet by shading the buble cotaining the letter ( B, C, D or ) corresponding to the correct answer. For the other short questios, write your answer on the answer sheet and sade the appropriate bubble below your answer. No steps are needed to just your anwer. Each question carries mak. No caculatos are allowe.
PLEASE D NOT TRN OVER UIL Y ARE TLD T D S
20
Mutple Choice Questions
1.
. F mdhevueo f (A)
B (C)
D (E)
2.
(123)+ (2 46) (69)+··· (3356701005) (136)+ (2x 6 (918) ·· + (335 1005 2010)
1
-
6
4
9
I , , de re numers suchh
b++ ++ b b+ = = = =, b c findhevalue of . (A) 3 (B) 1 (C) -1 3 or D (E) 3 or-
3.
; I O< x< nd sin x-cos x 7 nd 2 4 a pstive integr, in thevue o b (A) (B)
C) . D (E)
8 32
34 48 50
c.
tnx
b;
A
(B) (D)
()
5.
3/ 2 J
(B)
2J
D)
()
2
D A
Find t nubr o ordered pairs x ), where x is a integer and is a prfect sque uch that = x-90) -4907. (A)
0
(D)
1 2 3
(
4
(B)
7.
336 676
In te figure beow C is an isoscees tiange inscribed in a circe wit centre 0 nd diameter , with A = AC. intersects C a tE and is the midpoint of O. Given ha is pe toC d C= 2 m fd the lngt of CD in m. (A)
6.
104 224 312
LetS= {1 2, 3, . 910 }. A nonemty subset ofSis consdeed "Good i te number o even negers in te subset is more than r equal o te numbe of odd integers in te same subset. For eampe the subsets {48} 3, 478} and 1, 3, 6, 810 ae "Go. How man subsets fSe "Good? .
(A
(B) (D)
()
.
482 507 575
637 667 2
8.
gah of a quadaic ncion (x) 2 + bx + c a i 0 passes ro to disinc poins , k) and (s, k), wa is j( + s ? I he
(A) (B) (C ) D E)
9
2k c k-c 2k-c None of e above
Fin he number o posiive inges k < sch ht 236n ) + k (2n +l) - is isibe by 7 for ay posiive inteer n. (A) B) C) D E)
.
=
0 12 13
4 6
90° , s shown in 13 A M CM em ad e gure belw . Given t s e ipoin of i 2 17 m, fm he ea of e apeziu C i c . C + C+ A Let
ACD be
trezm ith alel to BC nd
LC
(A) B) C D) (E)
26 28
A
3
33 35
B
2
c
Shot Questions
11.
The ea of a ectane emins unchanged when eithe its legt is neased y 6 unis and idth deceased y 2 units, o its lengt decease 12 units and its wid inceased 6 units If te peiete of the oigial ectangle is x units, find th value of .
12.
Fo r = 1
u Find the vlue of 3 100 1 2 +3, 2let +=1+ 23+...+r. +···+ + +···+ +
+
13.
If
2010!=Mx10k,
14
I
a > b > 1 ad
15.
.
+
whee Mis an intege not divisile 10, find he value of k.
og a b
+
1og a
=
, find the alue of ·
rl
1 og b
1
og a
Fo ny eal nume X, et X denote te smalest intege that is geate than o
equa to and denote the lges intee at is less tan o equa to (fo example, J1.23l = 2 nd L23 =1). Find the vlue of
J
l J ·
2010 _ 2o10
k
2o1o
1.
Le f ( )
17.
f a, ad c e posie eal nuers suc tha
=
2o1o +
(1_)2010• Fd e vlue of
ab + a b = b c ca c a = 35 )(b 1)(c 1). + +
+ +
ind the value of a+
2
8
he figure below, A and CD e pllel cords of a circle wi cee 0 d radius rem. I is gien a A = 46 m, CD 8 m and LAO= 3 x LCOD Find e v f r. A
B
9
Find e number of ways ha 200 ca be wrien as a sum of one or more posiive iegers in nondecreasing order suc a e differnce beween e la e and e rs e is a mos 1
20
Find e lges pssible value of n suc ha ere exis n consecuive posiive · inegers ose sum is equa o 2010
2
Deeine he number of pairs of positive inegrs n nd suc ha 1! + 2! + 3 +
22
.
+ n! = 2•
Te gure below sows a circle wi diameer A. C d D e poin on he circle on e same side of suc ha B isecs CBA Te cords AC nd inersec a E. I is give a = 169 m and EC = 19 m . ED x m, n e value of x.
25
3
ind te number of ordred pairs(m, n) of positive iteers i nd n suc at m + n = 19 ad m d n e retivey pime
24
Find the east possibe vaue of j(x)
25
Find te umber of ways of anging 13 identic bue as and 5 identica red bas on a saiht ine such that between any 2 red bas tere is at east 1 bue ba
26
LetS= {1 2 3 4 1000} in the east ossie vaue o k such tat any subset A ofS wi IAI = 21 cotains to dstinct numbesa ad wi Ia- : k
27
Find the nuber of ways of taveing om A to B,as shown in the figure beow if you re ony aowed to wa east or nor aong the grid and avoiding a te 4 poins med x.
9 25 + were x ranges 1+cosx 1-cos over a real numbers for which j(x) is defined =
Nort
L
East
A
28
Two circes C1 and C2 of radii 1 m nd 8 m respectivey e tangent to each oter inteay at a point A. AD is te diameter ofC1 and P and Me points on C1 d Cz especivey such tat PMis tgent to Cz as shown in the figure beow I PM= . m and LA nd e vaue of x =
6
29
Le , nd cbe iegs wit a b c > 0. f b an c e relaively prime + is mulie of a, ad a+ cs a multipl of b, erine e vale o .
30.
ind e umber of sbses < b - 1 c-3.
31.
Le f) dene numer O's in deiml resnin e psiive intee . For expl, /(10001 12 3 d /(124567 0 Le
{a, b, c} f {1 2 , 4, ... 20} sh
M
/(1)xfI j(2)x2<> /()x2<3> .
·
+ (99999)x2�.
Find e vlue f M- 100000.
32.
Deermine e odd prime number p suc t e sum f digis f e nmbe p -5 p 1 is e smlest ossible .
33.
Te e beow sows a ezum ACD in ic A II BC nd C 3A. F is e midint f A and E ies on C exened so a C CE. Te line segments EF nd C eet t te point G. It is give ta te e of iangle GCE is 15 cm2 nd te e f aezim AC s k m2• ind e vle of k. =
=
34.
c
E
a a1x a nx be polnoial i x wee e cefciens a, a, a ,•••, n e non-negaive integrs. f P( 25 and P(27) 1776 9, find te alue o a 2 3� (n )an. Le P(x)
·
·
·
7
35.
Let tree circl rl' r2' r with cntre t, Az, A3 rdi Tt, rz, r3 respetively be utualy tangent t each other exeall. Sppose tat te tangent to the circumcircle of the tiangle A13 a A3 ad he w extea common tanents of r1 and 2 meet at a cmmon point P, as shown in the gur belo. Given at r1 = 18 em, r2 = 8 e m and r3 = k m" n he lue of k
p
28
Singapore Mthemicl Society Singapore Mthemtic Olymid (SO)200 Senir Sectin Sutns
1
Aswer: (A)
(1x2x3) + (2x 4x 6) + (3x6x 9) + ··· (335 x 670x105) (1x 3x 6) + (2x 6x12) + (3x 9x18) + ···+ (335 x1005x200) +2 +3
=
2
1x2x3 1 = 1x3x6 3
Answer: (E) From e given equations, we obtai
a +b +c +d a( r +1), a + b + c +d = b ( r +1), a +b +c +d c(r +1), a +b +c+d =d (r +1) Addng tese four equations give
(a + b +c +d) (a + b + c + d)(r + 1),
that is,
(3-r)( a +b +c+d =0 Thus r 3, o a +b+c+d 0 . I a +b+c +d 0, we see from te srcinal given equations that r 1 Hence he value of r is eiter 3 or 1 =
=
.
Answer: (E)
1
xcos x= x-cos) =, whic h rmp1"es that s We ave (s Therefore we obtain
t+ ence
4
a+ +
1 sinx cosx 1 32 = = += tnx cosx sinx sinxosx 16-:
-
c 32+ 1+ 2 = 50 . =
Answer: (C) First we note tat 43
=
[n(n+ 1)] [(- 1) • Thus
9
16- 32
Terefore 14 15 + ... + 24 25 14 15 13 14 = 4 4 152 162
142 x152
4
4
24 X 25 4 25 X 26 =
4 25 x26
23 X 24
.... + +
�
4
2 2s 4 13 14
=
(25 x 13+ 13 x 7)(25 x 13-13 x7 )
4 4 = (32 X 13)(18 X 13) = 9 X 64 X 13 • Tus 5.
+15 +16 +.. +24 +25 = 38 13 312.
Answer: (B) Snce the diameter A perpendcully bsects the cord B, BE = EC = . Also, given at BD II FC, we ave LDBE = LFCE. Thus MDE s congruent to !CFE, so DE = FE. As F s the mdpont of OE, we ave O E ED Let OF = x. Ten AE = 5x. Usng Itersecton Chord Teorem, we av =
=
AExED = ExEC,
wch leads to 5x = 5. Consequently we obtan x = 1 Now CD2 = C£ ED = . gves CD = 6
Answer: (E) Let y = and (x0) = k where and k ae potve ntegers Ten we obtan k - =407=7701=1907, wc gves (k -)(k+) =7 701 or (k -)(k+) =1407. follows tat k = 7 and k 701 or k = 1 and k = 4907. Solvng tese two pars of equaton gves (k, ) = (354, 347) and (k, ) = (245, 2453). 30
Thror th ordrd pairs (x,y) that satisfy givn qution : (444, 347 ), (-264, 347 ), (24, 243 ), (-2364, 23 ). Hnc th answr is 4.
7.
Answr: (D) Let th nmber o evn inegrs in a "oo sbset o S b i, whr i 1, 2, 3, 4, 5, and th numbr o odd intgers in that subst , whr 1 2 ... , i. Thn te number o "Good sbses oS is =
tJ :} : +}� :) < () G =
...
. . .
=(1 + ) + 1(1 + + ) + + (1 + + 1 + 10 + + 1) =30 + 60 + 260 + + 32 =637
8
Answr: (B) L (x) (x)-k Thn gr = ()k k-k Similly, gs Thrfor and roots of th quadratic equation gx = x-k =0 from which
b w deduc that + s = - Hnc a
b j( s ) j(-) =a (- ) (-) . a a a 9
Answr: (D) W hav 2 (3 6n) k(2n+l) -1 2 (27n) +2k(8n) - 2( -1)n 2k(n) -1 (mod 7 ) =
=
2k 1 (mod 7) l + n + n k(2 Thus, for any positiv itegr n, 2( 3 ) ) 1 is divisibl by 7 if and only i 2k 0 (mod 7) s k< 10, it is clar that th congrunc holds for k = 3, 1 17 . .. 94. hs te reqired nmbr o positive integers k is 4.
1
Answr (C)
31
xtend DA an CM to meet atE s sow in the figure above. Snce A = MB, an L LB, we cnclude hat � is congret to t Thereore A = B and M=EM. Thus E =
=
=
2CM 3. _2 (DE)(D, we have (DE+DC) =DE +DC +2(D)(DC) =C£ +4S =13 +4S =169+4S. Now DE+D=DA+AE+D=DA+B+D=17. Hence 17 =169+4S, and it follows that = 30. 1. Answer: 12 Let the legth and with of the orginal rectngle be d respectively. Then LW= (L+6)-2) and L = (L-12) + 6). Let the ea of apezium AD be cm2 . Then
ad
a
W
Simplifying the above equtions, we obtain
L-W=12
and
3WL6. 132
8 and
Solvin the simultaneous equations, we get L= perimeter of the rctgle is units.
12.
Answer:
2575
) = 1+2+ 3 + ...+ 22 , we have =2- t_ t ( +1) (�- + 1) +1 . +1 (
r r
r
As
8. Hence the
W=
=
r�l ur
�l
r r
r�l
r
i
r
i
Hence, Sn : =
-/ =( J=i +12 )=_ ( (n2 ) n)=�4 n +3. �
� + u, � In pticul, OO
�
n
+
=255.
3.
501
Anser: Te numbr k is the number of the factor that occurs in give b the nmber o pirs of prim actors 2 and in ow between and there are: • • • •
0
2010 tegers with as a factor; 80402integers with 255 as a factor; 1 integes with 25 as a factor; 3 integers wit 5 as a fcto.
00!. This umber is 5 200!.
Therefor te ttal nuber of prime factor 5 i 210! is 02 + 80 + 6 + 3 =501. As there ae cey ore thn 501 pie factor 2 in 200!, we obtn k =501.
14.
Answe:
35
1 log
1 log
1 0. Then
1
First nte that snce > > 1,
log
log
= og -log = og a+ 1 ) log +1 log log
1 - 1 = log
·
og
= =
[�+
= =
15.
35.
1994 2010 Consider k = 1 2 .. 2010. I k 2010, then X: k 10 y k ' 010, then 0 < y:= < 1 so r l =1. Answer:
Since the prime factorization of 2010 is disinct divisors. Hece
16.
swer:
k
1005
Observe that
l Jl
=0, so rXl=O.If
'
2 x 3 x 5 x 67, we see that 201 has 16
201 = LJX+ 21y l k
=number o nondivisor of 2010 ong =2010 -16 =1994.
_
j(x) + j(lx
k
(- xyoO _ xoo -1. + 1-x + x20JO
x2o Jo + 1-x
I folows hat
3
=
17.
1005
Answer: 216 Addng 1 t both sdes of the gvn equaton 35, we obtan ( 1)( 1) = 36 Lkewse, addng 1 to te oter two gven equatons gves ( 1)( 1) = 36 and ( 1)( 1) = 36 Now multplng te three resultng equatons above leads to =
[(a+1)(b+1)(c+1)] 2 = 36 = 6•
It follows tat +1(+1)(+1)=6=1. 18.
Answer:
27
B
Let M and N be te midpoints f A nd CD respectivey an et CO=x Ten O=3x ad 23 x 9 = N = sx = sx =3_4si x 23 1 1 1 Tus s x = 3 g = , ad so s x " .
.
(- )
Hence CN 27 sinx
3
9.
Answer: 2010 Cosier ny inteer k where k 20. B ivsio lgorthm ere exists iue air o intees (q, r) such that 0 kq+ r ith 0: r: k- . We rewrite his as 2010 (k- r)q+ r(q+ ). hat is, k- r cpies of q and r copies of q+ add up to 0. Thus there s one desire expression for eac vlue of k, which is clely uniqe . Hence there ae 200 such exessions i all .
20.
nswer: 60 Let a be a ositive integer suc that the sum of n consecuive iteers a a+ 1, ... , a+ (n- 1) is 2010 that is,
()···( ) 2. =
n(2an ) =200 r ThIS gves
2 n(2 ) =2 xx 5x7.
(1)
Since n< 2+ n - e have
<2x 326.
2)
Now () and (2) imply ta E{2 S2S2 6. Snce and 2a+ n- have ifferent arities, it follows ht an we have In
2 .
4020 n
hve differ�nt arities. Consequenty,
ES2S06}.
60 then 2a =
4020 =67 so a 4. Ths te gest ossible value of n s 60 60
Answer: 2 First note that i n 2 4, hen
!2!!···! !2!!! (mdS) 2 (modS). Since te sque of any integr is congruent to eiher r 4 mdulo S, it follows that 1 ! 2! 3! n! i m2 for any integer m in this cse So we cnsider n< 4. Now we have ·
1!
·
·
+ 2!
! + ! + !
32•
ence we conclude that tere are two pairs of postie integers (n m , namely, ( ) and 3 3 such ha ! 2! 3! ! m2• ···
35
22.
Anse: 5 EC 119 = = . Thus we can let C =19 169 and BA 169 for some real umber . Sice LBCA 90°, we ha Since BE bisectsLCA we hae
BC
BA
2 AC2 B C2
+ (119 (169 y =(169+119 (169-119 X169+119)=(169+119 169+119 144 = 169-119 25 12 5
=
Hence, om tiangle C, we have BE
+ (119 =119 x
13 . 5
Finaly, note tat nd MC e simil, s we he E xCE 169x119 65cm. = ED= 119x 153 BE
23.
Answer 2 Fist we nd the number of odered pais (m, n) of positie integers m nd n such that m n 10 and m ad n e not reatively prime. To tis end, wte m = k and n = , where k, ad e positie integers with k > 1. Since m n 190, we see at k is a facto of 190 2x5x19 with k t 19. We conside si cases (i) (ii) (iii (iv) (v (vi)
k = 2. Then 95, and tere e 94 suc pairs (, ) f nd suh that the equation holds. k 5. Then 38, d tee e 37 such pairs (, ) f d such that the equation olds k = 19. Then 10, ad tere e 9 such pairs (, ) f and uch tat the equation holds. k = 10. Then 19, nd thee ae 8 such pairs (, of a and b such hat he equation holds. k 38. Then 5, and tere e 4 such pars (, ) of and such tha the equation holds. =
=
kthat Then hods. 2, and tere is 1 such pair a f and such = 95. the equaion
It follow m the above cases that the number of ordered pairs m, n of posiie integers m nd n such tat m n 19 and m n e not relativey prie is 931841 117.
6
190 189
Sinc th total nmbr of ordred pairs (m, n) sch that m + n = is w onclude that th required number of ordered pairs (m, n) where m and n rlatiely prie is = 7
1 24.
19
2
32
Answer: For all ral values of x for which] (x is defid, e have
9
25 9 _2 25
9
25
j(x) = = cos 2x 1-cos 2x 2cos x 2sin x = � t x ) ( cot x )
_2
= 17
� 17 = =
_2 � t
25
25 co ) X
(MGM Inqulit)
17 + 2 32.
25)
Note at j(t
25.
X
2
=
32. Thus th last possible value ofj(x) is 32. .
202 5
1
9
9
5
4 2
9
9+ -!J (� J c:J �
26
2
nswer: First we place the rd balls on a saigt line and then place blue ball betwen adacent re balls. With is ngment fe, t condition of the question is satise. We ae now left with blue balls W can place the remaining blue bals ito the spaces befor, aer or in between th red balls The number of ways that his can be one is the nswer to e question . Including te two ns, tere r + = 6 spaces into which thse blue balls can be pace Te numbr of ways of distibuting the blue blls into th 6 saces is 4 �
�
9
2002.
4
nswer: Consider the following subsets of S:
{, 2, 3, ..., , }, S2 ,, 0...,, 9...,, 149 10,}, }, s3 {{1,, 2102 Szo {1, 2, 9, ..., 9, 1000. In othr wors, Si 5i - 49 50 i- 8, ..., 0i for 1, 2, ..., 2. e hat is ptitioned into these subsets t Sz, S 3 , Szooo. S1
=
37
By Pigeonhole Pinciple, fo any suset A of S ith AI= 2010 thee xists i, he i: 20, such that AnSi I� 2 Let ab EAsi I is clea that a-bl 49 To sho that 9 is the least possibe alue of k, e find a subsetA � S with A = 200 such that a I -b 9 o any distinct ab EA Let
={ j+: j= 2 . 20}={50 8 98} Then A is a subset of S with IAI= 200 and a-b 49 fo any distinc ab EA
27
Anse: 2
Fige
A
c
We obseve that to aoid the fo ponts maed x, the path mst coss eithe C, D oE as shown in Fige aboe. Fthe, the paths that crossC, D oE e exclsie, that is, no path can coss bothC and D o D andE, o C andE. C
Thee is only way to A gettoom e ways to get from D A to 1 1
I
and om A toE 6 5
1
2
15 0
1
3
6
1
2
3
1
1
c
is easy to see that thee
B
Fige 2
To count the nmbe of as to t fom eitheC, D oE t B, we note that the nmbe of ways to get to a cetain jnction is he sm of the numbes of ways to get to the to jnctions imediatey pecedig it fom the left o fom belo (as shown in igue 2) Teefoe thee e2 ays to get fo C to B Similly, thee ae 2 ways to get fom D to and 7 way to get fom E to B. Hence the nmbe of ays to get fomA to B that •
C: num of ways fom A to C x nmbe of ways fom C to pass B = though X 2 = 2; pass togh D: numbe of ays fomA to Dx numbe of ays fom D to B= X 2 = 84; • pass toghE: numbe of ways fomA to E x nmbe of way fom E to B=1x7=7. It folos tat the total nume of ays fomA to B is 2 + 84+7=2 •
38
28.
Answer: 60
0 be e cenre f C2 and let P intersect 2 aB The hmhety cened at mapping Cz t C1 has similitude rai _ I mapsB t . Thus _ (This 1 1 cn als be seen by cnnecing P te cene 0' C1 s a the iangles O Le
and O' simila he per P wih respec t C2 is M . Thus
A ! ,we 1 ban =8 and = 10 . Cnsequenly, iange O is equilaerl,and hence 6°
· ,r equivaleny (P - )P =20 . Tgeher ith
29.
Answer: 6 We shall shw ha = 3, =2 nd =1. Ne ha 2 > + As + is a mutiple f , it fllws a = + . Le + =k . Then k = + + + ,s 2 = k- 1) . Since < ,we must have k =2 nd erefre =2. Since and e relaively prime,his implies a c = 1 ad =2. hus = 3. Hence = 6. =
30 .
Answer: 680 Fr any 3eement subse { ,,},dfine a mappng by ( {,,}) ={, - 1,c- 3}. Now bserve {,,c} s a ue {1, 2 3 4, ... 20 i < - 1< - 3 if d nly if j { ,b, } ) is a 3eemet su se 1 2 3, ... , 17}. Hene e nswer is
C:)
680 .
9
31.
Answer: 2780 Note that 0 ;() or 1 99999. Fo 0, 1, 2, 3, , let enote the number of integers hr 1 99999 such haj () k. Then M = L k 2 = L k 2 . O By consiering the numbe of igit, 3igit, 4git an 5igit positve ntegers wih ectly one 0 in their ecimal representation, we obtain � = 9 + 9 X 9 X 2+ 9 X 9 X 9 X 3+ 9 X 9 X 9 X 9 X =282 . Similly, we have =9+ 9 X 9 X
( �)
9 X 9 X 9 X
(�
=626,
=+9 x=333, = 9. Hence =1x28602x22x66x2 +3333x2
+x9x2 =10278,
an it follows that M- 100000= 2780.
32.
Answer: 5 Let = 5 +13. When = 3, we have 9 an so he sum of igits is 13. When 5, we have 513 te resulting sum of igits is9 Now let > 5 be a prime. We have = 5 +13=( 2)( 1)(+1)( 2)+9. Since - 2, - 1, , + 1 an + 2 are ve consecutive integers, at least one of tem is ivisible by 5. Since 5, mus ha 5 divides on o th nmbrs - 2, - 1, + 1 and + 2, so 5 ivies the prouct - 2)- 1) + 1) + 2). bserve that at least one of e nmbs + 1 and + 2 is even. Therefore we see that - 2)- 1) + 1) + 2) is ivisible by 10. It folls that for ny pime > 5, the nmber n has at least two igits an te units igit is 9, so the sum of the igits is greater tn 9 Conseqently th smallest possible sum of igits is 9, an this value is attine when 5.
40
33
Answer: 360
ExtendB andCD to meet at X .LetH be the point onCD such tatFH Let A
CE
a. Then
BC
1 (A +
ECG,
a, andFH
y the simility of trianlesFHG and FH x ea of (i) ea of FH (ii)
HG
CE
L
FH
-
CG CE
BC
2 we have
L ECG
(iii)
X A FH
1
2 ea of XDF
(iv) ea of XFH
, so tha
HG
-
.
� x60 90 cm .
ence the ea of tapeziumBCD
and
F and hence
=
2y cm;
y.
It follows from (iii ) and (iv ) that the ea of triangl e FDH Since the ea of triangleFDH is 90 cm we get y 45 . inally by the simlity of triangles andB
LXBC
3 2
2x ea of FH
BC A 2
BC.
60 cm ;
2 be cm . y the simility of triangles
- X L (A) y
ea of
2a.
2 , so thatHG = 2CG andDH =HC =
I follows from (i) and (ii ) that the ea of triangeFDH Now, let ea o triangle XFH , we have
y 9 y
8y = 360 cm•
41
4 y-2y
2y cm .
34.
nswe: 75 Fist we note tat evey positive integem can be witten uniquely (in base 27) as = + 27+b227 +···+27, whee and b0, bp b2, , br (depending on m) e non-negative integes with bi < 27fo i = 0 1 2 •••
Since of n.P(x) aebynon-negatie integesthendpolynomial P(1) = 25,(x) we see ;25te coefficients 27 fo 0 k Thus he above ema, is that uniquely detemined by the value P(2)= 771769 Witing 1771769 in base 27, we obtain 1771769 = +7+97+327• Theefoe (x) =2+x+x+3x Hence +3 +···+(+) =2++9+ 375 35
nswe: 2
p
Fist we shall show that N Let P be the point of concuence of the tangent to the cicumcicle of the tiangle � at A and the two eteal common tangents of nd • Note that he line joining and � aso passes hough P. = PA =. That is, P= Fist we have P =P ·PA, so p p 2 p On the other hand, PA sL� P s� 1 +3 -= = = p� sin P s +3 Thus ' + ' • Solving for , we obtain 2N. Subsituting 8 ,
-
=
' 2 =8, we get =. +
2
ingapore Mhemaial Sciey Od () 2010 ( Senior Secton, Round 2) Saturday 25 June 01
0930-1230
INSTRUCTONS T NTESTANTS 1 Answer AL 5 qesions. Show al the stps in your working 3 Each ueston caries 10 mark.
4. No calculators are allowed.
1. In the triangle ABC with AC > AB,
D is he foo of he perpendicular from A no BC and E is the foot of he perpendicular fromD onto AC. Let F be the poin BD DE Prove that is perpendicular on the line DE such that EF DC to BF =
·
2. The numbers
t, �, ...
·
are written on a blackboard. A student chooses any
,
two of the numbers, say x, y, erases hem and then wries down x y + xy He continues to do this until only one number is le on the blackboard.+Wha is his number?
3. Given
a1 2 1 and ak+ ; ak + 1 for all k 1, 2, , , show tha af + + + a � 2 (a1 + 2 + + an) 2 . =
·
·
·
· · ·
· · ·
4 An innite sequence of integers, a , a1, a3 , wih a0 > 0, has the propert that for any 2 0, an+l an - bn, where bn is the number having the same sign as an, but having the digits written in the reverse rder. For example f a0 1210, a1 1089 and a 2 -8712, ec. Find the smallest value of a0 so that an i 0 for all ;. .•
=
=
=
=
5. Let p be a prime number and let a1, a2 , .. , ak be dsint inteers chsen from 1, 2, ... p 1. For : i : k, let r n) denote the remainder of the integer upon n division by p so 0 : r} ) < p. Dee
}
-
S
=
{ : 1 : : p - 1,
Sho that S has less tha
elements.
43
rin)
<
·
·
·
kn ) } .
Singapore Mathematical Olympiad (SMO)
2010
(Senior Section, Round 2 solutions)
1. Since we are supposed to prove LAFB=go, it means that the 4 points A, B, D, F are concyclic. Note that AC > AB implies that LB > LC. If TD is the tangent to the circumcircle w of the triangle ABD ith B and T lying opposite ides of the lne AD, then LADT = LB > LC = LADE so that w intersects the interior of DE at F. erefor F can onl be in the int erior of DE. Now observe that the triangles ADE and DCE are simila r so that AD/AE =DC/DE. y the given condition, this can be written as AD/ AE = BD / EF. his means the triangles AB D and AFE are similar. hus LABD = LAFE. his shows that A, B, D, F are concclic. herefore LAFB=LAD= go A
2
See Junior ection Question 5
3 We will prove it by idctio. ist, it is le that suppose it is tue o n terms. he
ar af ice a1 2 1
n+l 2 a+l L n a� � a+l (Lak n ) La� k=l k=l k=l n+l n (Lak) a+1-a1-2an+Lak. k=l k=l 2
=
4
xt,
To complet th induction, we'll no show thata�+l-a;+l-2a+l=lak 0. Since ak+l- ak 1, we hve a+l- a ak+l ak. Summing up oer k 1, ... , n, and using ar - a 0, e have a+l- at a+l 2 L ak - a kl
. Thus a has at least 2 digits. If a ab 0ab, then a 1 9(a-b) which is ivisible by 9. it follows that ll subsequent terms re divisible by 9. Checking all 2-digit multiples of 9 shos that eventually 9 appears (Note that b and ba give rise to the same sequence, but wih opposite signs): 81 63 27 45 9. 4.
If ao has a single digit, then a
-
-
-
-
If ao abc, then a 99(a - c). Thus if suces to investigte3digit multiples of i.e., 198, ..., 990. Here we nd hat 9 will eventually appear:
99,
990 891 63 29 495 - 99. If ao abed, then a 999(a-d)(b-c). If b, are both 0, then a and all subsequent terms are multiples of 999. However, if such numbers appear in the sequence, eventually 999 will appear: 999 8991 693 2997 -4995 999. -
-
-
-
-
-
-
-
For 1010, we get
909 and for 1011 we get 90. For 1012, we get 1012 -1089 -8712 6534 2178 -6534 and the sequence becomes periodic thereaer. Thus the smallest a 1012. -
-
-
-
-
5.
Let
rbn)
0 and Tk�1 p. Set =
S' Note that
=
{n: 1 n
=O
S' ·
Since for
=
P
i
·
·
- dn)
rk�1. n(ai+l- ai) (mod p
r)1- rn), 1: n: p- 1, are all distinct. k � ( I lri�- r"l �
j � i=O jl
I t lri�- ri"l � EB' i=O i=O EB'
Therefore
Tbn) : rin) :
n E S', l r) - rn)
and p f (ai+l- ai), the numbers
Pl8'1
=
k
L lr- rn) Thus lSI
k p - 1, lr�- r�n)l p}. i=O
S <
45
Therefore +
).
Singapore Mathemtical Society Singapore athemaical Olymiad SO ) 201
( Open Section, Round 1
090-00 hrs
Wednesday, June 00
Instructions o contesas 1 Answer ALL 2 questions. 2 Wrie your answers in he answer shee provided and shade he appropria bubbls below yor answers. 3 N seps ar needed o jusif yor answers.
4. Eac uesion crries 1 mark. 5. No calculaos are alloed.
PLEASE DO NOT TURN OVER UNTIL YOU E TOLD TO DOSO 46
1.
.
Let S be the set of all tegers such that the value of L ll.
8n3-96 +60-400 2n-7
is an integer Find
nS
2.
etermine the largest value of x for wch
lx2 -4x- 39601 � lx2 +4x - 9601 . .
Given that
L/3j + 2/3j + L/3j + ... + L7999/3j, nd the value of L J, where LYJ denotes the gr eatest inteer less than or equal t y = (For eample, L2.1 2, L0J = 0, L-10.5J = -11.) X=
4. etermine the smallest postive ineer C such that n. : C for all positive· intgers . 5. Let CD be a chord o a circle r and B a diamter of per pendicular to C at N wih > NB. A cir ce 2 entred at C wit raius CN intrsets r at point P and Q, and the segments PQ and CD intersect at M . Given that the r adii of r and 2 are 6 1 and 60 respectivey, nd the length of M. 50
6 etermine the minimum value of
Xk, L kl
wher e the summation is done over all possible
50
positive numbers x, . . . X5o satising
_ = 1.
X kl k
7.
Find the sum of all positive inteers p uch that he epression (x- p)(xepressed in the for m ( + ( +r for ditinct integers and r.
x q x )
�- -
q
1) +4 can be
8. Let Pk = 1 + wherek is a oitive intger Find the least poitive integer n Pn eceeds 2010. such that the product P P
23
·
·
9.
Le B be a oint on the circle cntred t 0 with diameter C and let D and E be the cir cumcentr es of the tr iangles OB and OBC respectively iven that sin BOC = and C = 24, nd the area of the triangle BDE.
10.
Let f be a real-valued function with the rul f(x) = x3 + x +6x+1 dened for all r eal value of It is given that a and are two r ea umber s such that f() = 1 and f() = 19. Find the value of a +
11.
g
x
2
(
4
)2.
If cot + cot +cot =
- 45, tan
+ tan + tan' =
cot cot +cot cot' +cot' cot =
17
B and
1 7 nd the value of tan( -
47
.
+ +')
12 Th gure below shws a road map connctng two shoppng malls A and Bn a certan cty. Each sde of the smallst square n the gur rpresnts a road of dstance 1km . Regons C and D rpresnt to large resdental stats in the town . Fnd the number of shortest routs to travl from A to B along the roads sown n th gue. 8 km
c 10 km
D
13. Let a
=
an> 2
1, a2
=
2 and for al 2, anl
=
1 an 1 1 an-· It s known that 2
for all m where m is a postve integr. Fnd the least value of m
14. It s known that
8sin50 ° = + sin
for eactly one se of postv ntegers (a, ), where 0 < < 90. Fnd the value of 15. If a s a ral root of the equaton x -x + x 2 the least positve integer not ecedng x .
=
+ .
a
0, nd the value of L a6 J, where Lx J s
16. If a positve ntger cannot be wrtten as the dirence of two square numbers, then the integr is calld a "cute integer. For eampe, 1, 2 and 4 are the rst thre "cut integers. Find the 201th "cut nteer. ( Note: A square number is the square of a postve nter . As an llustration, 1, 4, 9 and 16 are the rst four square numbers.) 17. Let f (x ) be a polynomial in x of degree 5 . When f (x ) s dvided by x - 1, x 2, x -3 , x -4 and x 2 -x 1, f (x leaves a remainder of 3, 1, 7, 36 an x 1 espectively. Fnd the suare of the remainder when f (x ) is divded by x 1.
48
18. Determine the number of orered pir s of positive ntegers (a, b) satisfying the eqatio 10 +b= ab- 10.
Note: As an illustraion ( 1 2) and ( 2 1 re consiered s two disinct ordered pairs. ) 19. Let = a +b If a, b ad are all prme what s the value of ?
.
20. Determine the value of the following exprssion: 11 11 X 2 11 3 1 X 4 11 X 2009 2010 + 210 + 2010 + 201 +... + 210 ' wher e LYJ denotes the greatest integer les than r equal to .
l l l J
l
21. umbers 1 2 . . . 2010 are placed on the circumference of a circle in ome order . The umbers i and j, wher e i = j and i,j E { 1 2 . . . 2010} form a endly pair if ( i) i nd j are not neighbours to each other and ( ii) on one or oth of the ar cs connecting i nd j along the circle all numbers in etween them are greater than boh i ad j. Determine the minmal number of enl pairs. 2 2. LetS be the se of all non-zero realvalued nctions f deed on the set of all real umbers such that (x2 +yf(z= x ( x) +z(y for all real numers x, y and z Find the maximum value of£( 1 2345) where f E S.
. 23. All possie 6digit umber s i each of wich the digits occur in non-increasing order om le o r ight ( e. g. 966541) are wr itten as a sequence in increasing order ( the rst three -digit numbers in this sequence are 100000 1100 11100 and so on) . If the 21 number in this sequence is denoted by , nd the value of L J, where LxJ denotes the gr eatest iteger less tan or equl to x 24. Find the number of permutations a1 a2a3a4a5a6 of the six integers fr om 1 to 6 such that for all i om 1 to 5 ai does not exceed ai by 1. 25. Let A=
( ) c� )) ( ) C )) ( ) c )) �( - G� ) ) ; 201 0
o
2
+
21 1
O10 o
2
+
1 1
20 0 2
2
2
+... +
Determine the minimum nteger
s
such that
40 20 201
sA �
( Note: For a given positive integer, ther vlues of r dne
r
= '
·
!
' for r = 1 2 3 r. ( n _ r .
= . )
49
·
·
·
, ; and for al
Singapore Mathematical Society Singapore Mathematical Olympiad (SM) 210 (Open Section, Round 1)
1. Answer: 50 8n3- 96n2 + 60n- 400 Note that
. 4n2 4n + 61 + 27_ . Sce 4n2 4n + 61 is n 7 an integer for all integers n, we must have that 7 divisible by n- 7. Hence, n 7 ,1,,,9,9,-7,7, so that n ,4,,5,-1,8,-10,17. Hence the required sum equls 0. 2n _7
=
=
=
.
Answer:
99
By direct computation,
lx2 4x960 1
�
lx2 + 4x 9601
(x2 4x 9601) 2 (x2 + 4x 9601) 2 � 0 (x 2- 9601)(8x) 0 - x(x + 199)( x 199): 0, we conclude that the largest possi ble value of x is 199. -
.
Answer:
1159
Note that X
-
L1/3j
+
L/3j
+
/3j
+ ... +
7999/3 j
L Lk/3j + k/j + L k/3 j + ... + L k/3j 1 + + + ... + 19 +
+
3 13 3 3 43 3 19(8ooo) k3 = 2 19(8000) 19 0 115900
)
(
X
:. LlOOJ
=
159.
0
+ ... +
+
18 193 183
19 03 193
4. Answe: 65 6 ene f() =I fo =1, 2, 3, ···. It is ca that f() =6, f(2) =18, f(3) =36, f) =54 n. and f(5) =64.8. Fo all ; 6, f() =
6 6 6 6 6 6 � X X X X X 6 =6 X ! � l 2 J L S
3
X 2X
1 . 8 = 6 4 .8
5. Answe: 78 Extend DC meeting r2 at H. ote that DN =NC =CH = 60. Since M is of equal powe with espect to r1 and r2 Thus N ·MH MC ·MD . That is MN(MC60) . MC(MN +60) giving MN =MG. Thus MN =30. A
The pow of N with espect to r 1 is DN · NC = 60, and is also equal to N ·N B = N·(B -N)=N·(122-N). Thus N·( 22 -N)= 60. Solving this quadatic equation, we get N = 7 2 o 50. Since N > NB, we have N =7 2. onsequently M= +MN = +30=78. 6. Answe: 2500 y auchySchwaz inquality,
ad equaity holds if and nly if Xk =50 fok=, .. . , 50. Theefo the equie value is 2500. 7. Answe: 26 Let (x- )(x- 13)+4=(x+)(xr ) Substitutig x=- into the above idntity yields (- - )(- - 13) =-4, which becomes (+)(3) =-4. Since and ae integes, we must have the following cases: ( a) ( b) (c) ( d)
+4, +3=-; +=-4, +13=; += 2+3=-2;or +=- 2, +3= 2 5
For case (a ), we obtan q = 14,p= 8 and hence
(x- p)(x- 1)+4=(x 14)(x 17); (x- p)(x 1)+4 = (x- 9)(x 1); q = 15,p= 17 and hence (x p)(x 1+4 = (x - 15)2 whch
-For case (b ), we obtan q = 1, = 8 and henc For case (c ), we obtan s OT what we want;
For case (d ), we obtan q =- l l,p= 9 and hence (x p)(x- 1)+4 = (x 11)2; whch s also NOT what we want. Hence the two possbe values of are 8 and 18, the sum of whch s 6.
8.
Answer: 808 rst, note that
2 ) (1+ .k 2= (k - 1)(k+) k3
(
! ] _ ! P = 1+ k k 2 k3= 1 k Therefore,
1 32 42 3 5 2
· · · P 2P3··· P=� ····· Next, observe that
(n+1)2 4n
>
(n- 1)(n+1) 2 n3
=
(n+1) 2
4n
010 s equvaent to n288n+1 > 0,which s equvalent
to n(n 808) > -1. Snce n s a potve nteger, the last nequalty s satsed f and only f n 2 808. Consequently, the least n requred s 808.
9.
Answer: 45 Let d = = 4. Frst, t s not dcult to see that DEB = and EDB = B, so that the trangles DE and B are smlar .
Let M and be the feet of the perpendculrs m B onto DE and respectely. As M s the mdpont of B, we have BM = Also B = B snB = � x = · Therefore DE = x Thus the area of the trange BDE s = d = ! x BM x DE=! x x 58d = Substitutng d = 4, the area of the trange DE s
45.
10 nswer 4 We note that f(x) = (x + 1) +3x+ 1 ) 10. Let g() = + 3, which is strictl incresing odd fuctio Hence f() = 1 iplies g(+ 1 =-9 nd f(b ) 19 imles g(b +1)= 9. Thus, +1=-(b + 1) iplying +b =-2, so tt (+ b)2 =4. =
11. Answer 1 Let x=tan , =n nd z=ta' Then x+z+z xz x++ x++ xz
4 -5 17
6
-
17 5
- and x + +zx=�· It follws tht
om the bove thr ee equations, xz= tn(a + +')=
x + +z- xz 17/6- (-5/6) = =11. 1- 2/3 1- (x+z+zx)
1 2. Answer 2 20 23 Include the oints E F G H, I, J, K, LM, Nnd Pin the diagrm s shown nd consider ll possble rotes: km
B
F G
10 km
c
H
I
J
M
L
D A
For the route A E - B there is 1 wy For the route A- F- B, there are For the rote A- G- B, thee r e
C2 · (D c ) � °
2
53
·
=80 wys. =160 ys
� (D ! !! c ) el
For the route
A� H �I� B there are
·
For the roue
A� J �I� B there are
·
For the route
A� J � K � B there are
For the route
A� L �I�B there are
For the route
A� L � K � B there are
For the route
A� L � M B there ae
For the route
A� N � B there are
For the route
A� P �B there are
·
1 = 385 ways. 2
·
=
11 ways.
22023 ways.
4021
anl a=
n 3, we have
an an-1 =
n
an >
= 4410 ways.
= 490 ways.
·
n 2
= 7350 ways.
·
·
)
-
for
= 1176 ays.
·
Rearanging the recurrenc relation yields or
= 5880 ways.
·
·
Hence, by adding up, there are altogether 13. Answer:
= 980 ways.
-
n (an an-1 . ) for n 2. Thus, n+
n- 2 n 3
n (n-1 an-2 = n · n 1 ( an-2 an-3 ) n- 2 n 3 2 1 ··· = · n -1 ··· 4 · 3( a2 a1) 2 2 - - n(n 1 n 1 n
3. Using the method of dierence, we obtain an= 3 - � or n 3. Given that n 2009
2+
2010
, w have
3
2
2009
2 3 - > 2010 2010' yielding n > 4020. Hence the least vaue of m is 4021. 14. Answer:
14
We have
9 8 sin 50°
8 8 sin 0°- 8 sin 1° -8 sin 50° 9 + 8 sin 1° - 8 (2 si 0° cos 20°) 9 + 8 sin 1° - 8(1 - 2 sin2 1° 16 sin 2 1 + i 10° + 1 = (1 + 4 sin 1° 2 8 sin 506 1 + 4 sin 0°. .·. b+c = 14. hus, a = 1, b= 4 and c = 10, and ence a 54
15.
Anse:
3
It can be easly seen that he gven equaton has exactly one real root a snce (1) all olyomal equatons of egree 5 has at least one r eal r oot, and 2 he functo (x) = x- x + x- 2 s strctly ncr easg snce f' (x) = 5x4- 3x + 1 > 0 for all r eal vlues of x. It can also e checked that f �
6 <
<0
and (2)
>
0
1
so that 2 <<2. Ths s equlent to
snce
6 <
-
4- a + <
-
< 2-a5a++2 2 <0 1 - < a <2. -
In addton, e clam that
6 : 3 nce
a6 : 3
4- + 2a 3 - + 2 - 3 : 0 2 - a + 2 : 0
the last inequalty s alays tr ue Hence 3 16. Answer :
: a6 <
thereby showng that L 6J =
3.
8030
Any odd numr geater than 1 can be tten as 2k+1 whr e k+1 = (k+1)-k. Hence al odd neges greate than 1 ar e o "cute ntgers. lso, snce m = (+1)-(m-1? o that al ntegers of h fo rm m, wher e m > 1 ar e not "cute. We clam tha all ntegers of the fr m m + 2 ar "cute. Suppose m + 2 for 1) s not "cute, then
or some eers postve teers x nd . oever, x + and x - ave e sae par ty, so that x - y and x + must all be eve snce m + 2 s evn. Hence m + 2 s dvsble by hc s absrd. Tus, m + 2 s "cte. Th r st fw "cute nteers ar e 1 , 6, 10 For n > 3 the nth "cute neger s n- 10. h, he 00t "cute nteger s (2010)- 10 = 8030. · · ·.
17 Anser :
81
We ha f (1) =
3 f(2) = 1 f(3) = 7, f( 4 ) = 36
and
f(x) = x-x-1) x) + (x- 1) wee g(x s a olnoa n g(4) = 3. Thus
fx Tus,
=
f-1
[
x
(X _ _ 1). (3) .
o gee 3. Hene
(
x-
(1 = 3 g(2 = 0 g(3)
9 so
2)- 3) 4 + . (x- 1)(x- 2)x- 4)
2-3 (x-1) x)(x + (3. + (X _ 1 ()(2
=
=
ta ts sqr s .
5
((1)()
•
ad
18 Answer: 18 Since a and b are positive, we must have a > 100
Solving for b, we get b =
= 100 + Let a = 100 + m, where m is a positive integer. Thus b = Therefore m must be a fctor of 1010 = 101 x 2 2 x 5 2• Converely, each factor r of of the equation 100(a + m determines a unique solution (a, b) = (100 + r, 100 + b) = ab -10 0 There are 18 = (1 + 1)(2 + 1) + 1 factors of 10100 Consequently there are 18 soutions of the given equation. In fact, these 18 solutions can be found to be (a, b) = (101, 10200), (102, 5150), (104, 2625), (105, 2120), (110, 1110), (120, 605), (125, 504), (150, 2), (200, 201), (201, 200), (02, 150), (504, 125), (605, 120), (1110, 110), (2120, 105), (2625, 104), (5150, 102), (10 20, 101) 19 Answer: 17 If both a, b are odd prime numbers, thenp is even and � 4, contradicting the condition tat p is prime. Thus a =2 or b=2 Assume that a = 2 Then b i 2; otherwise, p =8 which s not prime. Thus b is an od prime number. Let b
2k + 1, where k is an integer greater than 1 Thus
We shall show that k=1 Suppose that k � 2 If k = 1 ( mod ), then b > and
b =2k + 1
=
O ( mod ),
contradicting the condition that b is prime. If k ¢ 1( mo d ), then
a contradiction too. hus k =1 and b= 3. Thereore
20 Anser: 10045 The number of grid points (x, y inside the rectangle with four corner vertics (0, 0), (11, 0),
(0, 2010) and (11, 2010) is (11 -1 )
X (010
-1) =20090
There are no gid ponts on(x, theydiagoal etween between excluding these two ( 0) and(0, (11, 0) 2010) 2010), we points, since for ay point on his segment and (11, have
2010x y=-· Bu for an integer x with 1 : x : 10,
is not an integer.
Hence the number of grid points (x, y inside the triangle formed by corner points (0, 0), (0, 2010) and (11, 2010) is
11 -1 )
X
(2010 -1)/2 56
=
20090/2 =10045
(11, 2010),
(0, 0), (0, 2010) and
For any grid point (x, ) inside the triangle formed by corner points we have
1
2009,
: X<
11y 2010'
1
Thus, for any ed positive integer , tere are the number of grid points satisfying the condition that : x < is
2009. Tus the number of grid points (x, ) inside 0), the tr iangle for med by cor ner points and is (0, 11 + 11 2 + 11 3 (0,+2010) 11 4 +(11,... +200)11 2009 . l2010J l 010 J l 2010 J 2010 J l 2010 Therefore the anser is 10045. 21. Answer: 2007 3. We shall show by inuction that the number Consider distinct numbers where of friendly pairs is always 3 for 3. Hence the minimal number of friendly pirs is also 3. If 3, then there are no friendly pais as each number is adjacent to the other two. Thus the number of friendly pairs is 0. Assume that the number of friendly pairs for any arr angement of distinct numbers on the circle is 3 for some 3. Consider +1 distinct numbers on the cir cle. Let be the largest number. Now consider the nubers a
is not an integer wen
1
J
:
:
2
2
=
2
on cirdo clenot after . The numbersAny adjacent which a friendly pair fordeleting m a friendly pairtwo anymore. other to friendly pairorriginally emains for a frmiendly pair when is deleted. n the other hand, any fr iendly pair a er m is deleted was or iginally a fr iendly pair. By induction hypothesis, there are - friendly pairs after is deeted. Therefore, there are ( + friendly pairs srcinally
3
1) - 3
22. Answer: 12345 We are given the equation f(x2+f(z))
0 0 (1), (1),
=
xf(x) +zf().
(1) 0. (2)
Substituting x = = into we get zf(O) = f(O) for all real number z. Hence f(O) = Substituting = into we get xf(x) Similarly, substituting x =
Substituting =
0 (1) f(f(z)) in
we get
1 into (3) we get
f(f(z))
for all real z. Now, using
=
(2) and (4), we obtain 57
zf().
(3)
zf(1)
(4) (5)
Substituting y =z =x into (3) also ields f(x(x)) = xf(x).
(6)
Comparing (5) and (6 it follows that x2f(1) = xf(x), so that i x is non-ero, then f(x) =x for som constant . Since f(O) = 0, we also have f(x) =x for all real values of x. Substituting this into (1), we have that (x2 + yz) = x2 + yz. This implies that 2= , forcing = 0 or = . Since f is non-ero, we must hve = 1, so that f(x) =x for all real valus of x. Hence f(345) =1345. 3 Answer: 864 The number of was of choosing objects from n dierent tpes of objects of which + rpetition is allowed is n � ) . In prticular, if we write -digit umbers using n digits allowing for repetitions with the condition that the digits appar in a non-increasing + ordr, there are n �- 1) was of dong so. Groping th givn numbrs into dierent ctegories, we have the following tbulation. n order to track the enumeration of these elements, the cumulative sum is also computed:
n+�
Numbers Beginning with
Digits used other than the e part
n
1
1,0
5
=6
6
,1,0
3
5
=1
7
3
3,,1,0
4
5
56
83
4
4,3,,1,0
5
5
=16
09
5
5,4,3,,1,0
6
5
=5 461
6
6,5,4,3,,1,0
7
5
=46 93
7
7,6,5,4,3,,1,0
8
5
=79 75
6
5
= 5 1967
4
4
Fom 800000 to ,4,3,,1,0
Cumultiv
855555 Fom 860000 to 3,,,0
=35
00
863333 The net 6-digit numbers are: 84000,84100,86410,864111,86400,8640,861,8640. Hence, the 00 6-digit number is 8640. Therefore, x=8640 so that l J =864. 4. nswer: 309 Lt be the set of permutations of the si intgrs from to 6. Thn =6! =70. Dene P( i) to e th subset of suc that th digit i + fol lows immediatel i, i = ,,3,4,5. 58
Then
� IP(i)l
=
2
( �) 5 !
_I 1P(i)nP(i2) (� )4! . 1P(i) nP(i2) P(i3)1 (� )! =
21 <2
=
21 <2<23
_I . 1P()ni2)nP(i3)n(i) =
21 <2<2<24
IP()n()nP()nP(4)nP(5)1
=
( 2! (�) 1!.
By the Prnciple of Inclusion and Exclusion, the required number is
25.
Anwer: Note that
201 1 2
A =
2
2
(COo10 -( (2�0 + (201 0)(20010()(21+0 (20210 - C01 0 +...+ G� G )
2
=
_
�f k=O
k ) ( _
2
.
Observe that
is the coecient of
x00 in the expansion of the following expression:
We also have
20 20k 101 ) k 20k=OL (01k0) k - 20 20101 k (20k10 - ( L k k=O k=O x+)200 _ xx+)200 1 _ xx+)200 200 ((2010 2010 k 2010 2010)xk-! 2009 ( 2010 )xk+ X
=
=
and
k =- 1
k
=
X
X
=
x
+ k 1
=
k
k =O
1 (x+ 2010 x+)2010 _) X
59
=
X
L
k=-1
+ k 1
+)200.
1 _ 1/x)x
The coecient of x2 in the epansion of the following epression: ( - x)(x )2(1 - /x)(x + )2 = ( - /x - x)(x + )2 is
(400) - (400) - (400) = (400) - (400) . 00
0
Hence A=
009
0
(400) (400) · _
2009
2010
Consider he iequaity: sA 2
(400)
2010 '
( - 00) 2 , 2 0 Hence the anser is 0
60
09
Singapore Mthematcal Socety Od (O)
010
(Open Section, Roud 2) Saturday, 26 June 2010
0900-1330
INSRUCONS TO COESANS 1 Answer ALL 5 questions. Show all the steps in your woring. 3 Eah quetion caries 10 mar.
4. No calculators are allowed.
1. Let CD be a chord of a circle r1 and AB a diameter of r1 perpendicular to CD
at N with AN> NB. circle r2 centred at C with radius CN intrsects r1 at points P and Q. The line PQ intersects CD at M and AC at K; and the extension of NK meets r2 at . rove that PQ is perpendicular to A. , , . . . be two sequences of integers dened by a1 for n; ,
2. Let an, bn, n
=
an1
=
7 an + bn + 6
bn1
=
4an + 7bn + 3
=
, b1
=
0
and
rove that a� is the dierence of two consecutive cubes. 3. Suppose that a1, ..., a15 are prime numbers forming an arithmetic progression with
common dierence> 0. If a1> , prove that> 30, 000. 4. Let n be a positive integer. Find the smalest positive integer k with the property that for any colouring of the squares of a 2n x k chessboard with n colours, there are columns and rows sch that the 4 squares in their intersections have the
same coour. 5. Let p be a prime number and let x, y, z be positive integers so that 0 < x < y <
z < p Suppose that x, y nd z3 have the same remainder when divided byp show that x2 + y2 + z2 is divisible by x + y + z
6
Singapore Mathematical Olympiad (SMO) 2010 Ope Sectio Roud
2
soutios
1. Extend DC meeting r2 at H. et the radius of r2 be r Note that D = C = CH =r Since M is of equal power with respect to r1 and r2. Thus M MH = MC ·D That is M(MC r =MC(M r giving M =MG. Thus M is the midpoint of C. ·
A
r1isand r2, te s K =LAG= lies on the radical points C, , A, are Thus LAC thatofA tangent to r2. Itfollows that AC concyclic. is perpendicula go soaxis to at K, and hence M=MC =MK.
Now let PQ intersect A at . We have LAK = LKM = LKM = LK and similarly LK = LAK. onsequently 2LK = 2(LAK LAK) LAK LAK LK LK= 180°, which means LK=go 2 First we shall prove that a� i the dierence of two cosecutive cubes. To do so we shall prove by induction that a� (bn 1)3- b. When n = 1, this is true. Suppose for n 1, this is true. We have
(bn+l 1)3 b+l =b;+l bn+l 1 =(4an 7bn )2 (4an 7bn ) 1 =48a; 147b; 168anbn 8an 147bn 7 =(7n 1bn 6)2 (b; + bn 1) a; a;+ (bn 1)3 b �=a+l =
where the last eqality is by inuction pothesis.
62
Lemma: Suppose is prime and a1, ... , ap are primes formng an A. P. with comon dierence d I a1 > , we claim that d
3.
Proof Snce is prime and every ai is a rime > , does not divide ai for any i By te pigeonhole principle, there exist 1 � i < j � so that ai aj ( mod ) . Now aj - ai = (j -i)d, and does not divide j - i So mst divi de d Apply the Lemma to the seqences at, ... , ak fork= 2,3, 5, 7,11 an 13. Then all sch k's are factors of d o d > 2 3 5 7 13 > 30, 000.
4.
The answer is
·
2n -n + 1
Consider an coloring o the 2n x k chessboard. A verticalpair is a pair of sqares in the same comn that are colored the same. In every colmn there are at least n verticalpairs. Let P be the total nmber of verticalpairs and Pi be the nmber of verticalpairs with color i Then P = P1+ + Pn nk Ths there is color i wth Pi k. There are e;) = 2n n pairs of rows. Ths if k 2n n 1, there s a pair of rows that contains two verticalpairs wth color i
Next for k = 2n - n, exhibt an coloring where no sch sets of 4 sqares exists. Note that it sces to nd sch ancoloring for the 2n x (2n-1) board. We can then rotate the coors to obtan n of these boards which can then be pt together to obtan the reqiring coloring o the 2n x (2n-n) board. For each i = 1,2,,2n-1, let
Ai = {(i,2n-1+i),(i+1,2n-2+i),,(n1+i,n+i)}, where 2n+k, k > 0, s taen to bek. Note hat the pairs in each Ai gve a partition of {1,2,,2n} Moreover, each pair o eeents appears in exactly one Ai. Now color the sqares of colmn i sing n colors so tat the two sqares in each pair of Ai receve the same color and the colors the 2n pars are mtall dstnct. This gives ancolorng of the 2n board with the reqired property and we are done.
5.
x
2n-1
It is given that
x3y3= (xy)(x + xy + y)
oreover,- < x-y < Oand so f -. Ths and I x + z z Hence
x+xy+ySimilarly, y+yzz
p x+ xy -yz - z = (x z)(x + y + z) Since x -z, x + y + z. Note hat 0 < x + y + z < 3. So x + y + z = or 2p We will show that x y + z Assming this for the oment, the proof is complete if x+y+z = . If x+y+z= 2p, then x+y+z is even and hence x+y+z is even. So bth 2 and divide x y + z Moreover, > 2 and so 2 and are relatively prime. Ths 2 divides x + y + z and the proof is also complete in this case.
63
