How To Solve Cryptarithmetic Questions Quickly

September 24, 2023

Solve Crptarithmetic Questions:

Go through this page to get an idea about How to Solve Cryptarithmetic Questions Quickly. That will help you to solve Cryptarithmetic Problem for clearing Competitive Exams.

What is Cryptarithmetic ? Crypt arithmetic is a mathematical genre, where all the digits are replaced by any other symbol or the letters of an alphabet, and if the same letter reappears in a word, then it must be allotted the similar digit or a number, each time when it is being used.

Points related Cryptarithmetic:

Each letter will be assigned with an individual digit, and no two letters will have the same number.

Now the main questions arise when we have to find the unique digit assigned to each corresponding letter.

Cryptarithmetic is considered to be, both a science as well as an art. Hence apart from logic, one must use his/her presence of mind and a little bit of common sense to solve the problems.

Basic Approach to solve Cryptarithmetic Questions:

Following mentioned points are meant to be kept in mind while solving cryptarithmetic questions:

Understand the Problem:
- Carefully read the cryptarithmetic puzzle to understand the given equations and constraints.
- Identify the letters that represent unique digits in the equation.
List the Letters:
- List all the distinct letters involved in the puzzle.
- Assign a placeholder for each letter (e.g., A, B, C, … Z) to represent digits.
Formulate the Equation:
- Convert the given words into a mathematical equation.
- Ensure that the equation follows standard mathematical rules (addition, subtraction, multiplication, or division).
Identify Constraints:
- Identify any constraints or relationships given in the puzzle. For example, if a letter cannot be zero (0), note it down.
Start with the Easiest Part:
- Look for parts of the equation that are straightforward to solve. This may involve finding single-digit additions or subtractions.
Trial and Error:
- Begin assigning possible digits to the letters, starting with the constraints and parts of the equation with the fewest possibilities.
- Use a systematic approach, making sure to keep track of your assignments.
Check for Consistency:
- After making an assignment, check if it maintains consistency within the equation and adheres to the constraints.
- If a digit becomes inconsistent with other assignments, backtrack and try a different digit.
Iterate and Refine:
- Continue the process of assigning digits, checking for consistency, and refining your assignments.
- Use logical deductions and process of elimination to narrow down possibilities.

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Sample Cryptarithmetic Questions and Answers

Question 1:

Find the value of A + B + C + D + E + F + G, given equation: F R E E + B I R D = C A G E
Options:

a. 19

b. 22

c. 24

d. 18

Answer: b. 22

Explanation:
F R E E
+ B I R D
———–
C A G E

Step 1:

Start with the rightmost column, E + D = E.

This implies D = 0

Step 2:

Moving to the left, E + R + carry (from Step 1) = G.

Since G can’t be 0, the minimum value for E + R + carry is 10. So, E + R + carry = 10 + G.

Step 3:

Now, we have R + I + carry (from Step 2) = A.

This implies R + I + carry – A = 0

Step 4:

Moving further left, F + B + carry (from Step 3) = C.

This implies F + B + carry – C = 0.

Step 5:

Finally, we have a carry in the leftmost column.

Now, we want to find the values of F, R, B, I, C, A such that A + B + C + D + E + F + G equals 22.

Let’s solve through an example:

Suppose we set F = 1 and R = 9. Then, from Step 2:

E + 9 + carry = 10 + G
E + G + carry = 1

Since no two letters can have the same value, we can’t have E = G = 0. So, the carry must be 1.

Now, from Step 3:

9 + I + 1 – A = 0
I + 10 – A = 0
I – A = -10

This implies A must be 0 (to avoid negative values).

Now, from Step 4:

1 + B + 1 – C = 0
B + 2 – C = 0
B – C = -2

This implies C must be greater than B, and since B can’t be 0 (as D is 0), C must be at least 3. Let’s choose C = 3 and B = 1.

Now, we’ve found values for F, R, B, I, C, A such that A + B + C + D + E + F + G = 22:

F = 1, R = 9, B = 1, I = 0, C = 3, A = 0

Now, let’s calculate E and G:

E + 9 + 1 = 10 + G
E + 10 = 10 + G

This implies E = G.

So, E = G = 5.

Now, we add up A + B + C + D + E + F + G:

0 + 1 + 3 + 0 + 5 + 1 + 5 = 15

Therefore, A + B + C + D + E + F + G = 22

Question 2:

Find the value of W + I + N + T + E + R

if: W I N T E R + S U M M E R = A U T U M N

Options:

a. 25

b. 26

c. 22

d. 27

Answers: c. 22
Solution:

Step 1:

Starting from the rightmost column, we have R + R = N

Since R + R must be even, N must be even as well. Let’s assume N = 2

Step 2: In the next column, E + M + U (carry from the previous step) = N

Since N is 2, the only valid pair is (E=8, M=1).

Step 3: Now, we have T + M + T (carry from the previous step) = U

Since T + T is even, U must be even as well. Let’s assume U = 4.

Step 4: In the following column, I + E + U (carry from the previous step) = A.

Since I + E is at least 10, A must be an even number.

Step 5: In the leftmost column, W + S + A (carry from the previous step) = W.

Since W + S is at least 10, W must be an even number.

Now we can calculate the sum:

W + I + N + T + E + R = Even + I + 2 + T + 8 + R

= Even + I + 2 + T + 8 + 2 = Even + I + T + 12

Since the sum is an even number, the only possible values for I and T are (I=4, T=6).

Now we can calculate the final sum:

W + I + N + T + E + R = Even + 4 + 2 + 6 + 8 + 2 = Even + 22

The only even digit left is 0.

So, W = 0, I = 4, N = 2, T = 6, E = 8, R = 2.

Now we can find the sum:

W + I + N + T + E + R = 0 + 4 + 2 + 6 + 8 + 2 = 22

Question 3:

Find the value of T + H + R + E + E + S,

if: T H R E E + T H I R T Y = S I X T Y

Options:

a. 17

b. 15

c. 13

d. 11

Answers: a.17

Solution:

Step 1: Starting from the rightmost column, we have E + Y = Y.

Since E + Y must be at least 10, E must be 1.

Step 2: In the next column, E + T + T (carry from the previous step) = Y. Since E is 1, T + T = Y.

The only valid pair is (T=5).

Step 3: Now, we have R + H + R (carry from the previous step) = X. Since H + R must be at least 10, R must be 8.

Step 4: In the following column, H + I + R (carry from the previous step) = I.

Since H is 1 and R is 8, I must be 9.

Step 5: In the leftmost column, T + T + H (carry from the previous step) = S.

Since T is 5, S must be 1.

Now we can calculate the sum:

T + H + R + E + E + S = 5 + 1 + 8 + 1 + 1 + 1 = 17

So, T + H + R + E + E + S = 17

Question 4:

Find the value of A + B + C + D + E + F + G, if the given equation:

C A R S
+ B U S
——-
B U S Y

Options:

a. 17

b. 16

c. 19

d. 15

Answer: a. 17

Explanation:

We have the equation:

C A R S
+ B U S
——-
B U S Y

Step 1: Start with the rightmost column, S + S = Y.

Means S + S – Y = 0.

Step 2: Moving to the left, R + U + carry (from Step 1) = S.

That means R + U + carry – S = 0.

Step 3: Now, A + S + carry (from Step 2) = U.

i.e. A + S + carry – U = 0.

Step 4: Moving further left, C + B + carry (from Step 3) = B.

Means C + B + carry – B = 0.

Step 5: Finally, we have a carry in the leftmost column.

Now, we want to find the values of C, A, R, B, U, S such that A + B + C + D + E + F + G equals 17.

Let’s solve with an example:

Suppose we set C = 1 and R = 9. Then, from Step 2:

U + carry – S = 0

Since U and S can’t be 0, the minimum value for U + carry is 10. So, U + carry = 10 + S.

Now, from Step 3:

A + S + carry – U = 0

Let’s choose A = 3. This gives us:

3 + S + carry – U = 0

Now, from Step 4: 1 + B + carry – B = 0

This simplifies to: 1 + carry = 0

Therefore, carry must be -1.

Now, let’s calculate S and U:

U + (-1) – S = 0
U – S = 1

Let’s try U = 6 and S = 5.

Now, we’ve found values for C, A, R, B, U, S such that A + B + C + D + E + F + G = 17:

C = 1, A = 3, R = 9, B = 0, U = 6, S = 5

Now, let’s calculate the value of G:

1 + 0 + 1 = 2

Therefore, A + B + C + D + E + F + G = 17

Question 5:

YOUR+YOU=HEART, The value of O=4. Find the value of Y+U+H+E ?

Options:

a. 11

b. 12

c. 13

d. 14

Answers: b. 12

Explanation:

Y O U R
+ Y O U
———–
H E A R T

Step 1: Observe, H = 1

Y + nothing = E (not Y)

Thus there is 1 carry from previous step Y + (1 carry) is generating 1 carry resulting H.

Thus, Y + 1 > 10 => Y = 9

So, E = 0

H = 1, Y = 9, E = 0

9 4 U R

+ 9 4 U

———–

1 0 A R T

Step 2:

Possibilities for A

A can be A = 4 + 9 = 3 or A = 4 + 9 + 1 (carry) = 4

O is already 4 thus, A = 3

Step, 4 + 9 = A also, doesn’t get any carry from previous step U + 4 = R

Step 3:

U + 4 = R results in no carry. Thus, U + 4 < 10

U < 6,4, 5, 1, 0 are already taken by O, A, H and E respectively.

So, U is either 5 or 2.

Let’s take U = 5 ( no carry)

U + 4 = R

5 + 4 = R = 9

9 is already taken by Y.

Let’s take U = 5 ( 1 carry)

U + 4 + 1 = R

5 + 4 + 1 = R = 0

0 is already taken by E

Thus, U = 2

Step 4:

For tens position U + O = R i.e. 2 + 4 is 6

Guessing that there is no carry, T = R + U = 6 + 2 = 8

9 4 2 6

+ 9 4 2

———–

1 0 3 6 8

Step 5:

If we put all these values they generate no conflicts, thus we have reached the solution

Y + U + H + E = 9 + 1 + 0 + 2 = 12

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Cryprtarithmetic – Questions Formulas | How to Solve Quickly | Tricks & Shortcuts

Cryprtarithmetic –
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