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 be the two-digit number, be the three-digit number. Putting together the given, we have . Using SFFT, this factorizes to , and .
Since , we can use trial and error on factors of 1000. If , we get a non-integer. If , we get and , which satisifies the conditions. Hence the answer is .
Solution 2
As shown above, we have , so . must be just a little bit smaller than 9, so we find , , and the solution is .
Solution 3
To begin, we rewrite
as
and
This is the most important part: Notice is and is . That means is also . Rewrite as .
Now we have to find a number that divides 1000 using prime factors 2 or 5 and is . It is quick to find there is only one: 125. That gives 14 as and 112 as . Therefore the answer is .
-jackshi2006
See also
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 | ||
All AIME Problems and Solutions |
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