Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.0 Junior Find the answer

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

Pick one

Solution

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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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.