1997 AIME Problems/Problem 3

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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 $(9x-1)\left(y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}$ , and $(9x-1)(9y-1000)=1000$ .

Since $89 < 9x-1 < 890$ , 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 $112 + 14 = 126$ .

Solution 2

As shown above, we have $1000x+y=9xy$ , so $1000/y=9-1/x$ . $1000/y$ must be just a little bit smaller than 9, so we find $y=112$ , $x=14$ , and the solution is $126$ .

1997 AIME (Problems • 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

From AoPSWiki

Contents

Problem

Solution

Solution 2

See also