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Difference between revisions of "1997 AIME Problems/Problem 3"

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Since <math>89 < 9x-1 < 890</math>, we can use trial and error on factors of 1000. If <math>9x - 1 = 100</math>, we get a non-integer. If <math>9x - 1 = 125</math>, we get <math>x=14</math> and <math>y=112</math>, which satisifies the conditions. Hence the answer is <math>112 + 14 = 126</math>.
 
Since <math>89 < 9x-1 < 890</math>, we can use trial and error on factors of 1000. If <math>9x - 1 = 100</math>, we get a non-integer. If <math>9x - 1 = 125</math>, we get <math>x=14</math> and <math>y=112</math>, which satisifies the conditions. Hence the answer is <math>112 + 14 = 126</math>.
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== Solution 2 ==
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As shown above, we have 1000x+y=9xy, so 1000/y=9-1/x. 1000/y must be just a little bit smaller than 9, so we find y=112, x=14, and the solution is 126.
  
 
== See also ==
 
== See also ==

Revision as of 21:38, 4 March 2008

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 $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \Longrightarrow 9xy-1000x-y=0$. Using SFFT, this factorizes to $(9x-1)\left(y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}$, and $(9x-1)(9y-1000)=1000$.

Since $89 < 9x-1 < 890$, we can use trial and error on factors of 1000. If $9x - 1 = 100$, we get a non-integer. If $9x - 1 = 125$, we get $x=14$ and $y=112$, which satisifies the conditions. Hence the answer is $112 + 14 = 126$.

Solution 2

As shown above, we have 1000x+y=9xy, so 1000/y=9-1/x. 1000/y must be just a little bit smaller than 9, so we find y=112, x=14, and the solution is 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
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