Today we’ll discuss all cryptarithmetic questions and also cryptarithmetic questions with solutions. But, before discussing the questions we’ll make you know about Cryptarithmetic. So,
Table of Contents
What is Cryptarithmetic?
Cryptarithmetic is the art and science of solving and creating cryptarithms. Cryptarithmetic is also a genre of mathematical puzzle where digits are replaced by the letters of alphabets or other symbols.
This art previously was known as verbal arithmetic or letter arithmetic. Cryptarithmetic is originated in ancient China. Similarly, in India, arithmetic restorations or skeletons were originated. In which all of the digits are replaced by asterisks.
The well-known Cryptarithmetic question SEND + MORE = MONEY was created by H.E Dudeney and was first published in the year 1924. It was published in Strand Magazine which was associated with the story of the kidnapper’s ransom demand.
Importance of Cryptarithmetic Questions
It takes about 10 to 15 minutes to solve Cryptarithmetic questions in exams. If you’re familiar with solving Cryptarithmetic questions then you can easily answer most of the questions only in 4 to 5 minutes.
Cryptarithmetic questions in exams are based on basic mathematics like finding the value of A+B+C, 2A+C, or triangle-related questions. If you want to solve maximum questions in an exam then you’re at the right place. By solving the below questions you’ll be able to attempt maximum questions in a short period of time.
Rules For Solving Cryptarithmetic Questions
- Each and every letter should have distinct and unique value.
- Each and every letter represents only one digit throughout the entire problem.
- Numbers should not start with zero. For example 7575(correct) and 07575(wrong).
- In Cryptarithmetic problem you’ve to find the value of each and every letter.
- There is always one solution for one entire problem.
- When all letters are replaced with digits, then the resulting arithmetic operations should be correct.
Now, let us learn how to solve any Cryptarithmetic questions with answers:
Q.1 HERE = COMES – SHE (Assume S=8). Find the value of R + H + O?
Ans: Option 2
Explanation: Another way of writing HERE + SHE = COMES is
H E R E + S H E --------- C O M E S
Now let us assume C=1, O=0, H=9, E+E=S=8,2 E=8, & E= 4. So, COMES – SHE = HERE, 9454 + 894 = 10348.
R + H + O = 5 + 9 + 0 = 14. So the final answer is 14.
Q.2 If AA + BB + CC = ABC. Then what is the value of A + B + C = ?
Ans: Option 3.
Explanation: AA + BB + CC = ABC can also be written as,
A A + B B + C C ----------- A B C
Here, the digits are positive and distinct. So, let’s focus on value A. If we add 3 two-digit numbers then most of you get answers in 200s. For example: AA + BB + CC (99 + 88 + 77 = 264) so the possible largest value of A can be 2. Either value of A will be 1 or 2 (A=2 or A=1).
We get the value of A, Now let’s focus on the value of B. Let’s take the unit digit value of the given question. A + B + C = C (units). So, this can happen only if A + B = 0 (units) A & B add up to 10.
There are two possibilities: 22 + 88 + CC = 28C (1) and 11 + 99 + CC = 19C (2).
Now take equation number 1, 110 + CC = 28C. Let us focus on ten’s place, 1 + C = 8 where, C=7 then the equation becomes, 22 + 88 + 77 = 187. So the equation 1 is not the possible solution.
Now take equation number 2, 11 + 99 + CC = 19C, 110 + CC = 19C. 1 + C =9 where, C=8 then the equation becomes, 11 + 99 +88 = 198, Hence solved. Therefore, A = 1, B = 9 and C = 8, A + B + C = 18. The answer is 18.
Q.3 MAC + MAAR = JOCKO, Now find the value of 3A + 2M + 2C?
Ans: Option 3.
Explanation: MAC +MAAR = JOCKO can be written as,
M A C + M A A R ----------- J O C K O
In the question J is carry so, let us consider J = 1, O = 0 with carry 1 and M = 9, C + R = O, 0 with carry 1. Therefore C = 2 and R = 8, M + A = C,2 with the carry 1. A = 3, A + A + 1 = K, Now put value of A in A + A + 1 = K. So the equation becomes, 3 + 3 + 1 = K i.e K = 7.
932 + 9338 = 10270 So, the value of A = 3, M = 9 and C =2, After putting values in the equation 3A + 2M + 2C = 9 + 18 + 4 = 31. So the answer is 31.
Q.4 SEND + MORE = MONEY, Now find the value of M + O + N + E + Y?
Ans: Option 1.
Explanation: SEND + MORE = MONEY can also be written as,
S E N D
+ M O R E
M O N E Y
Here, the only possible value of M is 1 because it’s only the carry-over possible from the sum of 2 single-digit numbers in column 4. So, M = 1, S + 1 = two-digit number. Therefore S = 1 and O can be 0 but not 1.
Here E and N are consecutive. So, with the hit and trial possibility, we get SEND = 9567, MORE = 1085, and MONEY = 10652.
i.e M + O + N + E + Y = 1 + 0 + 6 + 5 + 2 = 14. So, we get the answer as 14.
Q.5 NO + GUN + NO = HUNT, Now find the value of HUNT?
Ans: Option 2.
Explanation: NO + GUN + NO = HUNT
N O + N O + G U N ----------- H U N T
Now, consider the value of H as 1 and from the column NUNN we must have carry 1 so, G = 9 and U = 0. We have carry as 0, 1, or 2 from the column ONOT. Similarly, we’ve N + U = 8, 9, or 10, but duplication is not allowed. Therefore N = 8 with the carry 2 from column ONOT. Hence, O + O = T + 20 – 8 = T + 12. Now after doing testing for T = 2, 4 or 6 only T = 2 is acceptable, O = 7.
Hence, we’ve 87 + 908 + 87 = 1082. So, we get the answer as 1082.
Q.6 TOM + NAG = GOAT Now find the value of G + O + A + T?
- Cannot be Determined
Ans: Option 4.
Explanation: TOM + NAG = GOAT
T O M + N A G ----------- G O A T
After adding only two single digits numbers, the maximum carry we can get is 1. Therefore the value of G is 1 i.e G=1. Consider the equation O + A = A and from this equation, we can say that O = 0. T + N = O ( O must be a two-digit number that will end with 0, then only the value of G will be 1).
The sum of T & N should be 10. After substituting the values equation becomes, T(6) + N(4) = 10. From the equation M + G = T we get value of M as 1 i.e M = 1. So, we get values as O = 0, G = 1, N = 4, M= 5, T = 6 and the left out numbers are 2, 3, 7, 8, 9. Therefore A can take any value from this. A don’t have any definite value. So the answer is cannot be determined.
Q.7 FORTY + TEN + TEN = SIXTY, Now find the value of T + E + N?
Ans: Option 3.
Explanation: FORTY + TEN + TEN = SIXTY
F O R T Y
+ T E N
+ T E N
S I X T Y
Consider the rightmost column where we get an equation as Y + N + N = Y, where N = 0. Now take the next column T + E + E = T, in which T should not be 0 but E + E, if we get, the sum of ten then, 10 + T we get unit digit as T. Therefore, 2E = 10 and E = 5.
The letter O must have carry then only we’ll get the value of I i.e O + carry = I. The value of I should be a two-digit number because the leftmost column needs a carry (F + carry = S) then only we’ll get the value of S.
The addition of O + carry = I, here the value of I should be a two-digit number. If we want to get the value of I as 2 digit number then, O must have to take the value as 9. Let the value of carry be 1 (9 + 1 = 10). Here the value of I is a unit digit, but I should not take the value as 0 because N has already taken the value 0. Therefore we’ve to keep the carry as 2 so, 9 + 2 = 11 then the unit digit 1 is the actual value of I.
Here the ones digit will be the carry for the next column i.e F (F + 1 = S).
Now, R + T + T + carry = X ( the addition of these three numbers must give you the value in the range between the 20s because there is carry to the next column). Therefore R & T must be taken as the maximum value to get a number in the range of 20s.
Now, let the value of R = 7 and T = 8. R + T + T + carry = 7 + 8 + 8 + 1 = 24. Then the equation becomes T + E + E = T ( 2 digit number ) i.e 8 + 5 + 5 = 18.
Here, the value of O = 9, T = 8, R = 7, E = 5, X = 4, I = 1 & the left out numbers are 6, 3, 2. Now, we know that F + 1 = S i.e 2 and 3 will be taken by F and S. Therefore at last, the value of Y will be 6.
After substituting the values in S + I + X + T + Y = 3 + 1 + 4 + 8 + 6 = 22. So, the answer we get is 22.
Q.8 YOUR + YOU = HEART (Assume O = 4). Now, find the value of Y + U + R + E?
Ans: Option 3.
Explanation: YOUR + YOU = HEART can be written as,
Y O U R + Y O U ----------- H E A R T
In the question itself, the value of O is given as 4 i.e O = 4. If you see in the question the value of Y and E cannot be the same, that’s why there should be a carryover 1. Now consider the equation 1 + Y = 10, here E will take the value as 0 and H will take 1. O + Y = A (4 + 9 = 13) i.e A = 3.
Here, U + O(4) = R (only a single-digit number), but U cannot take the value like 5, 3, 4 so the only possible value of U will be 2. And if the value of U is 2 then the R will become 6.
After substituting all the values in R + U = T, 6 + 2 = 8 where T = 8 we get Y + U + R + E = 9 + 2 + 6 + 0 = 17. Therefore we get answer 17.
Q.9 If EAT + THAT = APPLE, then what is the sum of A + P + P + L + E?
Ans: Option 1.
Explanation: EAT + THAT = APPLE
E A T + T H A T ----------- A P P L E
From the given data we’ll get the value of A as 1 because it’s only the carryover possible number from the sum of two single-digit numbers. T can take only maximum number 9 and there must be a carryover for T to get the sum as 2 digit number.
Here we get T = 9, P = 0, A = 1. T + T = 18 and the value of E = 8 and 1 will be a carryover to the next column. So, 1 + A + A = L = 3 and finally the value of H becomes 2 i.e H = 2.
Hence, 819 + 9219 = 10038. Now substitute the values of A, P, L and E in A + P + P + L + E then we get, 1 + 0 + 0 + 3 + 8 = 12. Therefore, the final answer is 12.
Q.10 If POINT + ZERO = ENERGY, then E + N + E + R + G + Y = ?
Ans: Option 2.
Explanation: POINT + ZERO = ENERGY
P O I N T + Z E R O ----------- E N E R G Y
From the data the value of E = 1 because it is the only carryover possible from the sum of 2 single-digit numbers. Now consider the values of E = 1, P = 9 and N = 0, N + R = G, 0 + R = G. Here R = G is could not be possible but 1 + R = G is possible. Here G & R are consecutive, G > R, 1 + I = R that’s why I and R are consecutive.
Now, O + T should give carryover & O + Z can also give a carryover. So, here O is the biggest number. Now consider the values of G, R and I as 8, 7, 6 or 7, 6, 5 etc and check trial and error possibilities. We get POINT = 98504, ZERO = 3168 and ENERGY = 101672 so, E + N + E + R + G + Y = 1 + 0 + 1 + 6 + 7 + 2 = 17. Hence we get the answer as 17.
Q.11 USSR + USA = PEACE, then what is the value of P + E + A + C + E?
Ans: Option 2.
Explanation: USSR + USA = PEACE
U S A + U S S R ----------- P E A C E
Here, we can clearly see that the value of P = 1, U = 9, and E = 0 so first replace all the values in the equation. Now consider the last column A + R = 0, This is possible only if A = 5 and R = 5. but both the values cannot take the same values. Therefore it’s possible only with (8, 2), (7, 3), (6, 4), (3, 7), (2, 8). Here we’ve to implement the hit and trial method and take (2, 8) and replace it with the new values.
Now we’ll assume value of S to be 3 i.e S = 3. We’ve S + S + 1 (carry) = C, After substituting the values in S + S + 1 = C becomes 3 + 3 + 1 = C i.e C = 7 which is unoccupied. Here all the values are founded but we just need to verify.
Here the values are replaced and all the operations are working fine. So, the values of P + E + A + C + E = 1 + 0 + 2 + 7 + 0 = 10. So we get the answer as 10.
Q.12 TO + GO = OUT, Then find the value of T + G + O + U?
Ans: Option 1.
Explanation: TO + GO = OUT
T O + G O ----- O U T
Here the value of O is clearly 1 but T + G is generating O which is carry so the value of O is 1. After substituting the values T = 1 + 1 we get the value of T as 2 i.e T = 2. Now 2 + G is greater than 10 which is resulting in the carry 1 on next. Here, the possible values of G to get carry at the next step is 8 or 9.
If the value of G is 9 then, U = 2 + 9 = 11. So, the value of U becomes 1, and 1 goes to the carry. Here, the value of O is already 1 so, the value of U cannot be 1. Now, the value of G has to be 8. U = 2 + 8 = 10. So, the value of U becomes zero and there are no conflicts.
Therefore, the final values are T = 2, O = 1, G = 8 and U = 0. After substituting these values in T + G + O + U = 2 + 8 + 1 + 0 we get the answer as 11.
So, in this way we’ve learned how to solve any Cryptarithmetic question in a well and defined manner. If you solve all the cryptarithmetic questions given in this post then you’ll be confident while solving any problems in the exam. So, if you’ve any doubt then please feel free to contact us.