1997 AIME Problems/Problem 3

Problem

Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?

Solution

Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \Longrightarrow$ 9xy-1000x-y=0 $. Using [[SFFT]], this factorizes to$ (9x-1)(y-\dfrac{1000}{9})=\dfrac{1000}{9} $, and$ (9x-1)(9y-1000)=1000$.

Since$ (Error compiling LaTeX. Unknown error_msg)89 < 9x-1 < 890 $, we can use trial and error on factors of 1000. If$ 9x - 1 = 100 $, we get a non-integer. If$ 9x - 1 = 125 $, we get$ x=14 $and$ y=112 $, which satisifies the conditions. Hence the answer is$ 112 + 14 = 126$.

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1997 AIME Problems/Problem 3

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