ngapo Mahmaal Olympad (MO)
0
no on (Fs Rond)
sday
May 0
0000 hs
Instructions to contestants
Answer ALL questions Enter your answers on the answer sheet pvided. For the multiple choice uestions, enter your answer on the answer sheet by shading the bubble containing the letter (A, B, or E) corresponding to the correct answer. 4. For the other short uestion s, write your answer on the answer sheet and shade the ap ppriate bubble below your answer.
No steps are needed to justify your answers. Each uestion carries mark. No calculators are allowed. Thughout this paper, let denote the greatest integer less than or eual to example, 2.j = 2 3.9J = 3 PLEASE DO NOT TURN VER UNTI YOU ARE TOD TO DO SO.
1
For
Mlpl Cho Qsons
a be the rots o the quaratic equaton vaue o f? s
et
(A) ;
2.
;
(C)
2;
() ;
x2
+ 2bx + b = The smaest ossibe
4
It s ow that n202 n200 is vsibe by or some ositive iteger olowig umbers is not a ossibe vaue or ?
+
(A)
2; •
;
(C) 35;
n.
Which o the
9
(D) 47;
Usig the vertces o a cube as ver tices , how may triagular yrami ca you orm?
8;
(A) 4;
4
(C) 6;
7
() 64;
AB is a chor o a circe with centre CD is the iameter ereicua r to the chor AB with AB coser to C than to D iven that AOB = 9°, the the quotiet area o ABC area o AOD (A)
;
2
-
(C)
2
ABCD is a araelogram an both AE a BE are straight G be the intersectios o BE with CD a AC res ectivey ive that BG = EF the quotient DE AE
The iagra m beow shows that ies et
F
a
2
6 Four circes each o raius a a square are arra ge withi a cir ce o raius
8 as show
i the oowig gure.
What is the rage o
<< 4; 0 < < 8( + ); ) 4 < < 8( );
4 << 4; 4 << 4 + ).
(A)
(C)
H e ites to ABC AB = 5 BC = CAA =to a oit D o sie BC such t hat AD bisects BAC Fi the area o the art o the e ABD
7. Aam has a triaguar e with 8 a searate the e ito two arts by buiig a straight ece rom
(A)
4 . '
(C)
5 '
)
oe o the above
8 For ay rea umber , et be the arg est iteger ess ha or equa to a et
a be rea umbers with such that
Whih o the oowig statemets is icorrect? (A) I is a iteger the
is a iteger;
I is a o-zero iteger the is a iteger; (C) I is a ratioa umber the
is a ratioa umber;
) I is a ozero ratioa umber the is a ratioa umber;
I is a eve umber the is a eve umber. 3
= - J.
9. ive that
=
is a itege r. Which of the foowig is icorrect
x x x ) (A)
(C)
ca amit the value of ay o-zero iteger; ca be ay ositive umber; ca be ay egative umber; ca ae the value
y ca tae the value
10
uose that
2; -2
A B C are three teachers worig i three ieret schools X, Y Z a se
cializig i three ieret subjects Mathematics, Lati a Music. It is ow that
i A oes ot teach Mathematics a B oes ot wor i schoo Z; (ii) The teacher i school
Z teaches Music;
(iii) The teacher i schoo X oes ot teach ati; (iv)
oes ot teach Mathematics.
B
Which of the followig statemet is correct
B wors i school X a C wors i school Y; A teaches Lati a wors i school Z; (C) B teaches Lati a wors i schoo Y; ) A teaches Music a C teaches Lati; (A)
oe of the abo ve.
ho Qsons
11 12
Let
a b be rea umbers such that b, 2 + 2 = 7 a 2 + 2 = 12 Fi the
value of
2
. . . . . F t he sum o f a l ost ve teger s
x
suh t hat
xx--1xx++ 1201 S
(
)(
)
. a teger.
13
Cosier the equation
- 8x + 1 + 9x - 24x - 8 =
.
It i s ow tha t the larges t root o f the equati o is -k times the smallest root. Fi
14
Fi the four-igit umber
(For examle, if
15 x 16 n uose
Let
ab
satisfyig
2(ab) + 1000 = ba. a = 1, b = 2, = = ab = x + - 22x - 20 + 221 = 0 a
4, the
4.)
a are real umbers satisfyig
an
k
Fi
x
be ositie itegers satisfyig
n + 876 = 4n + 217n
Fi the sum of all ossible valu es of
17 18
x, x x x X, x , , X
For ay real umber
let J eote the argest iteger less tha or equal to
value of J of the smallest
uose
x - x = 100 J
J
Fi the
X + 2x + + 49x ·
Fi the miimum value of
+ 20- + + 21 - z + + 20 - + 20 Let
+ 21-x
be a 4-igit iteger . Whe bot h the rst igit (left- most) an the thi r igit are
n a the seco igit a the fourth igit are ecrease by umber is n times A Fi the value of A icrease by
21
x
are real umbers such that
Fi the maximum value of
19
satisig
Fn the remaier when
1021
is vie by
5
1023
n,
the ew
22.
onser a lis of six numbers When he lrges number is remove from he ls, he average s ecrease by When he smalles number s remove , he average s ncrease by When boh he larges an he smalles numbers are remove, he average of he reminng four numbers s
23.
2
For each osive neger
2.
n , we ene he recursive relaion given by
uose ha
24.
Fin he rouc o he lrges an he smalles numbers
a aa Fin he sumaof he suares of all ossble values of a
A osive neger is calle riny f i is visble by he sum of is igis For exame, s frienly bu is no Fn he number of all wo-gi frienly numbers
E B, E, BE E B B. B0• n the iagram below,
an
le on the sie
an
B 2 B.
an lies on he sie
t s gven ha
where
such hat Fin ,
2.
B(2, B 9°. E
n he agram below, hat
of rangle
an
3 B. E.
lie on he -axs an
A rectangle
(,
lies on he -axis such
is inscribe in rangle
s , calculae he area of the recangle
y
Gven ha he area
c
6
X
ABCDEF e a regular eagon. Let G be a poin on ED suc tat EG = 3GD . If te AEF is 00, nd te area of te eagon ABCDEF G D E
27 . Let area of
F
A
B
28 . Given a pacge containing 200 red marbles, 300 blue marbles and 400 green marbles. At eac occasion, yo are allowed to witdraw at most one red marble, at most two blue marbles and a total of at most ve marbles out o f te pacage. Find te minimal number of witd rawals required to witdraw all te marbles from te pacage. 29. 3 red marbles, 4 blue marbles and green marbles are dstributed to 2 st udents. ac student gets one and only one marble. In ow many ways can te marbles be distri bted so tat Jamy and Jaren get te same color and Jason gets a green marble?
5
30. A round cae is cut into
n pieces wit 3 cuts. Find te product of all possible values of n
3. How many triples of nonnegative integers
x, y, z satisfying te equation
xyz xy yz zx x y z = 202? 32 . Tere are 20 2 stdents n a secondary scool. very stdent writes a new year card. Te cards are mied up an d randomly distributed to students. Supp ose eac stdent gets one and only one card. Find te epected number of students wo get bac teir own cards.
A
B
A
33. Two players and play rocp aperscissors contnuously until pl ayer wins 2 consecutive gaes . Sppose eac player is eqally liely to se eac andsign in every game. Wat is te epected number of games tey will play?
34. Tere are 2012 students standing in a circle; tey are numbered 1,2 , ,20 12 clocwie Te counting starts om te rst student ( number 1 and proceeds arond te circle clocwise Alternate students will be eliminated from te circle in te following way: Te rst stdent stays in te circle wile te second student leaves te circle Te tird student stays wie te fourt student leave s and so on Wen te counting reaces number it goes ac to number and te elimination continues until te last student remains Wat is te umber of te ast student?
2012,
1
3. Tere are k people and n cairs in a row, were 2 k < n Tere is a couple among te k people Te number of ways in wic l k people can be seated suc tat te couple is seate togeter is equal to te number of ways in wic te (k-2 people, witout te coupl presen, can be seated Find te smalest value of
8
n
ingaoe Matheatial oiety Sngapore Mathematcal Oympiad (SMO)
Junior Section (First Round)
Mlpl Cho Qsons
1. Answer ) Sin x + 2bx + (b-1 = 0, we ave = b-1 and + f = -2b. Ten = + - 4f = (-2b 3 4 3 = 4(b - b + 1 = 4[(b - 1 2-4( +b-1 Te equality olds if and only if b = 1 2 2 Answer () Not tat n 2o 2 + n20o
If If
=
n 20o (n 2 + 1 .
n = 2, ten 5 (n + 1 and 2 n . If n = 13 or 47, ten 10 (n + 1. If n = 59, ten 5 (5 9 + 1 and 5n .
n
=
35,
then
2 (n
+
1)
and
5 n 00 .
3. Answer Tre are 8 verties in a ube. Any 4 vertics form a triangular pyramid unless tey lie on te same plane. = 8 4. Answer Let e radius of te irle be 1. Ten sBc AE E AE CE 2 CE 2(1-1 ) SOD SOD AE = 1 = 2
!
_
9
c
5. Answer: A . DE . Ten = EF . Note tat Let AE EB DC DF EF + FG EF + EB-2EF + 1 B AB + FCAB EB GB EB 1 3 2V ( < < 1). Ten + - 3 = , and tus = 6 Answer: () onsider te etreme cases:
8-4
Ten i
2
= 4 - 2�
and Xmax
=
8
�2 + 1 = 8(� - 1 ).
=
7. Answer: (D)
B y Heron's formula,
s6ABC = s6ABD s6ACD
5) (12 - 8) (12 - 11) 4v . � X AB X AD X sin AB 5 � X AC X AD X sin AC 1 1
Then S6ABD =
=
5 5 + 1
X 4 =
5v
2
8 Answer: ().
Note that = b ·
� ( l�J { �} ) l� { �} = b
It follows from =
= b
+
+b
{} +
tat
·
= . He ne,
=
is an integer.
Obviously, is not neessay an integer even if is an integer.
9. Answer: ()
0, ten y = X = X x=X = 0 xX = 2x = -2 If x < 0, ten y = = ( x = X X X x an tae any nonero real number. If x
10 Answer: () e assignment is as follows: A: in Z, teaes usi; B in Y teaes Latin; C in
X,
teaes atematis.
ho Qsons
11
4 752 (2 + 2 )(2 2 ) = 2 2 + 2 Ten 2 2 = 417 = 4 1 Sine , we obtain 2 = 4 Answer:
Answer:
25(x 1)( x 1) = x x . t s an teger "f and ony . f (x2 1) 120 2 en x - 1 = ±1 2 3 4 5 6 8 10 1 2 15 20 24 30 40 60 120 Hene, x = 0 2 3 4 5 11; and 2 + 3 + 4 + 5 + 11 = 25 13 Answer: 9. 12
N
ote that
x 3 - x + 120
1 20
I
.
Let =
8x
Ten te equation becomes 1 = 3
1 = 3 . Squaring bot sides, we ave 3 - 1 = 9 6 ; tat is, 0 = 0 Ten = 2 or = 5 (rejecte because 0) Solve 3x - 8x = 2 . Ten x = 3 and x = 3 Hence, k = 3/3 = 9 14 Answer: 299 Rewrite te equation as te following: a b a b + 1 0 0 b a Since a is te last digit of 2, a is even; since 2a 1 9, a 4 So a = 2 or a = 4 If a = 4, ten 2a 1 = 9 and tus = 9; but ten te last digit o 2 would be 8 a, a Ten + 3 -
contradiction.
If =
ten
= and te last digit of
is
so =
Te equation reduces to
a 2, 2a 1 5
b 2 2; b 1 b Tere are 2 cases: eiter 2 = b and 2b = 10 , wic as no integer solution; or 2 + 1 = 0 b and 2b 1 = 10 + , wic gives b = 9 5 Answer: 10 omplete te square: (x - 1) + ( 10) = 0 Ten x = 1 and= 10; tus x = 10 Answer:
Rearranging,93 we ave
8 n 2 n 217 Ten ( n27) 8 It follows tat n27 = ±1, ±2, ±4, ±8 So n = 218, 2, 219 ,25 ,221, 2 3, 225 ,209, and n = 4, 12, 0, 8, 2, ,3,5 respectively. n 21 225 N ot e t at = s a post Ve teger. T en = = 8 or = = 75 n 12 3 mn - 2 7n = 4mn - 8 76
=
n =
12
4mn - 8 76 7
= 4-
17. Answer 50. Write x = xJ + Ten 100(xJ + -xJ = 2 xj -+ < 2xJ + . S o xJ 50 and x x J = 100 +50 = 2600. On te ote and, x = v is a solution. 18. Answer 35. (X + 2X + + 49X ) = 1 X + X + " " " + " X r (= 491 +X 250+X 1 += 49)35 . x + 2x + + 49x 2 1 Te equality olds if and only if x = = X = ± 35 19. Answer 58. As sown below,
+ (20 - ) + + (21 z) + + (20 - ) + + (2 1 - x) = AB + BC + CD +DA = A'B + BC + C + A" A' = 42 + 40 = 58. - " ,
,oI C' B
; .
"
;I
- z
\\\
\
\j
"
20. Answer 1818. Let te 4digit number be
A = ab . Ten 1000(a + n) + 100( b - n) + 10( + n) + ( n) = nA. It gives A + 909n = nA; or equivalently, n - 1 )A = 909n. Note tat (n-1) and n are relatively prime and 101 is a prime number We must ve (n- 1) So n = 2 or n = 4. If n = 4, ten A = 1212 , wi is impossible sine b < n. So n = 2 and A = 909 2 = 181 8. 13
21
4 Note tat 1024 = 2 1 (mod 1023) en 1021 ( _ 2) 2 2 1024 2 1 2 4 Answer:
(mod
1023)
22 Answer: 375. Let and M be te smallest and te argest numbers. Ten +5 80 + 1 = M +5 80 1 = +M6 + 80 Solvng te system, = 5 and M = 25. Ten M = 375. 23. Answer: 3. Let
a =a
Ten
+ 2a a = +1 a a = 1 a a = 3+2 + aa a= 3 +3a a = F + F a +a
. . . . In general,
where F = 0, F = 1 and Fn = Fn + Fn for all n � 1 . If a =
F o + F oa
Snce F
>
= a, then (a + a - 1) F o = 0.
0, we have a + a - 1 = 0. Let
a
and ( be the roots of a + a - 1 = 0. Then
24 . Answer: 23.
ga lO a + b = + 1 , we have (a + b) ga. a+b a+b Case 1 : If 3 f (a + b), then (a + b) a, and thus b = 0. We have 10 20 40 50 7 0 80. 3a 3a Case 2: If 3 (a + b) but g (a + b), then (a + b) 3a and 1 3. If = 1, then a+b a+b 3a 3a 2a = b, we have 12 24 48. If = 2, then a = 2, we have 21 42 84. If = 3, then a+b a+b b = 0, we have 30 60. .
Se
-
Case 3: If g
(a + b), then a + b = g or 18. If a + b = 18, then a =b= g,
which s mpossible.
If a + b = g , then we have g frendly numbers 18 27 36 45 54 63 72 81 g . Therefore, n total there are 23 2-dgit friendly numbers. 25. Answer: 108. Since DA = DE = F , LEF A = g .
Let LEBF = LEFB =
LB FC =g - x0 and L CBF = 2x0, LBA C =g - 2x0•
1
x0• Then LBCF =
. Ten x = 36. So x = 80 90x 902x = 3x = 08.
It is given tat 3x = 2 0x 26. Answer 468.
A = 3020 = 45. Ten te a rea of ABC. 20 + 452 30 = 975. Let te eigt of CGFbe h Ten h = 35 = 53 h = 3 30 975 30 h Note tat te rectangle DEFG as te same base as CGF Ten its area is
Note tat
35
2 2 = 468
27. Answer 240 .
DE
EG = 3 4. 1 + 2
We may assume tat = and deno te = + . + = mp es 8
EGA 1 EY
Note that
AB CDEF DEXY
=
AGEF DEY
6
8"
Then
8 1 + 2 6
AGEF
AB CDEF 12 If AGEF = 100, then ABCDEF = 10 0 x = 240 .
S
E
G
D
Fl I I I I I I I I I
X/ A
B
28. Answer: 20 0.
15
5 12
Since at most one red marble can be witdrawn ea time, it requires at least
200 witdrawals.
200 i possible. For eample, 150 . (1, 2, 2) + 20 . (1 , 0, 4) + 10 . (1 ,0,2) + 20 . (1 , 0, 0) = (200, 300, 400). 29. Answer: 3150. ase Jamy and Jaren bot tae red marble. So 1 red, 4 blue and 4 green marbles are ditributed to students: 9 = 630. ase 2: Jamy and Jaren bot tae blue marbles. So 3 red, 2 blue and 4 green marbles are disributed to 9 students: = 1260. ase 3: Jamy and Jaren bot tae green marbles. So 3 red, 4 blue and 2 green marbles are distributed to 9 tudents. Te number is te ame as case 2. 630 + 1260 + 1260 = 3150. 30. Answer: 840. On te oter and,
:
G
=
Wit tree cut, a round cae can be cut into at least
1 + 1 + 2 + 3 = 7 pieces. oreover,
31.
n = 4, 5, 6, 7
pieces, and at most
1 + 3 4 45 6 7 = 840.
are all possible.
27. x + 1) ( + 1) (z + 1) = 2013 = 3 11 61. If all x, , z are positive, tere are 3! = 6 solutions. If eactly one of x, , z is 0, tere are 3 6 = 18 solutions. If eactly two of x, , z are 0, tere are 3 olutions. 6 + 18 + 3 = 27. 32. Answer: 1. 1 . Ten te epectation For eac student, te proability tat e gets bac is card is 20 12 12012 = 1. of te wole class S. 2012 Answer:
16
33. Answer: 12. Let E be te epectation. If A
A does not win, te probabilty is
2/3 and te game restarts.
(1/3)(2/3) and te game restarts . e (1/3)(1/3). en E = 2 (E + 1) + 2 (E + 2) + 1 2.
If wins and ten does not win, te probability is robability tat wins two consecutive games is
A
Solving te equaton, we get
E = 12. 34. Answer: 1976. f tere are 1024 = 2 students, ten te 1024 student is te last one leaving te circle. Suppose 2012 - 1024 = 988 students ave le. Among te remaining 1024 students, te last student is (2 988-1) + 1 = 1976. 35. Answer: 12. (n - 1 )
x
2
x
(� = �
x
( k 2) ! =
(k : 2)
x
( k - 2) ! .
Then
2=
(n
_
k+
1
n
_
k + 2) "
at is, 2n -(4k6)n + (2k - 6k + 4 n) = 0. We can solve -7-5 (n k). Note tat te squ are of any od d number as te form 8k7. oose k so tat - 7-5 = 4, .e., k 11. en n = 12.
17
Sngapoe Mathematca Socety Singapore Mathematical Olympiad (SM)
(Junior Secti, Rond 2) 093-1230
Satrdy, 23 Jn 2012
1.
Lt be the centre of a paralllogra BCD and P be any point in the plane Let M, b e the idpoints o f P, BP, espect ively and Q b e the in tersection of MC and D Prove that P and Q are collinear
2.
Does there eist an integer such that each of the ten digits 0, 1, eactly once as a digit in eactly one of the nubers , 2, 3.
3.
n BC, the eternal bisectors of LA and LB eet at a point D Prove that the ci rcucentre of BD and the poits C, D lie on the sae straight line.
.
Deterine the values of the positive integer equaions has a solution in positive integrs each sch
for which the following syste of
x1, x2, , Xn Find all solutions for
1Xl + 1X2 +· · ·+ Xn 1 = 116 = + + . ·+ Xn
X1
X2
9 appears
(1) (2)
Suppose = a1, a2, , a15 is a set o distinct p ositive integers chosen fro 2, 3, ... , 2012 such that every two of th are coprie Prove that co ntains a prie nuber (ote" Two posit ive integers are coprie if their only coon factor is 1 )
18
Sngapoe Matheatca Socety Sngapore athematcal ympad (S)
(nior Sction, Rond 2 soltions)
1.
Sinc MN AB CD, w hav MQN" CDQ. nc MN AB/ CD/.
Thus Q is a dian an through QM ACP, Q dividsQ.CM in th atio 12. CQ/. PO passs Thus is th cntroid.nnc th CM dian
p
2
2.
3
totand al nu digits in o fA,digits is total in 2, 32this 11 A and A ust th to talof nu in 20A, 20is2,10,20th solutiondigits 32, 32 Sinc 9 annur satis 21 A 31. Sinc th unit digits of A, A2, A3 ar distinct, th unit digit of A can onl 2, 3, 7, 8. Thus th onl p oss i valus of A ar 2 2, 23 , 27 , 28 on of th has th dsid propt. Thus no such nu ists . 3. ot th at CD iscts LC. f CA CB, thn CD is th p pnd icula r iscto of AB. Thus th circucn t of ABD is on C.
f C A CB, w a assu that CA > CB. Lt a point on CB tndd and th point on CA so that C C B. Thn, sin CD is th ppndic ular iscto of
19
B, we have A B BA hus AB is a cyclic quadril ateral, i e, is on the cir cuci rcle of A he cicu centre lies on the perpendicular bisector of B which is C
. Without loss of generality, we ay assue that S S S X· f 1 then fro (2) and () cannot be satised hus � 2 f 2 then 2 and again () cannot e satised hus � 3 Siilarly, X � 4 hus + + X S with � 4 hus 4 s (i) o solution (ii) 2 he oly solution of + _2 is 2 whic doesn 't satis () hus there is no solutio (iii) 3 he only solutions of + 12 + are , , (2 3 6) (2 4 4) and ( 3 3 3) hey all do not satisfy () (iv) 4 ccording to the discussion in the rst paragaph, the solutions of + + X 6 are , , ,
(2 3 4 ) (2 3 6) (2 4 4 6) (2 4 ) (3 3 4 6)(3 3 )(3 4 4 )(4 4 4 4)
Only the ast one satis (2) hus the syste of equations has a soution only when only solution s X X 4
4 and for this the
Suppos e, on the contrar y, that co ntains no pries For eah i, let P be the sallest prie divisor of a hen , , , are istinct since the nue rs in are pairwise coprie he rst pries are 2 3 1 13 9 23 29 3 3 4 43 4 f Pj is the larg est a ong , , , , then Pj � 4 and aj � 4 4 239 > 22 a contradiction hus ust contain a prie nuber
2
Singpor Mhmil Oympi (SMO)
0
Snior Sion (Fir Ron)
y
My 0
0000 hr
Inrion o onn
Answer ALL questios. Enter your aswers o the answer sheet pvided. For the multiple choice questions eter your answer on the answer sheet by shading the bubble containing the letter (A or E) correspoding to the correct aswer.
4- For the other short questions wrie your aswer on the answer sheet and shade the appropriate bubble below your answe.
No steps are needed to justify your aswers Each question carries mark. No calculators are allowed. Throughout this paper let x deote the greatest intege less than or equal to x For example
2
2
3
3
PLEASE DO NOT TURN OVER UNTIL YOU RE TOLD TO DO SO
21
Mltipl Choic Qstions
. Suppose a
are real uers that satis the equatio
J+
x + 2 i the value o
A C E
J+ 0
x+
+
J+ J - - 8
() 8 -
+ J - J+ J - 1 - 8
(D) Noe o the aove
2. i the value o 20
202 - 203
203
202
A C E
+ + + ! ! ! + - - ! ! ! + - - ! ! !
() (D)
204 - 205
+ - - ! ! ! + + + ! ! !
3. The icreasig sequece T 2 3 5 6 7 8 0 cosists o all positive ite gers which are ot perect squares . What is the 202th ter o T?
A
C
() 2056
2055
E
(D) 2058
2057
2059
4. Let e the ceter o the iscrib e circle o triagle ABC a e the poit o AC with AC AB 0 AC 9 BC C
A
4
(B)
4.5
(C)
5
(D)
22
5.5
E
6
5. id the value f
A
()
cs 75°+ si 75°+ 3 sin 75° cs 75° cs 75°+ sin 75°+ 4 sin 75° cos 75° ·
C
(D) 1
E
cs 75°+ si 75°
6. If the rts f the equat i x + 3 - 1 = 0 are als the rots of the equatio x+ ax + bx+ = 0 d the value f a+ b+ 4.
A
13
() 7
C
5
(D) 7
E
1
7. id the su f the dgs f all ubers i the sequ ence 2 3 4 . . . 1000
A
4501
() 1215
C
13501
(D ) 4500
E
Ne of the abve
8 id the uber f real sl utis to the equa ti
X s,
100 here x is easured i radians.
A
30
() 32
C
62
(D) 63
E
64
. that I theB triagle AC, A of AC, AC ad thetpi s AC that such is the agle bisectr If B the tpit such = AC = is40°eteded DE= AD , d CA A
A
20°
() 30°
C
E
40°
50°
10. a e ad are psitive, d itegers such thatpo If the beidetical > value 2012 the sallest ssible f last +t hree digits f 2012
A
8
() 100
C
(D) 104
102
3
E
Nne f the abve
Shor Qsons
11 Let a c and d be four distint positive real numbers that sais the equations a2 m 2
and
_
c2 o1 2) (a2 01 2
(b 2 o1 2 c2 o1 2) (b 2 o1 2 of (cd ) 2 01 2 - (ab ) 2 01 2 . _
Find the value
_
_
d 2 0 1 2) d2 o1 2)
=
=
2011
2011
12 Det ermine the tot al number of pairs of integers x and that satisf t he equation
1
1
+2
1
' 2
1 , a olletion of subsets of is s aid to be ntersectng 13 Given a set 1 2 if for an two subsets A and B n we have A B What is the maimum size of? 14 The set ontains all the integral values of m suh that the polnomial 2 ( m 1)x m m+ 12)x +6m has either one rep eated or two distint integral roots Find the number of elements of 1 Find the minimum value of
sin x+ os x +
osx - sinx os2x
16 Find the numb er of was to arran ge the etters , , , , suh that there are no onseut ive idential letters
1 7.
Supp ose
x
=
is an integer Determine the value of
C C D and E in a line x
18 Let x) e the po lnomial (xa )xa ) (xa )xa)(xas) where a a a a and a are dist int integers Given that (1 4) 2012 evaluate a+a+a+a+a 19 Suppose x y z and are positive real numbers suh that yz 6x xz 6y xy 6z x+y +z 1 ind the vaue o (xyz)
20 Fid the least value of the epression (x + y)( + z), gien that x, y, z are positive real ubers satising the equatin
xyz(x+y+z ) 1 21 For eah real nuber x, let (x) be the iiu of the nu bers 4x + 1 x + 2 and - 2 x +4 Deterie t he aiu v alue of 6(x) + 2012 22 Fid the nu ber of pairs (A, B) of distit subsets of 1 2 3, 4 6 suh that A is a proper subset of B. Note that A a be an ept set 23 Fid the su of all the ite gral values of x that s atisf
+ 3- 4 + + 8- 6 1 24 Give that
I x+ 4x+
- x+ 2x +
for real values of x, nd the aiu value of .
2 Three itegers are seleted f ro the set 1 2 3, 19 20 Fid the u ber of seletios where the su of the three iteger s is divi sible b 3 26 I the diagr a below ABCD is a li quadrilateral with AB AC The line FG is tangnt to the irle at the point C, and is parallel to BD f AB 6 and B 4 d the value of 3AE A
2
27 Two Wei i teams A and B, eah omprising of 7 members tae on eah oter in a ompetition The plaers on eah team are elded in a ed sequene The rst game is plaed b the rst plaer _f eah team The losing plaer is eli minated while he winning plaer stays on to pla with te net plaer of the op posing team This ontinues until one tam is ompletel eliminated and the surviving tem emere s as the nal winner thus ielding a possible gaming out ome ind the tot al number of possible gaming outomes
28 Given that = os B) + sin B) j and = sin B) + os B) j where and j are the usual unit vetors along the xais and th e ai respetivel and B ( 2 f the le ngth or magnitude of the veto m + is given b + = nd the value of os + i + 29 Given that the real numbers x and z satises the ondition x + + z = 3 nd the maimum value of x z ) = +13 + + + +12 30 Let x be a polnomial of degree 34 su that (k) = k (k+ 1 for all integers k0 1 2 34 valuate 42840 3 31 Given that is an aute angle satisfing 480 sin - 2 = 0
360os + nd the value of 40 tan
32 Given that a c d e are real numbers suh that
a+ +c+d+e = 8 and
a 2+b2 + c2+d2 + e2 =
Determine the maimum value of
16
33 Let L denote the minimum value of the quotient of a 3digit number formed by three distint digi ts divided b the sum of its digts Determine 10 £ .
34 ind the last 2 digits of
3 Let ( n) be the integer neare st to
ind the value of
26
Sngapoe ateatca Socet Sngapor Mathmata Oympa (SMO) 202 Snor Ston (Frst on Soltons)
Mltpl Cho Qstons
.
nswer (D) Note th at a
+ + + _ + a+ + 6 + +v+ - + 6 J
a
and a
J
) 3a(a ) ( af)
(J
2. nswer ()
27
3 swer Note that 44 = 1936 4 = 202 ad 46 = 2116 So 2 3 2012 has at ost 2012 - 44 ters For the 2012th ter we eed to add the 44 ubers fro 2013 to 206 ut i doig so we are outig 4 = 202 so the 2012th ter shol be 2012 + 44 + 1= 207 4 swer
B AC is taget to the irle at D, by ostrutig E ad F as show we have CD= CF, A= AE ad BE= BF. Solvig for the uows give CD=
swer (D) os x + si x+ 4 si x os x = ( os x + si x si x + os x si x os x + 4 si x os x = si x+ os x + 3 si x os
6 swer We have the fatorizatio
x + 3x - 1 x + x - = x + ax 2 + x + oparig oeiets give 3 + = 0 -1 + 3= a ad -3 = We a solve these equatios to obtai a+ + 4= 7 7 swer og all ubers with 3 or le ss digit s eah i i= 0 1 2 9 appe ars eatly 300 ties Thus the su of the digits of all the ubers i the sequee 1 2 3 4 999 is 300 1 + 2 + + 9 = 1300 ad so the aswer is 13 01
28
8 swer: (D) Sie -1 si x 1, we have -100 x 100 We also observe tha 31 < 100 < 32 = si x has eatl two sol utios i For eah iteger k with 1 k 16, [2 (k1) (2k - 1) ], but i t has o solutios i (( 2k - 1) 2k) Thus this equa tio has eatl 32 oegativ e real solu ios i e x = 0 ad eatl 3 positive real solutios The it also has ea tl 31 egative real solutios givig a total of 63
9 swer: ostrut apo it F o C suh that F= A. Sie LA LF A is ogruet o F. Thus F = A = E ad LF = LA = 60 We also have LEC = LA= 60, whih implies that LFC= 60 ad CF is ogruet to CE. I olusio LE= LFC= 40 A
10 swer: (D) We wat to solve 2012 _ 2012 mod 1000) whih is equivalet to
Sie (12 - 1 ) is odd we mus have 8 12 so 2 It remais to h e that 125 12 - 1 i e 12 1 mod 125) Let be uler 's phi futio s (125) = 125 - 25= 100, b uler 's theorem we ow that he small est - must be a faor of 100 heig all possible fators we a olude - = 100 ad so he smal lest po ssible valu e for + is 04 sie 2
29
Shor sons
ser: 0
ad = The A ad B are dstct root s of the equato (x - ) (x - ) = 0 Thus the prod uct of roots, AB = - 011 ad - AB = 0
Let
A
= a2 01 2 ,
B
= b2 01 2 , C = c2 01 2
ser: 6 Note tha t x ad must sats · 3 = ( ) e rst assume x 0, hch meas both ad are eve tegers ther 3 or 3 I th rst case , assumg > 0, e have = 3 · ad = (3 . ) = The ol ay for ths equato to hold s k = and x = 3 So (x, y, y+ 2) = (3, 6, 8).
I the case 3 , assumg > 0, have = 3· ad = (3· - ) = No the ol poss iblt s k = ad x = , so (x , , ) = (, 4, 6) . I the to prevous cases, e could also have both ad to be gatve , gvg (x, , ) = ( 3, -8 6) or ( 2, -6, -4) all, cosder x < 0 so 3 = ( ) I ths case ca ol have x = ad (x, , ) = (, , 3) or (-1, -3, - ) Hece poss ble ( x, ) pars ar (3 , 6) , (3 , -8) (, 4) , (, 6) , ( - , ) ad ( -, -3) 3 ser: If A F the the complemet S nA ca belog to F, that s
F. So at most half of all th subsets of F: =
qualt holds because e ca tae
S
F to be all subsets of S cotag
14 ser: 4 If = , the polomal reduces to x 6 = 0 hch has o tgral roots or 1 the polom al factorzes as ((x - ) ( ( - )x - 6) , th roots x = or tegral root s, must be eve ad must dvde 6 The ad x = ol poss ible value s are = - , 0, ad 4 So has 4 lemets
ser: Note that cos x = cos x-s x So cos x - s x = si x cos x s x cos x sx c os x csx Set w = s x cos x ad mmze lw �I· G equalt, f w s postve the the mmum of w s f w s egatve, th the mamum of w + � s - Therefore, the mmum of lw � s
3
·
16 nswer 2220 e shall us the Priniple of Inlusion and lusion There are was to arrange he letters withou t restriti on There are was to arrange the lett ers suh th at both the s our onseutivel (Note that this is the sae nuber if we use or instead of ) There are was to arrange the etters suh that both s and s are onseutive (gain this nuber is the sae for other poss ible pairs) Finall there are was to arrange the letters suh that s s and s our onseutivel For there o be o onseutive idential etters total uber of was is -3 _ + 3 26 5 - 2220 22 222 17 nswer 9 Taing logarith we get log x = + log x Let y = log x The onl possible solution for y = + y is 2 Therefore x = 3 = 9. 18 nswer 17 The prie fa torization of 2012 is 2 503 Let b = 104 If are distint so are b ie b ( a) , b 3 b 4 and b 5 us be eatl the integers 1 1 2 - 503 Suing up we have
5( 104) -
+a+ 3 + 4 +5 = 1 - 1 +2 2 +503
19. nswr 54 utipling the rst three equations b x, y and z respetivel we have
xyz = 6x = 6y = 6z Sine# 0 and x+y+z = 1, we dede that x = y = z = , x = y = z = V" and Hene
20 swer 2 Observe that 1 (x + y)(y +z) = y + xz + y+ yz = y(x + y + z) + xz = xz + xz 2 were te equalit holds if and onl if xz = 1 Let x = z = 1 and y = 1, then we have the iiu value 2 for (x + y)(y + z)
31
21 nswer 2028 Let represent the three lines 4x + 1 x + 2 and -x + 4 respetive Observe that and interset s at G D and intersets at 3) and and intersets at (�, ). Thus
< x r ' << (x) x + 2 ' 8 + 4, > -2x 4x 1
7
Thus the maimum value of (x) is and the maimum value of 6(x) + 2012 is 2028 22 nswer: 665 Sine B annot be empt, the numbr o elements in B is between 1 to 6. er piing B with k eements, there are 2 - 1 possible subsets of B whih ualies for A as A and B must be distint Thus the tota number of possibilities i
tm
(2 1)
(2
m m m
2 = ( + 1) - (1 + 1) 3 - 2 665
23 nswer 45 The euation an be rewritten as If 3 it redues to 10 ; 2 it redues to 2 If
- 2) + 3) 1 - + 3 = 1 ie 3 giing + 3 - 1 ie = 2 giving 5 If 2 < <3 ie 5
24 nswer 4 S
I
+ 2) + (0 - 1) -
32
1
+ 1) + (0 - 2)
Let = (x, 0), A = (-2, 1) an B = (-, 2), ten repreent te ierence between te lengt A an B i maimum wen te point , A an B are collinear an tat oc cur wen = (-3, 0) So
Tu te maimum value of S 4 = 4.
25. nwer: 384 Partition ito tree ubet accor ing to teir reiue m oulo 3: = 3, 6, . . . , 8 , 81 = 1, 4, . . . , 19 an 82 = 2, 5, . . , 2 In orer for te um of tree integer to b e iviible by 3, eiter all tree mut belong t o eactly one Si or all tree mut belong to ierent si Hence total numbe of uc coice i + 2 G + 6 7 7 = 384.
26. nwer: 10 Since B an i tagent to t circle at C, we ave BC = BE = C = BC = BAC rtermore BEC = BAC+ ABE = BE+ AE = ABC = AC e can then conclue tat E = C = C = 4. lo, ABE i imilar to CE If we let AE = x, E AE 2 E = x. = C AB y te Interecting or Teorem , AE · EC = BE· E, ie x(6- x) = 4(�x), wic give x = 13°, o 3AE = 3x = 0.
27. nwer: 3432
e ue a1, a2, • • • , a7 an b1, b2, • • • , b7 to en ote te player of Team A a Tam B, repectively poible gaming outcome can be repreete by a linear equence of te bove 14 term or intanc e, we may ave a1a2b1b2a3b34b5a4bb7a5a6a7 wic inicate player 1 followe by player 2 from Team A were eliminate rt, an the tir player eliminate wa player 1 rom team B However Team A emerge the nal winner a all even player of Team B get eliminate wit a5, a an a7 reaining uneliminate Note a6 a a7 neve actually playe Tu, a gaming outcome can be forme by cooing 7 out of 14 poible p oition for T eam A, wit te remaining lle by Team B player Terefore , te to tal number of gaming outcome i given by C 74) = = 3432.
33
28. nswer: 1
=
+ 2 cs - 2 J sin
m+ = =2
1+ cs
= 2 J cs
+
I i i
Since < ( < 2, we have 7 < + i < that cs + i = - and hence
5 cs
(
7.
Thus, cs
+ i < 0
This ipies
+5= 1
+
29. nswer: 8 sing the MM inequaity, we have f(x
y
z) =
+
3
1 3. V +
=
2+3 + 4
-< =
+ 5+
+
42
+ 12 5 . {. { +
3y+5 + 2+ 2
+
4
1 29 (x+ y+ z )+ 4
=8
The equlity is achieved at x =
3
,
8z+12
12 � �
+ 2 + 2+ 2
4
y
= 1 and z =
30. nswer: 40460 Let = 34 and Q(x) = (x+ 1)P(x)-x Then Q(x) is a pynia degree + 1 and Q(k) = 0 r a k = 0 1 2 Thus ther i s a cnstant C such that
Qx) = (x+ )P(x)- x = x(x- 1)(x- 2 (x-
Letting x = -1 gives Thus C = P(x)
S
1 = C(-1)( -2 (-1)n+l + 1)! and
( - - 1) = C(-1)n+l (+ 1)!
Q( )) _ ( x+ 1 x x x+ 1 x
(-1)n+lx(x- 1)(x- 2 ) ( + 1).
· · ·
(x- n )
+ (1) 1 tl = P( n+1) = +1 2 + 1+ (-1)n+ = ") (n+1+(-1 )
+ 2
4
+ 2
.
since
= 34 is even Hence 42840 P(35) = 34 35 36 3364 40460
31 nswer: 30 Let X =
- 480 sin Observe that
- 30 cs and Y =
X Y == 121512+ 20 - 2(12)(15) 2(12)(20) cs cs (90°- ) and 15+ 20 = 25 = (X+ Y) , s we can cnstrct a rightanged triange AB as shwn
A
X
15 2
In particar ABC = and CB = 90°act siiar t AB S A B = and
e can chec that ACB is in
15 40t = 40 x 20 = 30 3 swer: 3 e sha appy the wing ineqaity: 4(a2+ b2+c2d2)
�
(ab+c+ d ) 2
Sice a b+ c + d = 8- e and a+ b+ c+ d = 16- e, we have
4(16-e ) (8- e) ie e(5e- 16) 0 Ths 0 : e : 16 5 Notthtifa = b = c = d = 6 5wve = 16 5 Hence LeJ = 3 33 swer: 105 threedigit nber can be epressed as lOalOb, and s we are iniiing
lOb+ c F(a b c) = lOa a+b+
3
Observe that wit h distinct digits a b c, F(a b c) has the iniu vaue when a< b
+ lOb + c 99a + 9b F(a b c) = lOOa a+b+c = l+ a+b+c e bserve nw that F(a b c) is iniu when c = 9 F(a b 9) = 1+
9 9a + 9 b
= 1+
a+b+9
9(a + b + 9) + 9 0a- 81
a+b+9
= 1 0+
9(10a- 9)
a+b+9
Nw F(a b 9) is iniu when b = 8
9) 9(a +17) - 1611 = 1 1611 F(a 8 9) = 1+ 9(1aa+17 = 1 + a+l7 a+17 which has the iniu vaue when a
1, and s
lO = 1
34 nswer: 59 Let = 17
and f = 1
Since 13
F(l 8 9) = 1 5 and
is dd f -1 (d 1
6) be uer 's phi unctin, (lOO) = 4 and 4) = 16 y uer 's there,
Nw et
= 17 17 33 (d 4) where the ast cngruence can be cacuated by the etended ucidean agrith Thus by repeated squaring, we have 19 19 9 (d 1)
35 nswer: 5 Nte that ( + ) = +n+ s ( +n) = but ( +n+1) = n+1 S each the sequences ( () ) ' = ( 1 1 2 ) and ( + ()) = (2 3 6 ) increases by 1 r every increent by 1, ecept when = m+m = m+, we have () = m and ( + 1- ( + 1) ) = m, s the rer sequence has every perect square repeated n ce On the ther hand, i = m + m, we have + () = m+2m but ( + 1 + ( + 1)) = +2 m+ 2, s the atte r sequence its every perect square Thus
o
( ) n-f(n) + ( ) n+f(n) G) f(n) (�) f(n) 3 3 (r n m n m 0 + + - o
+
_
o
n=O
m=
=
36
n=
m=
Sngapoe Mathematca Socey Sngapore Mathematcal Olympad (SMO)
(Snior Sction, Rond 2) 0900-1300
Satrday, 23 Jn 2012
1.
circe thrugh the incentre f a triange ABC and tangent t AB at A, intersects the seg ent B C at D and the etensin f BC at Prve that the ine C intersects at a pint such th at MDM
2.
Deterine a psitive integers n such that n equas the square f the su f the digits f n
3.
f 46 squares are cred red in a 9 9 b ard, shw that there is a bc n the b ard in whic h at east 3 f the squar es are cred red
an anl be a nite sequence f real nubers satisfying and
ao anl ak-1 ak +akl
Prve that r =
n+ ak
r k
n
= k(n + - )
Prve that fr any rea nubers a b
d with a + b + d ,
37
Sngapoe Mthematca Socety Sngapore athematca ympad (S) (Snio Scion, Rond 2 solions)
1. oin A, I, IA nd AE Let IE intersect AC t N We hve LIANLIAB LIEA so tht the tr inges NIA nd AlE re siir. hus LAN!LEAl LIB so LCI LN! herefore , the tr inges C! nd NCI re congruent. ence LICLIC ipying MM
M
Let denote the su of the digits of . Suppose is positive integer such tht . Let k so tht k . hen k k. Let 10 � k r where r is posit ive intege r. ht is k hs ecty r digits o 10 � k, we hve r � ogk + 1. o k 10, we hve k 10 so tht k hs t ost 2 digits. heefore, k � 9 2 18 � 18og k + 18 which is less thn k if k 50. hus the eqution k k hs no solution ink if k 50. Let k < 50 nd k k. Ting od 9, we get k k od 9) husk 0, 1 2.
od h 28, 36, 37fo , 45, k 1 n k 9, isk.k 9, 10, 18, 1, 27,soutions we hve9) k he1,corresponding r 46.reOny when 1 or 81.
3.
Suppose tht t ost 2 squres re colored red in ny 2
2 squre .
hen in ny
9 2 b oc , tere re t ost 10 red squres. oreov er, if there r e 10 red s qures,
then ther e ust be 5 in ec h row.
38
ow et the nuber of red s qures in row i of the bord be ri. hen ri+ri1 , : i: 8. Suppose tht soe ri with i odd. hen (rr2)+· · ·+(ri-2+ri-1)ri+· · ·(rs+rg): 4 1 45. On the other h nd, suppos e tht r1, r3, r5, r7, rg 6. hen the su o f ny consecutive 's is : . So (r1+r2)+· · ·+(r7rs)+rg 4 + 4.
=
.
hen bo bnl nd bk-1 bk+bkl for k Let bk , . .. . Suppose there eists n inde i such tht ai > bi, then the sequence a0 b0, ..., an1bn1 hs positive ter. Let j b e the inde such th t aj-1b-1 ab nd aj bj hs the lrgest vue. hen (aj-1 bj-1)(ajlb j1) (aj bj)· Using
ak-1ak +ak1 nd bk-1 bk+bk1 for l k
we obtin
(aj-1 b-1)+(aj1 bj1) (ab j) contrdiction. hus ak: bk for k. Siilry, we cn show tht ak bk for k nd therefore ak : bk s required.
First note tht (ae bd)(ad be) (ab ed)2. o see this, w y ssue e d b since a+b e+d. hen edab . so we hve the two obvious inequities ae bd edab nd adbe edab. utipying the togeth er we get (ae + bd)(ad be) abcd) 2. et
a
(a2e2)(a2+d2)(b2+e2)(b2+d2) (ae+bd)(ad+be)(abed) 22+(abed) 2 + (a+b)2(e+d)2 (abed) (ae+bd)(ad+be)(abed) 22+6(ab ed)2 < + adbe)' - ( oh- cd 2 16(ab cd)2 by )
r
+ =
+d)'
(ab- cd)2 +16(ab-
2. 4 (abed) 22+6(abed)r his n epression is n incresing function of (abcd) 2. he rgest vlue of (ab ed)2 is when (a, b, e, d) (, , ), (, , , ) ( , , ), (, , ). onsequenty, 4 - (bcd) 22+6(ab ed)2: 2 proving the inequlity
39
Od O 2012 O Fs Rud
Wednesday,
May
hrs
nstrutions to ontestants
nswer uestions. Write your answers in the answer sheet pvided and shae the apprpriate bubbles below your answers.
No steps are needed to ustiy your answers. 4. ach uestion carries mark. No calculators are allowed.
EAE O OT TR OVER T YO ARE TO TO O O
4
L
hroughout this paper, let xJ denote the greatest integ er less than or equ al to x ) For eample, 2 1 = 2, 39 = 3 his notation is used in Questions 2, 10, 16, 7, 18 and 22
J
J (
1 he sum of the squares of 50 onseutiv e odd inte gers is 300 85 0 Find the largest od d integer whose square is the last term of this sum 2 Find the value of
3
log k J .
k=3 2 ) Given that f ( x is a polynomi al of degree 20 2 , and that f ( k ) = k for k = ,2 ,3,
f( 204 )
nd the va lue of 20 14
· · · 203,
( )
4 Find the total num ber of sets of posi tive intege rs x, , z , where x , and z are posit ive integers, with x z suh that
< <
x + + z=2 03 5 here a re a few integers su h that + + 1 divides + 61 Find the sum of the squares of these integers 6 It is given that the sequene
for all = 2, 3, 4,
· · ·
(n)�=l' with = 2 = 2 , is given by the reurrene r elation
Find the integer that is losest to the value of
k l . 2 k= k
7 etermine th e largest even positive integer wih annot be epressed as the sum of two omposite odd positive integers 8 he lengths of the sides of a tiangle are suessive terms of a geometri pogression Let and be the smallest and largest interior angles o the triangle respet ively If the shortest sid has length 6 m and sin - 2 sin + 3 sin 19 sin - 2 sin + 3sin nd the perimeter of the triangle in entimetres
9'
9 Find the least positive integral value of for whih the equation
y
x +
(
has integer solutions x , x , x ,
X
�+
· · ·
+
· · · X ) · 41
X
2002
is a poitive integer be a real root of the ubi equation x + 2x - 0, where 1 If f ( + 1 for 2 3 4 · · , nd the value of 1006
10 Let
A
B AB
LA LB
11 In the diagram below, the point lies inside the triangle suh that and 90 Given that 5 and 6, and the point M is the midpoint of nd the value of 8 DM
A , LB
12 Suppose te real numbers x and satisfy the equations x - 3x + 5x 1 and - 3 + 5 Find x +
5
y
13 ishe t of twthe o o fvalue the fo of ur roots 32produ etermine k of t he uart i equ ation x - 18x +kx+200x - 1984
0
14 ete rmine the smallest int eger wit h 2 suh that
is an integer 15 Given that f is a realvalued funtion on the set of all real numbers suh that for any real numbers and , f f Find the value of f 2011
( ( ))
( )
16 he solu tions to the equation x - 4 x 5 where x is a real number , are denoted by for some positive integer k Find x
J
17 etermine the maim um integer solution of the equati on
l!J l! l!J 1!
+
2!
+
3!
42
+ +
l_J 10!
1001
18. Let A, B, C be the three angles of a triagle Let be the maimum value of sin 3A + sin 3B + sin 3C . etermine 10J. the number of sets of solutions ( x, y, z, where x, y and z are integers, of the 19. theetermine equation x 2 + y 2 + z 2 = x 2 y 2 . an nd sets of 13 distint positive integers that add up to 2142. Find the largest 20. We possible greatest ommon divisor of these 13 distint positive integers. the maimum number of dierent sets onsisting of three terms whih form an 1. etermine arithmeti progressions that an be hosen om a sequene of real numbers , 2 , · · · , ,
where
22. Find the value of the series
k
k=O he sequene
23.
(
.
is dened reursively by
xn�=
) Xnl = 1 - Xn(2(2-- J Xn
+
with x = 1. etermine the value of X - X4 ·
'
24. etermine the maimum value of the following epression where x, x 2 , · · · ,
- X 2 X 3 X4 - X 2 4-
x 24 are distint numbers in the set
25. valuate 12011 Lk=O 6 (
k k 2012 .
-1) 3
2k
43
-1, 2, 3 , 4, · · · , 2014
Od O 2012 O Fs Rud us
1 swer: 121 olution. et the i tegers e X + 2 X + 4 X 100 The X 2 + X + 4 2 + + X + 100) = 30080 et y = X + 1 d regroupi g the terms, w e oti [ y 4 2 + y + 4 2 ] + y 47) + y + 47) ] + + [ y 2 + y + 1) ] = 30080 which simplies to 5y2 + 2 + 2 52 + 72 + + 49 ) = 30080 Usg the fct tht 1 + 3 + + · + (2n1) 2 = 4 n 1 n, we ot y = 72 ece X = 21, so tht the r equired umer is 2
2 swer: 7986 olution. ote tht 2 2 = 2 , d tht 2 < 2 1 i d oly i log J = k The the requires sum ( deoted y S c e otied y 1 9 s = 2 2 log + log 3 - 02 log t=lOOl 92k9< 2kl 10 k2 + 1 t2 l 9 8192 + 1 - 23(9) = 7986
{
D
3 swer: 4 olution. xf(x) = 0,2, wehecemutg(xhve ) is poly omil of degree 2013 Sice g() g(2) = g(3) et= g( x=) g(=2013) g(x) = >(x-)(x-2)(x-3) (x2012)( x-2013) or some > lso, g(O) = 2 = > 2013 !, we thus hve > = 20132 Hece, g(2014) = (2013) = 2014 (201 4) - 2 cocludi g tht 201 4 (2014) = 4 44
=
4
swer: 333 3 2022 = 202(2201) = 20301 ps.t.ve eger . sets Soution. irst te tht there re x ythreez) whivlcuhesstireseul the .giIv xe=euti These sluti sets iclude thse where tw y, the 2x +z = 203. y emertig, z = 1 , 3, 5, · · · 201. heThere uisSicehex <rmy <xz,xthez).reuiSimriledrlswe y , therer isre 101 slutis the frm xreyxthusd1 01xslyy.
5. swer: 62 Soution. Sice 3 - 1 = (n7 - 1)(n + n + 1), we w tht n + n + 1 divides 3 - 1. ls, sice 213 - 1 = 3 6 1 - 1, we ls w tht n + n + 1 divides 2 13 - 1. s 2 2 n 0 13 + 61 = n 0 13
_
1 + 62 ,
we must hve tht n + n + 1 divides 23 + 61 i d ly i n + n + 1 divides 62. se (i) : I n + n + 1 = 1, the n = 0, -1. se (i ) : n + n + 1 = 2, there is iteger sluti r n. se (i i) : I n + n + 1 = 31, the n = 6, -5. se (ilv): theI niteger + n +v1u=es62, nthere s -1, 6,tege-5. rHece slutithe rsumn. sures is 1 +36+25 = Thus, re 0, 62.
6. swer: 3015 Solution. The recurree relti c e writte s = n 1Summig r n = 2 t N, we ti
1
-
·
shwig tht 1 N l 3 1 - 2 N - N + 1 . Summig this up r N = 2 t N = 2011, we ti 211 ak-k = �2 (2010) - ( �2 - 2012 = 3014.5 + 2012 shwig tht the itege r clsest is 3015. 45
D
7. swer: 38 Soltion et n e eve posit ive iteer. The ech of the folowi epresses n s thet esum of twoof nodd- 15iteers nn=- (35n -is 15)divis+i15le y(n 3- so25)tht+ 25it oris composi (n - 35) te,+35. oecethtn st oe n 25 d he c eed epressed s the esumepressed of twoscomposite it c e veri tht 38 cot the sum ofoddtwoumers composiiftenodd 38.p osiIdeed, tive i teers. swer: 76 Soltion et the eths e i= ceti e tres thee 16,peri16meterd of16 the (where 1). The 1- -of22the+3si3de s of199thesotritht 32 . mHece, tri le = 16 1 + + = 76cm 9. swer: 4 Soltion Sice 2002 = 4(mod9) 4 = 1(mod9) d 2002 = 667 3 + 1, it foows tht 2002 serveoe tht offortheposifoltivowie iteers modulo 9=for4x re =04(±1mod9) . Therefore, x, the possie residue s + � + � + c hve residue of 4 moduo 9 . However, sice 2002 = 10 + 10 + 1 1, it folows tht 2002 2002 3 +) ( 10 3 . (210002 1 3 + 1 3 ) (2002 667) 3 (10 . 20026 67)3 + (10 . 20026 67)3 + (20026 67) 3 + (2 002 667)3 vThiuse shows of n stht 4. x + x� + x + x = 2002 is ideed sovle . Hece the lest iterl
D
10. swer: 2015 et + 2x - n. It is esy to se e tht f is strictly i cresi ctio for n = f(x) 2 3 4= nx· ·. rther, -n=
( - n2 + n + 1 ) < O
for n 2. so, f(1 ) = 2 Thus, the oy rel root of the equtio nx + 2x - n = 0 for n 2 is lo cted i the iterv 1 1 n. We(nmust hve n 1 )
=
k
=
. 2012 (2
2013) = 2015
46
.
Answe: 22 Soltion tend C D to E sch that CD = DE
, ' ''' '' \ '
It is clear that CD B and EDB are congr ent Hence EB = CB = and BED BCD hs, BE D = BC D = BD implies that the poin ts B , D, are E are concyclic Given that BDC = , hence EDB = BDE is a cyclic qadrilater al with EB as a dimeter hs, EB = In the ightangled triangle EB, we have
Sice D and M re the midpoints of EC and C respective ly, DM = 8DM = 22
.
6
E =
hs, D
Answer: 2
x 3 - x 2 + x = , we have x - ) 3 + 2 x - ) = -2, and from y3 - y 2 + y , we have ( y - ) 3 + (y - ) = 2
Soltion From
hs
x - )2) 3 + 2 x - )() + (y) - () 3 + )2)(y - ) = x + y - 2 x - - x - y - + y - + 2 x + y - 2 )( ) ( )2) )2 = x + y - 2 2 + x - - x - y - + y - .
For ny real nmbers
0
and
we lways hve
x x -y, )2 - x - ) (y - ) (y - ? 0 ad ths x + y - 2 = 0, implying that x + y = 2
47
D
13 Answe 86 Soution Let a, b, , be the fo root s of x -18x + kx + 200x - 1984 = 0 sh that ab = -32 Then aab++ba+ + +a+= b18,+ b + = k, ab +=ab-1984 + a + b = 200, ab Sine ab = 32 and ab = -1984, we have = 62 Then, fom ab+ab +a +b= we have 200 -200 = -32 - 32 + 62a 62b = 32( + ) + 62(a + b) Solving this equation together with the equation a + b+ + = 18 gives that a + b = 4, + = 14 om ab + a + a+ b +b+ = k, we hae
{
= 30 + (a b)( + ) = 86
D
14 Answe 337 Soution Assme that
and so
(n 1)(2n + 1) = 6 Thus 6 (n + 1)(2n + 1), implying that n = 1 or 5 mod 6) Case n = 6k + 5 Then = (k +1)(12k +11) Sine (k + 1) and (12k +11) are elatively pr ime, both mst be squaes So thee are p ositive integes a and b sch that k + 1 = a and 12k + 11 = b Thus 12a = b 1 t, as 12a = mod 4) and b + 1 = 1 o 2 mod 4), thee a no integes a and b sh that 12a = b + 1 Hene ase 1 cannot happen Case n = 6k + 1 Then = (3k + 1)(4k + 1) Since 3k + 1 and 4k 1 ae elatively pime, both must be squaes So there ae positive integes and suh that and Then = (2apowe 1) (2 aSo+1)neithe Obseve a in2atheb+ lefthand 3kpime +evey 1 othe = pime a thanfato 4k3+eept 1 = b 3 has an3beven 2a-1 thatnor 1 can be aside, Now we consider positive integes a sch that neithe 2a 1 no 2a + 1 an be a prime othe than 3 If a = 1, then b = 1 and n = 1 So we onside a 2 The net smallst suitable value for a is 13 When a = 13, we have 3b = 25 27 D and sob= 15 , implying that k = 56 and so n = 6k + 1 = 337 48
15 Answer: 2011 om the recurrence relaton, = f(1) and f(f()f(1)) = · 1 Hence, = Also, from the gven functonal f(1) y lettng = (1), we obtan f(f(1)) = (f(1)) equaton, we have f(f(1)) = 1, henc ((1)) 1, followng that (1) s ether 1 or 1 D Hence f(2011) = 2011
f(f(1)f(b)
b
f(b)
f(b)
16 Answer: 19
< Note that f : there wll be n o soluton as xx--4xJ x xx-4x = x(x -4)x :3,3(5) = 15 Also, f x 2, there wll be no solu ton as x -4x J < x -4(x 1) = x(x -4) + 4 4
Soltion. Note that
1
Hence the soluton must be n the nterval ( 2, 3) If x = 2, then x = -3, gvng x = N, whch s a soluton If x = 1 , the n x = 1, gvng x = 1 whc contradcts wth x = 1 f xJ = 0, then x = 5, hence there s no soluton If xJ = 1, then x = 9 Snce 2 < W < 3, there s no soluton If xJ = 2, then x = 13 Snce 2 < m < 3, = m s a soluton hus, the requred answer s -3 + 13 = 10
D
17 Answer: Soltion. It s clear that
584
+ + + + s a monotone ncreasng functon of x, and when x = 6!, the above epresson has a value larger than 1001 hus each soluton of the equaton +...+
+
+
= 100 1
x s a soluton, then + + + + = + + + +
s less than 6! and so f
As
x < 6!, t has a unque epresson of the form 5!
where f then Snce
4! C 3!
2! e
+ b wth+ 5,+b 4, + 3, � 2, e 1 Note that a b, , , e are nonnegatve nteger = 5! + b 4! + C 3! + 2! + e +
+
+
+
= 206a
+ 41b + 1 0c + 3d +
e.
41b + 10 + 3 + e � 201 we have 0 0 206 10 01 and so = 4 hus 41b + O + 3 + e = 177 , 49
which implie that
b = 4 and o on, giving that c = = and e = 0 h X = 4 5! + 4 4! + 3! 2! + 0 = 584
A 58 4 i the only int eger o ltion, th e anwer i 5 84
8 Anwer 25 Solution e hall how that -2 i n 3 + in 3B + in 3C 3 /2
0 It i cl ear that in 3B - and me hat B C hen 60° h in 3 in3C -1 h in 3 + in 3B + in 3C -2 Let B = C hen B = C = /2 If i very mall, B and C are cloe to , and th in 3 in 3B + in 3C i cloe t o -2 Now we how tha t in 3 + in 3B +in 3C 3/ 2 Firt the pper bond can be reach ed when = B = 20° and C = 40° Let X = 3, = 3B and Z = 3(C - 20°) hen X + Z = 80° and
:
Y
Y
in 3 + in3B + in 3C = in X + in
Y Y
Y
Y + in Z
Sppoe that X, Z ati th e condit ion tha t X + + Z = 8 0° ch th at in X + in in Z ha the maimm vale e can then how that X = = Z Ame that Z If X < Z, then
Y
Y+
X-Z X+Z X+Z X + Z = 2 2 co 2 < 2 2 , impying that
< � X +2 Z +� +� X +2 Z which contradict the amption that in +in Y + in Z ha the maimm vale Hence X = Y = Z = 60° , implyin g that = 20° , B = 20° and C = 40° and X� +� �
in 3 + in 3B + in 3C = 3/2 Since
� 732, the anwer i then obtained
D
Anwer
x = 0, y 0 and z = 0 i a oltion of thi eqation e hall how that thi i it only oltionbyby2,proving forof any thi keation and whenever are diviible they arethat aloifdiviible x, y, zinteger x, y, zbyi a2oltion 0 ' ' k Let x = x , y = 2 and z = 2 z hen x + y + z = x y i changed to Solution Note that
x', y ' , z '
It i eay to veri that only when are all even, x'2 + y '2 + z 12 and the ame remainder when divided by 4 h are diviible by 2 1 .
5
x, y, z
k
k x y' have
20.
Answer:
21
Sotion. Let d b e the greatest commo ivisor
1.
13 2142 = 2
Answer:
··· ··· = = 2142. 1 2 · · · 1 = 91 2142, 51 = 102, 2 ··· = (1, · · · 12,24),
···
51,
= 21
21.
gc of these istinc t psitive integers. Then these integers ca b e reresente as d , d2 , d3 , where gc , 2 , 3 Let S enot e + 2 + + 3· Then Sd In ore r for d t be th e largest pos + + 3+ + 3 sible , S must be the smallest . ince S an that S ivies a that 3 7 te smallest possible value of S can be an 3, the arges t vaue of d is thus By chosing , 2 , 3 , 2 , 3 D we conclue that d is possibe.
21.
2500
Sotion. First, for the folowing particuar sequece, there are really three-term arithmetic progressios which can be chosen from this sequence:
s, , 2 - s for all integers
They are an t with
5 � < � 101. 2500 t
2500
ierent
1, 2, 3, . 101 . s, 1 � s < � 51 2 - , with
an
t t for al itegers
< <···<
Nw we show that for any give sequece of rea umbers 2 , there are at most ierent threeterm arithmetic progressios which can be chose from this sequece.
2 � i � 100.
Let , , repres et a threeterm arih metic rogressio n. It is clear that If the the rst term has at most choices, as must be a inex i 2 If then the thir term has at most choices, as t + + must be an inex in ,
1, �· · · � -51,1 .
52 � � 100, · · · 1, 2, 101 . - 1) + (10 1 - ) = 1 + 2 +
s
-1
101-
o the umber of ierent three-term arithmetic progressions which ca be chosen from this sequence is at most 00 5 = 2
22.
Answer:
20121
Sotion. Write
5 2
x = x - LxJ Then
+ 5 + + 2 + +
1
49 = 25
D
! <
i 2 2 otherwise
Thus, we have
Ayig the above result for
x=
,
kO
I articuar, when
n = 20121,
n the inite series cnverges to
1
20121.
D
23 Answe: 0
- = tan , it follows that an + tan 1 = tan n + XnI = tan = tan 2 - tan an tan I 2 ( So, X nl 2 = tan an + = tan an Xn , implying that this seqence has a peiod o 2 Obseve that 00 = 5(mod 12) and 401 = 5(mod 12) onseqently,
Solution. Let
Xn = tan n .
Since 2
( )
=0
=
XlQOI - X401 X5 - X5
D
24 Answe: 203 Solution. ist i is clea that the ans we is an intege between 0 and 20 4 t it cannot
be 204 as
· · · X I - X 2 - X 3 - X4 · · · X20 4 and X I + X2 + · · · + X 20 4 = 2 + · · · + 204 = 1007 205
have the same paity Now we jst need to show that 203 can be achieved o any intege k ( 4k + 2) (4k
+ 4)
(4k
+ 5)
- (4k
+ 3)
=0
hs 24
- 5 - 3 - - (4k + 2)
(4k
+ 4)
- (4k + 5)
- ( 4k + 3)
-20 0 202 20 3 - 20 - 2014 1 0 204 203 D
25 Answe:
w = cos J + i sin J
olution. onside the comple nmbe binomial theoem one has
(
Re cos + i sin
)
2 2 3
�(� V)c )
Re
+i 20I2 2I 2 o 1 2
2 0IO
_
+
2
+ .. + [ ( 20I 2 2 1
_
-
On one hand, sing the
202
)
52
2
( )+ 202
4
.
..
20 2 4
(�
+ 3 I 0 06
2 008
202 20 12
On the ther hand, sing the e ivre's there ne has
2012 2012 2012 2 2 [1 - (2012) (202) . . . (2012) 20 1 2 ] � [ 2021 2 2041 2 . 22010102 22012 1 Re( ) = Re cs
hs,
2 01
s that
2
3
2
3
+
sin
3
32
4
3
= cs
- 3 1004
- 3
3
3
20 1 0
3
+
= cs
3 100
3
6
3
2012
=
=
2
=
D
Sgapoe Maheatca Socey Singapre Mathematcal lympad (SM
2012
(Opn Scion, Rond 2) Strdy, 30 n 22
.
2.
0900-300
he incirce with centre I of the tringe ABC touches the sides BC, CA nd AB t D, nd F respectivey. he ine ID intersects th e segent F t K. Prove tht A, K d M re coiner w here M is the idpoi nt of BC Find fuctions : I I so tht
(x + y) ((x) - (y) ) (x- y)(x + y) for x y J 3.
.
For ech i 1, 2, . . , , et a, b, c be integers such tht t est one of the is od d. Show th t one cn nd integers x y z such tht xa + yb + zc is odd for t est 4 /7 dierent vues of i Let
p be n odd prie.
r ove tht
+ 2 + 3 +
+
� - 1
2 - 2
p
od p .
here re 2012 distict points in the pne ech of which is to be cooured usin one of coours so tht the nu ber of poin ts of ec h coour re distinct . set of points is si d to be u tico oured if their coours re distinct . Deterine tht iizes the nu ber of u tico oured sets .
54
Sngaoe Mathematca Socety Sngapore athematcal lympad (S)
202
(Opn Sion, Rond 2 solions)
1.
A
Let the line AK intersect C at M We shal prove that M is the dpoint of C Sinc K = and K = C , we have K K lso
K sin AK
Theefore,
sinK sinK
A sin AK
A sin AK sinK sinK
sin sinC
K K
K sin K A
sin sinC
onequenty, M AM s in AK sinC = = = C AM CM sin sinKA so that M = CM
l
2
Suppse that is a slutin. Let
a = 1 (! (1) (1) ), b = 1 (! (1) ! (-1) ) and g() = () a b. Then ( y)(g() g(y)) = ( y)g( y) and g(1) = g(1) = 0. Letting = 1 and= 1 abve give
( 1)g() = ( 1)g( 1) g( 1) = ( 2)g() . Thus
( 1)g() = ( 1)g( 1) = ( 1)( 2)g() fr all . S g() = 0 fr all . ence () = ab. We can chec directly tha any functi n f th s fr (fr se a, b E satises the given euatin. nsider all the 7 trples (, y, z ), where , y, z are either 0 r 1 but nt all 0 Fr each , a east ne f the nuber s a , b, C is dd. Thus ang the 7 su s abzc, 3 are ev n and 4 are d d. ence there are altgether 4 dd sus. Thus there is chice f (, y, z ) fr which a t east 4 7 f the crrespnding sus are dd. (Yu can thin 3.
f a table where rws (, arey,nubered and are the the clun s crrespn d t 1, 2, . .in ,rw 7 chices f the the triples 7 sus a yb the zc . z ) . The 7 entries Thus there are 4 dd nubers in each r w, aing a ttal f 4 dd su s in the t able. Since th ere are 7 cluns, ne f the clu ns ust cntain at least 4 7 dd sus. )
.
First, fr each = 1, 2, . . ,
()
2 = ( - 1) ( - 2) . . ( (2 - 1) ) 2 (2 1) ence
p-12
L i 1
( )=
_ 1
(d ) .
()
p-12 2 P _ 2 p-12 1 P p - 2 p 2 - 2 - L i 1 i 1 2 p1 - 2 d ) (by Ferat 's Little There. ) ( = 2i p i 1
p -2 =
'
(1) (2) ((2 - 1) ) (2 1)
()
The last suatin cunts the evensized nnepty subsets f a eleen t s et f which there are 2 1 1.
56
Let m1 m2 mn be the nu ber of points of each coour We ca m1, m2, , mn the coour distribution hen m1 + + mn 2012 and the nu ber o f uticoo ured sets is M m1m2 mn We have the foowing observations (i) m1 > For if m1 then m1m2 mn m2 m3 mn 1( + mn ) his eans if we use coours with coour distribution m2, m3, , mn_, ( + mn), we
obtain a a rger M. (ii) m1m : 2 for a i For i f there eists k with mk1 mk 3 then the coour distribution with mk, mk1 repaced by mk + mk1 yieds a arger M. (iii) m1m 2 for at ost one i For i f there eist i j with m1m mj1 mj 2 the coour distribution with m, m j1 repaced by m + mj1 yieds a rger M. (iv) m1m 2 for eacty one i For if 1m for a i then m1 + + mn m1 + nn2-1 202 4 503. hus (m1 + ) 8 503. Since 503 is prie, the parity of and 2m1 + are op posite and 2m1 + > , we have 8 and m1 248. he coour distribution with m1 repaced by two nubers 2 246 (using + coours) yieds a arger M. (v) m1 2. f mn mn -1 2 then fro (i), we have m1 + + mn m1+ nn2-1 + 202. hus (m1 + 2 20. Since 20 is prie , we get 2 and m1 005 which wi ead to a con tradiction as in (iv) hus mnmn 1 . m1 m 2 for soe : i : 2. Suppose m1 3. Let m' m2 2. hen m m' m1 with repacing m by 2 m' yieds a larger M. hus m1 2. o the above ana ysis , with 2 coours, we see that the coour distribution 2 3 i i + i + 2 , + + 2 with 3 : i : , yieds the aiu M. ow we have m � ( + ) ( + 4 i 2012. hus 2+ 4020 2i, 3 i : , ie, 2 + 5n 4026 and 2 + 3n 4020. hus 6 and i 3. hus the aiu is achieved when 6 with th e coour distribut ion 2 4 5 6. . . 63 .
)
57
