Numerical puzzle, №1
F = A B 2 + 2 × A × B + B A 2 A , B are non-zero digits, but not necessarily distinct . Can F be a multiple of 1 0 1 ?
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1 solution
Unless either A or B equals 0 . For example, 1 0 2 + 2 × 1 × 0 + 0 1 2 = 1 0 1 .
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True, except we're told 1 ≤ A , B ≤ 9 . (You're right, though, I should have mentioned that explicitly in my solution - edited now.)
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However, in my solution, A = 1 , and B = 0 , satisfying 1 ≤ A , B ≤ 9 .
Actually, the meaning of the inequality is that both A and B lies in the range [1,9]
Rewriting the expression, we have F = ( 1 0 A + B ) 2 + 2 A B + ( 1 0 B + A ) 2 . Expanding and collecting terms, this becomes F = 1 0 1 A 2 + 4 2 A B + 1 0 1 B 2 . The only way this can be a multiple of 1 0 1 is if 2 A B is a multiple of 1 0 1 . But since 1 0 1 is a prime, and since (per the problem) 1 ≤ A , B ≤ 9 , this is clearly impossible.
So F is never a multiple of 1 0 1 .