# Top 10 Cryptarithmetic Questions With Answers 2023

If you think you are super smart at maths and if you are always looking for challenging problems then you should always try solving the Cryptarithmetic problems!

## What are Cryptarithmetic Problems?

If you are new to this concept or if you don’t know what that means then we have the answer you are looking for!
Cryptarithmetic problems are mathematical puzzles. But to make it a little different than others, here the digits would be changes old to alphabets. These kinds of questions are mostly asked in the many tech-savvy exams for recruiting candidates.
In this article, we have picked some of the most commonly asked and some very challenging Cryptarithmetic questions for you. Do take a look at these questions and see if you can solve them it would be like a mini-challenge for all the mathematic enthusiastic out there.
Take a look at the following list of Cryptarithmetic questions and see if you can solve them or you need to practice more.
Rules:
-Each letter should have a unique and distinct value.
-Each letter represents only one digit throughout the problem.
-Numbers must not begin with zero. i.e. 0937 (wrong), 937 (correct).
-You have to find the value of each letter in the Cryptarithmetic.
-There must be only one solution to the problem.
-After replacing letters by their digits, the resulting arithmetic operations must be correct. [toc]

1.YOUR + YOU = HEART (Assume O = 4). Find the value of Y + U +R +E.
15
16
17
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.
Q2. TOM + NAG = GOAT, find the value of G+O+A+T.
15
12
14
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.
Q3. FORTY + TEN + TEN = SIXTY, find the value of T+E+N.
11
22
31
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
Q4. If AA + BB = ABC, then what is the value of A+B+C= ?
15
18
21
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
Q5. HERE = COMES – SHE, (Assume S = 8). Find the value of R + H + O.
15
18
14
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

Q6. N O + G U N + N O = H U N T, find the value of HUNT.
1082
1802
1208
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.
Q7. MAC + MAAR = JOCKO, find the value of 3A + 2M + 2C.
31
36
33
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
Q8. SEND + MORE = MONEY, find the value of M+O+N+E+Y.
15
14
16
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
Q9. If “EAT + THAT = APPLE”, what is the sum of A+P+P+L+E?
13
14
12
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.