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Difference between revisions of "2007 AIME II Problems/Problem 2"

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Subtracting the one, we see that <math>\displaystyle x + y = \{0,1,3,4,9,19,24,49,99\}</math>. There are exactly <math>\displaystyle n - 1</math> ways to find pairs of <math>\displaystyle (x,y)</math> if <math>\displaystyle x + y = n</math>. Thus, there are <math>\displaystyle 0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = 200</math> solutions of <math>(a,b,c)</math>.
 
Subtracting the one, we see that <math>\displaystyle x + y = \{0,1,3,4,9,19,24,49,99\}</math>. There are exactly <math>\displaystyle n - 1</math> ways to find pairs of <math>\displaystyle (x,y)</math> if <math>\displaystyle x + y = n</math>. Thus, there are <math>\displaystyle 0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = 200</math> solutions of <math>(a,b,c)</math>.
 +
 +
Alternatively, note that the sum of the divisors of <math>100</math> is <math>(1 + 2 + 2^2)(1 + 5 + 5^2) \displaystyle</math> (notice that after distributing, every divisor is accounted for). This evaluates to <math>7 \cdot 31 = 217</math>. Subtract <math>9 \cdot 2</math> for reasons noted above to get <math>199</math>. Finally, this changes <math>\displaystyle 1 \Rightarrow -1</math>, so we have to add one to account for that. We get <math>200</math>.
  
 
== See also ==
 
== See also ==

Revision as of 08:23, 5 April 2007

Problem

Find the number of ordered triple $(a,b,c)$ where $a$, $b$, and $c$ are positive integers, $a$ is a factor of $b$, $a$ is a factor of $c$, and $a+b+c=100$.

Solution

Denote $x = \frac{b}{a}$ and $y = \frac{c}{a}$. The last condition reduces to $\displaystyle a(1 + x + y) = 100$. Therefore, $\displaystyle 1 + x + y$ is equal to one of the 9 factors of $\displaystyle 100 = 2^25^2$.

Subtracting the one, we see that $\displaystyle x + y = \{0,1,3,4,9,19,24,49,99\}$. There are exactly $\displaystyle n - 1$ ways to find pairs of $\displaystyle (x,y)$ if $\displaystyle x + y = n$. Thus, there are $\displaystyle 0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = 200$ solutions of $(a,b,c)$.

Alternatively, note that the sum of the divisors of $100$ is $(1 + 2 + 2^2)(1 + 5 + 5^2) \displaystyle$ (notice that after distributing, every divisor is accounted for). This evaluates to $7 \cdot 31 = 217$. Subtract $9 \cdot 2$ for reasons noted above to get $199$. Finally, this changes $\displaystyle 1 \Rightarrow -1$, so we have to add one to account for that. We get $200$.

See also

2007 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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