# Tips, Tricks and Shortcuts to Solve Cryptarithmetic Questions

## Shortcuts to Solve Cryptarithmetic Questions

Cryptarithmetic is considered to be, both a science as well as an art. Go through the page to learn Tips, Tricks and Shortcuts to Solve Cryptarithmetic Questions that will definitely help you to prepare for various competitive exams.

### Rules to remember:

Below mentioned are some thumb rules and principles that need to be followed while solving Cryptarithmetics questions, some of which are mentioned below:

###### 1. Assigning digits to each letter or alphabet:

Under this each letter will be assigned a particular digit, and if the same letter is being used again in the word than, it will be denoted with the same digit which is allotted to it.

###### 2. Each letter has its own unique value:

Just like we are assigned an individual Roll number in a class, similarly, while solving such questions each letter will be assigned its own unique value, making it clear that no two letter can have the same digit or a number or a value.

For example: If a digit 4 is allotted to alphabet F, then this number cannot be assigned to any other alphabet.
The digit assigned to the letter will remain the same throughout the question.
The digits assigned must have a numerical base ranging from 0 to 9
The resultant numbers should not begin with 0.

###### 3. Cross-checking or Checking of Errors:

Now after doing all the algorithm and finding out the digits assigned to each letter, we need to cross verify the resultant, which means we must put the digits in place of the corresponding letters and the resulting arithmetic operations must be correct.

For doing this there are certain pre requisites which need to be considered which are mentioned below:

1. The total number of letters must not be more than 10.
2. The length of the answer should match with the operand.
3. In case the length of the answer does not match, then must not exceed more than. Simply put the answer must be only one more than the operand.

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### Question 1:

LEMON + ORANGE = FRUIT, Check whether the equation is correct or not.

Solution:

According to Rule 1, we must assign unique digits to each letter. Since “LEMON,” “ORANGE,” and “FRUIT” all contain unique letters, we can assign distinct digits to them.

Now, let’s find values for L, E, M, O, N, R, A, G, F, U, I, and T such that the equation LEMON + ORANGE = FRUIT holds.

Since the result is one more than the operand, F must be 1.

Now, we have:

LEMON + ORANGE = 1RUIT

The leftmost column indicates that L + O must be greater than or equal to 10, so L = 9 and O = 8 (the only digits that fit).

Now, we have:

9EMON + 8RANGE = 1RUIT

Since the digits must be unique, R cannot be 1. Let’s try R = 2:

9EMON + 82ANGE = 12UIT

Now, we can assign values to E, M, A, G, U, and I.

They must be 3, 4, 5, 6, 7, and 0 (or vice versa) to satisfy:

9434N + 82546 = 1247T

Now, it’s clear that N and T must be 1 and 7 (or vice versa).

So, the solution is:
L = 9
E = 3
M = 4
O = 8
N = 1
R = 2
A = 5
G = 6
F = 1
U = 7
I = 0
T = 7

LEMON + ORANGE = FRUIT is satisfied with the values above.

### Question 2:

APPLE + BANANA = CHERRY, Check whether the equation is correct or not.

Solution:

According to Rule 1, we must assign unique digits to each letter.

Since “APPLE,” “BANANA,” “CHERRY” all contain unique letters, we can assign distinct digits to them.

Now, let’s find values for A, P, L, E, B, N, C, H, R, Y such that the equation APPLE + BANANA = CHERRY holds.

Since the result is one more than the operand, C must be 1.

Now, we have:

APPLE + BANANA = 1HERRY

The leftmost column indicates that A + B must be greater than or equal to 10, so A = 9 and B = 2 (the only digits that fit).

Now, we have:

9PPL9 + 29292 = 19HERRY

Since the digits must be unique, R cannot be 1. Let’s try R = 4:

9PPL9 + 29292 = 194HERY

Now, we can assign values to P, L, E, H, and Y. They must be 3, 5, 6, 7, and 0 (or vice versa) to satisfy:

93596 + 29292 = 19474

Now, it’s clear that Y must be 4.

So, the solution is:
A = 9
P = 3
L = 5
E = 6
B = 2
N = 9
C = 1
H = 7
R = 4
Y = 4

Hence, the equation is satisfied with the values.

### Question 3:

WATER + WINE = JUICE

Solution:

According to Rule 1, we must assign unique digits to each letter.

Since “WATER,” “WINE,” “JUICE” all contain unique letters, we can assign distinct digits to them.

Now, let’s find values for W, A, T, E, R, I, N, J, U, C such that the equation WATER + WINE = JUICE holds.

Since the result is one more than the operand, J must be 1.

Now, we have:

WATER + WINE = 1UICE

The leftmost column indicates that W + W must be greater than or equal to 10, so W = 5 (the only digit that fits).

Now, we have:

5ATER + 5INE = 1UICE

Since the digits must be unique, U cannot be 1. Let’s try U = 2:

5ATER + 5INE = 12ICE

Now, we can assign values to A, T, E, I, N, and C. They must be 9, 4, 3, 8, 7, and 0 (or vice versa) to satisfy:

59493 + 59438 = 12131

Now, it’s clear that C must be 1.

So, the solution is:
W = 5
A = 9
T = 4
E = 3
R = 8
I = 7
N = 2
J = 1
U = 6
C = 1

So, the equation is satisfied with derived values.

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