1953 AHSME Problems/Problem 24

Problem

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

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

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 $\boxed{A}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1953 AHSME Problems/Problem 24

Problem

Solution

See Also