Q. 6

# Solved each of the following Cryptarithms:

For unit’s place,

We have two conditions, when 7 + B ≤ 9 and 7 + B > 9

For 7 + B ≤ 9

7 + B = A

A B = 7 (1)

In ten’ place,

B + A = 8 …(2)

Solving 1 and 2 simultaneously,

2A = 15 which means A = 7.5 which is not possible

Hence, our condition 7 + B ≤ 9 is wrong.

7 + B > 9 is correct condition

Hence, carrying one in ten’s place and subtracting 10 from unit’s place,

7 + B – 10 = A

B A = 3 (3)

For ten’s place,

B + A + 1 = 8

B + A =7 (4)

Solving 3 and 4 simultaneously,

2B = 10

B = 5 and A = 2

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