Solved each of the following Cryptarithms:\(\frac{\begin{matrix}A & B

1.	Solved each of the following Cryptarithms:\(\frac{\begin{matrix}A & B & 7\\+7 & A & B\end{matrix}}{9\,\,\,\,\,\,\,\,8\,\,\,\,\,\,A}\)
Answer» 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

Solved each of the following Cryptarithms:\(\frac{\begin{matrix}A & B & 7\\+7 & A & B\end{matrix}}{9\,\,\,\,\,\,\,\,8\,\,\,\,\,\,A}\)

Answer»

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