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
9xy-1000x-y=0
(9x-1)(y-\dfrac{1000}{9})=\dfrac{1000}{9}
(9x-1)(9y-1000)=1000$.
Since$ (Error compiling LaTeX. Unknown error_msg)89 < 9x-1 < 8909x - 1 = 100
9x - 1 = 125
x=14
y=112
112 + 14 = 126$.
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 |