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Cryptarithmetic Multiplication Problems with Solutions – Download PDF [FREE]Click Me & Subscribe to Receive Awesome Study Material Directly in your Mail. Many Aspirants find the cryptarithmetic problems in elitmus a tough nut to crack. Keeping this in mind, we have compiled cryptarithmetic multiplication tricks and list of cryptarithmetic questions and answers in this Post. A cryptarithmetic puzzle is a mathematical game where the digits of some numbers are represented by letters (or symbols). Each letter represents a unique digit. The goal is to find the digits such that a given mathematical equation is verified.
Rules For Solving Cryptarithmetic Problems Each Letter or Symbol represents only one digit throughout the problem When letters are replaced by their digits,the resultant arithmetical operation must be correct. Our aim is to find the value of each letter. No two letters represent the same digit (If A = 2, B cannot be 2 and so on). Number can’t begin with 0 Rules are: 1. 0 * n = 0 2. 1 * n = n 3. 5 * odd number = x5 like 5 * 1 = 05 (x = 0) 5 * 3 = 15 (x = 1) 5 * 7 = 35 (x = 3) 5 * 9 = 45 (x = 4) 4. 5 * even number = always 0 like 5 * 2 = 0 5 * 4 = 0 5 * 6 = 0 5 * 8 = 0 5.. even number * 6 = (x)(even number) where x can be 1/2/4 like 2 * 6 = 12 (x = 1) 4 * 6 = 24 (x = 2) 8 * 6 = 35 (x = 4)
Cryptarithmetic Multiplication Problem 1 This type of problems can be solved by backtracking technique after finding one letter (number).
Approach (Method): The best method which I follow is from bottom to top multiplication (You can do in viceverse), but I found it was easy. If you see the question, start from bottom to top like D*(ABC) and E*(ABC) It is given that E * C = C this happens only with numbers of 5 & 6 let’s see how?
Explanation: 1 * 5 = 5 3 * 5 = _5 (Actually 15 but concentrate on last digit) 7 * 5 = _5 (35) I am not using 5 * 5 because it’s already assigned to C we should not repeat. 9 * 5 = _5 (45) so, from this don’t conclude C = 5 & E= (1, 3, 7, 9). Because there is another number with these properties Ex: – number (6) 6 * 2 = _2 (12) 6 * 4 = _4 (24) 6 * 8 = _8 (48) From this don’t conclude E = 6 & C = (2, 4, 8). Let, note all the possible cases and try. For Example take C = 5 and E = odd number (1, 3, 7, 9) Now Replace C with 5 and redraw the following table. A B C And also place E as odd number (3, 7, 9) I didn’t write 1 here because it’s not going to be 1 Let’s take first E = 3 and fill the table. D E If you see the colored portion then sum 3 + 5 = B (8) lets assign 8 to A B 5 B F E C And redraw the table, if u put B = 8, let’s check from first D 3 3 * 5 = 5 (carry 1) D E C 3 * 8 = 24 + 1 = 5 (carry 2) but in the table it is showing 3 it is F 3 5 contradictory, then we can understand that E! = 3. H G B C So Next I will take E = 7 and again fill the table. D 3 5 If you see the red color portion A B 5 A 2 5 Sum of 7 + 5 = 2 (carry 1) so H G B 5 we got B = 2. D 7 D 7 Assign B = 2 and fill the table. Now see the row which I F 7 5 F 7 5 highlighted and think what value we can keep for D after D 7 5 D 7 5 clear Observation we can get D H G B 5 H G 2 5 = 3 satisfy the given condition and put D = 3. To get 375 in A 2 5 second row the value of A 3 7 is A = 1 so put F 7 5 value of A in table 3 7 5 If you keep A H G 2 5 = 1 it’s very simple F = 8, G = 6, H = 4. 1 2 5
8
3
7 Cryptarithmetic Multiplication Problem 2
7
5
W
H
Y
3
7
5
4
6
2
5
N
U
T
O
O
N
P
O
Y
P
Y
O
U
H
A
O
N
E
P
O
P
When you see second multiple (W H Y ) * U = (O Y P Y) So, The number for U = 6 and Y is (2 , 4) Lets take the value of Y is 4 and replace it in figure. W
H
4
N
6
T
O
O
N
P
O
4
P
4
O
6
H
A
O
N
E
P
O
P
Yes, this is some typical question we need more tries to reach solution. Clearly if you see the yellow portion O + 6 = N and there is no carry to the next digit so we can confirm that ‘N’ is single digit number. If ‘N’ is single digit number ‘O’ should be less than (O < 3) Let’s take O = 1 and redraw the table… W
H
4
N
6
T
1
1
N
P
1
4
P
4
1
6
H
A
1
N
E
P
1
P
If you observe the orange shaded region 1 + 6 = N & N + 4 = 1 so, we can clearly the value of N =7 So replace N = 7 and redraw the table. W
H
4
7
6
T
1
1
7
P
1
4
P
4
1
6
H
A
1
7
E
P
1
P
Take the third multiple (W H 4) * 7 = 1 6 H A à from this we get A = 8, and if you replace the H with 3 and w = 2 then the equation satisfies. Like (2 3 4) * 7 = 1 6 3 8 so redraw the table with particular values. 2
3
4
7
6
T
1
1
7
P
1
4
P
4
1
6
3
8
1
7
E
P
1
P
The remaining numbers are (0,5,9) It is simple to analyse we can’t get 5 in place of p because when you multiply with 4 we cannot get the last digit as 5. So T = 5, P = 0, E = 9 After redraw the table. 2
3
4
7
6
5
1
1
7
0
1
4
0
4
1
6
3
8
1
7
9
0
1
0
Cryptarithmetic Multiplication Problem 3 This type of problems can be solved by backtracking technique after finding one letter (number).
Approach (Method): The best method which I follow is from bottom to top multiplication (You can do in viceverse), but I found it was easy.
Explanation:
B
O
B
B
O
B
M
E
O
Y
M
I
L
O
M
E
O
Y
M
A
R
L
E Y
Here we cannot apply the logic B * B = B (5 * 5 = _5) or (6 * 6 = _6) So alternative b = (2, 3, 4, 7, 8, 9) From the second line B O B * O = M I L O O * B = O So, O = 5 and B is an Odd number i.e B = (3,7,9) Replace O = 5 and B = 3 and redraw the table 3
5
3
3
5
3
1
0
5
9
1
7
6
5
1
0
5
9
1
2
4
6
0
9
B = 3 O = 5 Y = 9 E = 0 M = 1 L = 6 R = 4 A = 2 I = 7
Cryptarithmetic Multiplication Problem 4 A
B
C
D
E
C
F
G
H
I
A
C
A
C
A
E
A
F
A
G
C
H
Solution: First if you see ( D E C ) * C from this we can confirm that two possibilities: 1) E = ( 3, 7, 9 ) and C = 5 2) E = 6 and C = ( 2, 4, 8 ) So now you are in little bit confusion which possibility I should proceed. If you are not choosing the correct possibility time will be killed off course time is very very precious in eLitmus exam. So, better step forward to estimate the correct possibility–)
If you see last multiple ( D E C ) * C = E if you assume C = 5 and E = (3, 7, 9 ) the only possible values of D = (2, 4, 8) if you keep any value for D then the value E = 0 that is contradiction. So it is very clear that first possibilities is wrong then go with the second possibility means E = 6 and C = (2, 4, 8) Let’s take C = 2 Then the table changes as: A
B
2
D
6
2
3
0
4
I
A
2
A
2
A
6
A
3
A
0
2
4
Clearly from the table H = 4 and G = 0 (because G + 2 = 2 means G = 0), and F = 3 (C + carry= F). After filling this values let’s take (A B 2) * 2 = 3 0 4. It happens only when B = 5 and A = 1 Replace the values and redraw the table. 1
5
2
D
6
2
3
0
4
I
1
2
1
2
1
6
1
3
1
0
2
4
From the table (1 5 2) * D = 1 2 1 6 It will happen only when D = 8 and from the line (I + 1 + carry = 1) so from that we can say I = 9. Replace all the values, 1
5
2
8
6
2
3
0
4
9
1
2
1
2
1
6
1
3
1
0
2
4
Cryptarithmetic Multiplication Problem 5 This type of problems can be solved by backtracking technique after finding one letter (number).
Approach (Method): The best method which I follow is from bottom to top multiplication (You can do in viceverse), but I found it was easy.
Explanation: M
A
D
B
E
M
A
D
R
A
E
A
M
I
D
(M A D) * (E) = (M A D) we can clearly get that E = 1 from the second multiple D * B = E => 1 so only options is 7 * 3 = _1 3 * 7 = _1 Let’s take B = 3 and D = 7 and E = 1 redraw the table. M
A
7
3
1
M
A
7
R
A
1
A
M
I
7
M + A = M then we can understand that the value of A will be 9 because (9 + carry) = 10 Now, A = 9. M
A
7
3
1
M
A
7
R
A
1
A
M
I
7
See the last column R + (carry) = A [ carry may be equal to 1 because two less than 9 numbers leads to carry (one) ] So R = 8. M 9 7 * 3 1 8 9 1 To get R = 8, M should be equal to 2(M = 2)
2
9
7
3
1
2
9
7
8
9
1
9
2
I
7
Finally:M = 2 A = 9 D = 7 B = 3 E = 1 I = 0 R = 8
Cryptarithmetic Multiplication Problem 6 Approach (Method): S
E
E
S
O O
E
M
O
M
E
S
S
M
I
M
E
O
Clearly if we see the first multiple (S E E) * O = E M O O From that we can get easily E = 1. So, replace E with (1) and redraw the table. S
1
1
S
O O
1
M
O
M
1
S
S
M
I
M
1
O
If we see the shaded region in above table M + S = M the value of M = 9 (I explained in previous example). So replace S = 9 and redraw the table. 9
1
1
9
O O
1
M
O
M
1
9
9
M
I
M
1
O
By the second multiple (9 1 1) * 9 = (M 1 9 9) = (8 1 9 9) So M = 8 and also see the shaded region O + 9 = _1 so the value of must be equal to 2 (O = 2). Replace the values of M and O and redraw the table. 9
1
1
9
2 2
1
8
2
8
1
9
9
8
3
8
1
2
After replacing M and O value the value of I = 3.
Cryptarithmetic Multiplication Problem 7 Approach (Method): A
I
D
A
D D
R
I
A
D
D
C
D
D
I
C
E
D
Clearly if you see D * D = D the numbers which satisfy the condition is 5 & 6 only, Let’s see how 5 * 5 = _5 (25) and 6 * 6 = _6 (36). So take D = 5 and rearrange the table.
Explanation: A
R
I
I
5
A
5
A
5
5
5
C
5
5
I
C
E
5
So see the shaded region: I + C = C so the value of I may be 9 (or) 0 Why 9? Because carry will come from A + 5 = _E I + carry = 10 so we are taking I = 9. If we take I = 9 then the value of A will be: (A 9 5) * 5 = R 9 7 5 So A = 7 keep the value of I and A and redraw the table. 7
9
5
7
5 5
R
9
7
5
5
C
5
5
9
C
E
5
So no need of finding the values by procedure just by multiplying 7 9 5 * 7 5 Then we will get all values. 7
9
5
7
5 5
3
9
7
5
5
6
5
5
9
6
2
5
Cryptarithmetic Multiplication Problem 8 H
E
I
S
H
U
B
O
F
L
A
B
This type of problems can’t be solved by tracking technique.
APPROACH: This is the toughest problem which I ever faced because there is no repeated words in the multiplication
table. And the procedure we should follow is from each and every number we should check, There is no particular method to solve this H = 1 E = 8 I = 3 S = 9 U = 6 B = 2 O = 5 F = 4 L = 7 A = 0
Cryptarithmetic Multiplication Problem 9 Approach (Method): C
U
T
I
T T
B
U
S
T
N
N
T
T
E
N
E
T
If you see the last multiple T * T = T this operation will be done only with the numbers of 5 & 6. Let’s take T = 5 and redraw the table. C
U
5
I
5 5
B
U
S
5
N
N
5
5
E
N
E
5
See the shaded region U + N = N it means that the value of U may be equal to 0 or it will be equal to 9. Two cases: If S + 5 = _E (if carry comes then value of U will be 9) S + 5 = E (No Carry then the value of U will be 0) Let’s take the value of U will be ‘0’ and redraw the table. C
0
5
I
5
B
0
S
5
N
N
5
5
E
N
E
5
5
And (C 0 5) * 5 = B 0 2 5 So, the value of (S = 2), Observe the last multiple (C 0 5) * I = 5 N N 5 means the value of I is odd (3, 7, 9) and it is greater than 5000 so take value of I is greatest odd digit = 9 and redraw the table. C
0
5
9
5 5
B
0
2
5
N
N
5
5
E
N
E
5
See the shaded region 2 + 5= E the value of E = 7, and (C 0 5) * 9 = 5 4 4 5 The value of N = 4 and redraw the table. C
0
5
9
5 5
B
0
2
5
4
4
5
5
7
4
7
5
To get B + 4 = 7 the value of B = 3 and to get (C 0 5) * 5 = 3 0 2 5 the value of c should be equal to 6 and redraw the table. 6
0
5
9
5 5
3
0
2
5
4
4
5
5
7
4
7
5
Cryptarithmetic Multiplication Problem 10 A
S
K
T
O
K
A
R
O
S
A
K
T
Y
S
O
L
L
Solution: Let’s see clearly (T O) * K = K from this there will be two possibilities 1) T = (3, 7, 9) and K = 5 2) T = 6 and K = (2, 4, 8) And there is no other clue to estimate which possibility is better to approach first. Let’s try with the first possibility take T = 3 and K = 5 and draw the table A
S
5
3
O L
5
A
R
O
S
A
5
T
Y
S
O
L
Next the only way to approach is 5 * O = L the value for O = (2, 4, 6, 8) if we keep O = 2 then at last O + carry = T ( 2 + 1 = 3) the value 3 is assigned to two letters it is contradictory. Let’s take the value of O = 4, T = 7(3 is contradiction) and proceed at last O + Carry = T ( 4 + 1 = 5) the value 5 is assigned to two letters it is also contradictory. So take value of O = 6 and draw the table. A
S
5
7
6 L
5
A
R
6
S
A
5
7
Y
S
6
L
From the table it is clear that 5 * 6 = L (5 * 6 = 0) L = 0 and T = 7, R + 5 = 6 so, R = 1. A
S
5
7
6 0
5
A
1
6
S
A
5
7
Y
S
6
0
(A S 5) * 6 = 5 A 1 0 it will happen only when S = 8 and A = 9 and from the table 5 + S = Y (5 + 8 + carry = 4), Y = 4 let’s replace the table with all values.
9
8
5
7
6 0
5
9
1
6
8
9
5
7
4
8
6
0
Cryptarithmetic Multiplication Problem 11 Approach (Method): R
E
D
A
S S
A
R
C
R
E
D
C
D
T
S
If you see the last multiple (D * S) = S clearly we can conclude that D is an odd number (3, 7, 9) and S = 5, and (R E D) * A = (R E D) So, A = 1. Fill the values in the above table. R
E
D
1
5 5
1
R
C
R
E
D
C
D
T
5
If you see (R E D) * 5 = 1 R C 5 see the most significant bit is ‘1’ then the value of R must be (2, 3). If you take R = 2, and see the shaded region in the table 1 + R = C => 1 + 2 = 3 so replace the values of R and C in the above table. 2
E
D
1
5 5
1
2
3
2
E
D
3
D
T
5
To get (2 E D) * 5 = 1 2 3 5 the values of E and D is E = 4 and D = 7 if you replace the values of E and D we will get the value of T = 10. 2
4
7
1
5 5
1
2
3
2
4
7
3
7
0
5
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