Difference between revisions of "1997 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
| + | Let the two digit number be ab, and the three digit number be cde. | ||
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| + | <math>ab=10a+b=x</math> | ||
| + | |||
| + | <math>cde=100c+10d+e=y</math> | ||
| + | |||
| + | <math>abcde=10000a+1000b+100c+10d+e=9(10a+b)(100c+10d+e}</math> | ||
| + | |||
| + | <math>1000x+y=9xy</math> | ||
| + | |||
| + | <math>9xy-1000x-y=0</math> | ||
| + | |||
| + | <math>(9x-1)(y-\dfrac{1000}{9})=\dfrac{1000}{9}</math> | ||
| + | |||
| + | <math>(9x-1)(9y-1000)=1000</math> | ||
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| + | 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 | ||
== See also == | == See also == | ||
* [[1997 AIME Problems]] | * [[1997 AIME Problems]] | ||
Revision as of 12:32, 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 nmber. 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
