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

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== Solution ==
 
== Solution ==
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Let the two-digit number be ab, and the three-digit number be cde.
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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 </math>9xy-1000x-y=0<math>. Using [[SFFT]], this factorizes to </math>(9x-1)(y-\dfrac{1000}{9})=\dfrac{1000}{9}<math>, and </math>(9x-1)(9y-1000)=1000<math>.
  
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<math>ab=10a+b=x</math>
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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$.
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<math>cde=100c+10d+e=y</math>
 
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<math>abcde=10000a+1000b+100c+10d+e=9(10a+b)(100c+10d+e}</math>
 
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<math>1000x+y=9xy</math>
 
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<math>9xy-1000x-y=0</math>
 
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<math>(9x-1)(y-\dfrac{1000}{9})=\dfrac{1000}{9}</math>
 
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<math>(9x-1)(9y-1000)=1000</math>
 
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9y-1000 must be positive, so y>111.
 
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Since x is greater than 89, we can try certain factors of 1000:
 
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100: 9x=101, nope.
 
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125: 9x=126, x=14
 
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Then 9y-1000=8, 1008=9y, y=112.
 
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112+14=126
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1997|num-b=2|num-a=4}}
 
{{AIME box|year=1997|num-b=2|num-a=4}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 17:50, 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 $x$ 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$(9x-1)(y-\dfrac{1000}{9})=\dfrac{1000}{9}$, and$(9x-1)(9y-1000)=1000$.

Since$ (Error compiling LaTeX. Unknown error_msg)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$112 + 14 = 126$.

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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