Solve Cryptarithms. A B 7 + 6 A B 9 8 A - Vidyadip

Question

Solve Cryptarithms. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\ \ \ \text{B} \ \ \ \ 7\ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \underline{+\ \ \ \ 6\ \ \ \ \text{A}\ \ \ \ \text{B}\ }\\\ \ \ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ 9\ \ \ \ 8\ \ \ \ \ \text{A}}$

✓

Answer

If $\text{A}+\text{B}=8,\text{A}+\text{B}\geq9$ is possible only if $A = b = 9$ But from $7 + B = A, A = B = 9$ is impossible. Surely, $\text{A}+\text{B}=8, \text{A}+\text{B}\leq9$ So,$ A + 7 = 9,$ Surely $A = 27 + B = A, 7 + B = 2, B = 5 So, A = 2, B = 5$

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