Difference between revisions of "1997 AIME Problems/Problem 3"
m (→See also) |
m (→Solution) |
||
Line 21: | Line 21: | ||
9y-1000 must be positive, so y>111. | 9y-1000 must be positive, so y>111. | ||
− | Since x is greater than 89, we | + | Since x is greater than 89, we can try certain factors of 1000: |
100: 9x=101, nope. | 100: 9x=101, nope. |
Revision as of 13:13, 21 November 2007
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 the two-digit number be ab, and the three-digit number be cde.
$abcde=10000a+1000b+100c+10d+e=9(10a+b)(100c+10d+e}$ (Error compiling LaTeX. ! Extra }, or forgotten $.)
9y-1000 must be positive, so y>111.
Since x is greater than 89, we can try certain factors of 1000:
100: 9x=101, nope.
125: 9x=126, x=14
Then 9y-1000=8, 1008=9y, 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 |