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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.
+
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>.
  
−
<math>ab=10a+b=x</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>.
  
−
<math>cde=100c+10d+e=y</math>
+
== Solution 2 ==
  
−
<math>abcde=10000a+1000b+100c+10d+e=9(10a+b)(100c+10d+e}</math>
+
As shown above, we have <math>1000x+y=9xy</math>, so <math>1000/y=9-1/x</math>. <math>1000/y</math> must be just a little bit smaller than 9, so we find <math>y=112</math>, <math>x=14</math>, and the solution is <math>\boxed{126}</math>.
  
−
<math>1000x+y=9xy</math>
 
  
−
<math>9xy-1000x-y=0</math>
+
== Solution 3==
  
−
<math>(9x-1)(y-\dfrac{1000}{9})=\dfrac{1000}{9}</math>
+
To begin, we rewrite <math>(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z</math>
  
−
<math>(9x-1)(9y-1000)=1000</math>
+
as
  
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9y-1000 must be positive, so y>111.
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<math>(90a+9b-1)(100x+10y+z) = 10000a + 1000b</math>
  
−
Since x is greater than 89, we ca try certain factors of 1000:
+
and
  
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100: 9x=101, nope.
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<math>(90a+9b-1)(100x+10y+z) = 1000(10a + b)</math>
  
−
125: 9x=126, x=14
 
  
−
Then 9y-1000=8, 1008=9y, y=112.
+
This is the most important part: Notice <math>(90a+9b-1)</math> is <math>-1 \pmod{10a+b}</math> and <math>1000(10a + b)</math> is <math>0\pmod{10a+b}</math>. That means <math>(100x+10y+z)</math> is also <math>0\pmod{10+b}</math>. Rewrite <math>(100x+10y+z)</math> as <math>n\times(10a+b)</math>.
  
−
112+14=126
+
 
 +
<math>(90a+9b-1)\times n(10a+b)= 1000(10a + b)</math>
 +
 
 +
 
 +
<math>(90a+9b-1)\times n= 1000</math>
 +
 
 +
 
 +
Now we have to find a number that divides 1000 using prime factors 2 or 5 and is <math>8\pmod9</math>. It is quick to find there is only one: 125. That gives 14 as <math>10a+b</math> and 112 as <math>100x+10y+z</math>. Therefore the answer is <math>112 + 14 = \boxed{126}</math>.
 +
 
 +
 
 +
 
 +
-jackshi2006
  
 
== See also ==
 
== See also ==
−
* [[1997 AIME Problems]]
+
{{AIME box|year=1997|num-b=2|num-a=4}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:34, 4 August 2021

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 $112 + 14 = \boxed{126}$.

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 $\boxed{126}$.


Solution 3

To begin, we rewrite $(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z$

as

$(90a+9b-1)(100x+10y+z) = 10000a + 1000b$

and

$(90a+9b-1)(100x+10y+z) = 1000(10a + b)$


This is the most important part: Notice $(90a+9b-1)$ is $-1 \pmod{10a+b}$ and $1000(10a + b)$ is $0\pmod{10a+b}$. That means $(100x+10y+z)$ is also $0\pmod{10+b}$. Rewrite $(100x+10y+z)$ as $n\times(10a+b)$.


$(90a+9b-1)\times n(10a+b)= 1000(10a + b)$


$(90a+9b-1)\times n= 1000$


Now we have to find a number that divides 1000 using prime factors 2 or 5 and is $8\pmod9$. It is quick to find there is only one: 125. That gives 14 as $10a+b$ and 112 as $100x+10y+z$. Therefore the answer is $112 + 14 = \boxed{126}$.


-jackshi2006

See also

1997 AIME (Problems • Answer Key • Resources)
Preceded by
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Problem 4
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All AIME Problems and Solutions

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