Difference between revisions of "1997 AIME Problems/Problem 3"
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Let <math>x</math> be the two-digit number, <math>y</math> be the three-digit number. Putting together the given, we have <math>1000x+y=9xy \Longrightarrow 9xy-1000x-y=0</math>. Using [[Simon's Favorite Factoring Trick|SFFT]], this factorizes to <math>(9x-1)\left(y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}</math>, and <math>(9x-1)(9y-1000)=1000</math>. | Let <math>x</math> be the two-digit number, <math>y</math> be the three-digit number. Putting together the given, we have <math>1000x+y=9xy \Longrightarrow 9xy-1000x-y=0</math>. Using [[Simon's Favorite Factoring Trick|SFFT]], this factorizes to <math>(9x-1)\left(y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}</math>, and <math>(9x-1)(9y-1000)=1000</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>. | + | 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 = \boxed{126}</math>. |
== Solution 2 == | == Solution 2 == | ||
Revision as of 22:36, 3 August 2011
Contents
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 
. Using SFFT, this factorizes to 
, and 
.
Since 
, we can use trial and error on factors of 1000. If 
, we get a non-integer. If 
, we get 
and 
, which satisifies the conditions. Hence the answer is 
.
Solution 2
As shown above, we have 
, so 
. 
must be just a little bit smaller than 9, so we find , , and the solution is 
.
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 | ||
