Difference between revisions of "1997 AIME Problems/Problem 3"
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| − | Let the two digit number be ab, and the three digit number be cde. | + | Let the two-digit number be ab, and the three-digit number be cde. |
<math>ab=10a+b=x</math> | <math>ab=10a+b=x</math> | ||
Revision as of 12:35, 11 October 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. Unknown error_msg)
9y-1000 must be positive, so y>111.
Since x is greater than 89, we ca 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
