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Cryptarithmetic Questions asked in Infosys and eLitmus

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Cryptarithmetic Questions asked in Infosys and eLitmus

Cryptarithmetic questions are most commonly asked in the Infosys recruitment and eLitmus exam. The most repeated Cryptarithmetic questions with answers are discussed here.

Cryptarithmetic is a mathematical puzzle which involves the replacement of digits with alphabets, symbols and letters. Only through certain practice, one can become an expert in solving the cryptarithmetic questions.

Infosys and Elitmus cryparithmetic questions

 

Rules to solve Cryptarithmetic Questions

  1. Each letter should have a unique and distinct value.
  2. Each letter represents only one digit throughout the problem.
  3. Numbers must not begin with zero. i.e. 0937 (wrong), 937 (correct).
  4. You have to find the value of each letter in the Cryptarithmetic.
  5. There must be only one solution to the problem.
  6. After replacing letters by their digits, the resulting arithmetic operations must be correct.

So, the problem is to find the unique digit corresponding to a unique letter.

 

Infosys and eLitmus Cryptarithmetic Questions with answers

1. If AA + BB = ABC, then what is the value of A+B+C= ?

  1. 15
  2. 18
  3. 21
  4. 12

Ans: 2

Explanation:

     A  A

     B  B   +

     C  C

-----------

 A  B  C

------------

The digits are distinct and positive. Let’s first focus on the value A, when we add three 2 digit numbers the most you get is in the 200’s (ex: AA + BB + CC = ABC à 99 + 88 + 77 = 264). From this, we can tell that the largest value of A can be 2. So Either A = 1 or A = 2. 

Now focus on value B, let’s take the unit digit of the given question: A + B +C = C (units). This can happen only if A + B = 0 (in the units) à A and B add up to 10.

Two possibilities: 11 + 99 + CC = 19C  à (1)   or    22 + 88 + CC = 28C  à (2)

Take equation (2), 110 + CC = 28C

Focus on ten’s place,   1 + C = 8, here C = 7. Then 22 + 88 + 77 = 187

Thus, Equation (2) is not possible.

From Equation (1), 11+99+CC = 19C à 110 + CC =19C à 1 + C = 9, then C = 8.

11 + 99 + 88 = 198  à hence solved A = 1, B = 9 and C = 8

A + B + C = 18

Answer is 18

2. HERE = COMES – SHE, (Assume S = 8). Find the value of R + H + O.

  1. 15
  2. 18
  3. 14
  4. 12

Ans: 3

Explanation:

HERE = COMES – SHE which can also be written as HERE + SHE = COMES

     HERE
+     SHE
 ---------
 COMES
---------
C = 1, O = 0, H = 9, E + E = S = 8, 2 E =8, And E=4.

So, COMES – SHE = HERE, 9454 + 894 = 10348
R + H + O = 5 + 9 + 0 = 14

Answer is 14

3. N O + G U N + N O = H U N T, find the value of HUNT.

  1. 1082
  2. 1802
  3. 1208
  4. 1280

Ans:  1

Explanation:

              N  O
    +   G  U  N

              N  O
     -------------
     H  U  N  T
     -------------

Here H = 1, from the NUNN column we must have "carry 1," so G = 9, U = zero. Since we have "carry" zero or 1 or 2 from the ONOT column, correspondingly we have N + U = 10 or 9 or 8. But duplication is not allowed, so N = 8 with "carry 2" from ONOT. Hence, O + O = T + 20 - 8 = T + 12. Testing for T = 2, 4 or 6, we find only T = 2 acceptable, O = 7. So we have 87 + 908 + 87 = 1082.

Answer is 1082

4. MAC + MAAR = JOCKO, find the value of 3A + 2M + 2C.

  1. 31
  2. 36
  3. 33
  4. 38

Ans: 1 

Explanation:  

            M A C
    + M A  A  R
     ------------------
      J O  C  K O
     ------------------ 

Here J is carry, J=1 when J=1, O=0 with carry 1 and M=9 C+R=O à 0 with carry 1. So, C=2 and R=8 M+A=C à 2 with carry 1, A=3, A+A+1= K, 3+3+1=K=7, 932+9338=10270 so, finally A = 3, M = 9, C = 2, = 3A + 2M + 2C = 9 + 18 + 4 = 31

Answer is 31

5. SEND + MORE = MONEY, find the value of M+O+N+E+Y.

  1. 15
  2. 14
  3. 16
  4. 18

 Ans: 2

Explanation:

         S E N D
    +  M O R E
    -----------------
     M O N E Y
    -----------------

From the given data, the value of M will be 1 because it is the only carry-over possible from the sum of 2 single digit numbers in column 4. M = 1, S + 1 = a (two digit number). So S = 1 and O cannot be 1 but 0. Also E and N are consecutive. With trial and error possibilities, we get SEND   = 9567, MORE = 1085 and MONEY = 10652. 

So, M + O + N + E + Y = 1 + 0 + 6 + 5 + 2 = 14

Answer is 14

6. If “EAT + THAT = APPLE”, what is the sum of A+P+P+L+E?

  1. 13
  2. 14
  3. 12
  4. 15

Ans: 3

Explanation:

           E A T
    +  T H A T
    ------------------
     A P P L E
    ------------------

From the given data, the value of A will be 1 because it is the only carry-over possible from sthe um of 2 single digit number. T maximum it can take only 9 and there should a carryover for T to give sum as 2 digit number. So T =9, P = 0, A = 1. T + T = 18, the value of E is 8 and 1 will be a carry over to the next column. That is 1 + A + A= L = 3. And finally H = 2. Hence, 819 + 9219 = 10038. A+P+P+L=E = 1+0+0+3+8 = 12.

Answer is 12

7. If POINT + ZERO = ENERGY, then E + N + E + R + G + Y = ?

  1.  17
  2. 13
  3. 19
  4. 18

 Ans: 1

Explanation:

           P   O   I   N   T
    +         Z   E   R   O
    -----------------------------
     E   N   E   R   G   Y
    -----------------------------

From the given data, the value of E will be 1 because it is the only carry-over possible from sum of 2 single digit number. E = 1, P = 9, N = 0, and N + R = G, 0 + R = G but R = G not possible. 1 + R = G possible. So R and G are consecutive. G > R. 1 + I = R, so I and R are consecutive. R > I. i.e., G > R > I. and G, R, I are consecutive. Now O + T should give carry over and O + Z also give carry over. So O is bigger number. Now take values for G, R, I as 8, 7, 6 or 7, 6, 5 etc. and do trial and error. POINT = 98504, ZERO = 3168 and ENERGY = 101672. So E + N + E + R + G + Y = 1 + 0 + 1 + 6 + 7 +2 = 17

Answer is 17

8. YOUR + YOU = HEART (Assume O = 4). Find the value of Y + U +R +E.

  1. 15
  2. 16
  3. 17
  4. 18

Ans:  3

Explanation:

       Y   O  U  R
+          Y  O  U
 -----------------------
   H  E  A   R  T
------------------------

Given the value of O as 4, Y and E cannot be the same, so there should be a carry-over 1. 1 + Y = 10, so E will take 0 and H as 1.  O + Y = A (4 + 9 = 13) i.e. A = 3. Now, U + O (4) =R (which can only be a single digit number), U cannot take 5, 3, 4 and only possibility will be 2. If U is 2, then R will be 6. R + U = T, 6 + 2 = 8(T = 8). Hence, the value of Y + U + R + E = 9 + 2 + 6+ 0 = 17.

Answer is 17

9. TOM + NAG = GOAT, find the value of G+O+A+T.

  1. 15
  2. 12
  3. 14
  4. Cannot be determined

Ans: 4

Explanation:

          T    O    M

     +   N    A    G

  -------------------------
    G   O    A    T
  ------------------------

Adding two single digit number the maximum carry it can have is 1, so G=1.

O + A = A, from this we can tell that O = 0. T + N = O (O should be a two digit number ending in zero, only then G will be 1). Sum of T and N should be 10 i.e. T (6) + N (4) = 10.  M + G = T, from this we will get the value of M as 1.  

O=0, G=1, N=4, M=5, T=6 and the left out numbers are 2, 3, 7, 8, 9, from this A can take any value. There is no definite value for A.

Answer is: Cannot be determined

10. FORTY + TEN + TEN = SIXTY, find the value of T+E+N.

  1. 11
  2. 22
  3. 31
  4. 24

Ans: 2

Explanation:

       F   O   R   T   Y

                  T   E   N

        +       T   E    N

----------------------------

      S   I    X    T   Y

From the rightmost column i.e. Y + N + N = Y, which means N =0.  And the next column T + E + E = T here T cannot be zero but E + E if it gives the sum of ten then 10 + T will give the unit digit as T. So 2E = 10, E = 5.

The letter O should have a carry to give the value I  i.e. O + carry = I, and in turn the I value should be a two-digit number because the leftmost column needs a carry (F + carry = S) to get the value S.

The sum of O + carry = I, where I  should be a two digit number. In order to get I as 2 digit number O has to take the maximum value 9, and let the carry be 1 (9 + 1 = 10) here the unit digit is I, but I cannot take the value 0 because zero is already assigned to N. So, let’s keep the carry as 2, 9 + 2 = 11 then unit digit 1 is the value of I, and one’s digit will be a carry for the next column i.e. F (F + 1 = S).

Next, R + T + T + carry = X (here the sum of these three numbers should give you the value in the range of twenty’s because 2 has been taken as a carry to the next column). So R and T should be taken the maximum value in order to get a number in the range of twenty’s.  

Let R = 7 and T = 8, R + T + T + carry = 7 + 8 + 8 +1 = 24. Then, T + E + E = T (two digit number) --> 8 + 5 + 5= 18.

The values: O=9, T=8, R =7, E=5, X=4, I=1 and the left out numbers are 6, 3, 2.  We know F + 1 = S i.e. F and S will take 2 and 3, and at last, the Y will take 6.

The value of S +I +X +T + Y = 3 + 1 + 4 + 8 + 6 = 22

Answer is 22

Cryptarithmetic questions are most commonly asked in the Infosys recruitment and eLitmus exam. Only by practicing more number of questions, one can become an expert in cryptarithmetic.

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