1953 AHSME Problems/Problem 24

If $a,b,c$ are positive integers less than $10$ , then $(10a + b)(10a + c) = 100a(a + 1) + bc$ if:

$\textbf{(A)}$ $b+c=10$ $\textbf{(B)}$ $b=c$ $\textbf{(C)}$ $a+b=10$ $\textbf {(D)}$ $a=b$ $\textbf{(E)}$ $a+b+c=10$

Multiply out the LHS to get $100a^2+10ac+10ab+bc=100a(a+1)+bc$ . Subtract $bc$ and factor to get $10a(10a+b+c)=10a(10a+10)$ . Divide both sides by $10a$ and then subtract $10a$ to get $b+c=10$ , giving an answer of $\boxed{A}$ .

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1953 AHSME Problems/Problem 24