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

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== Problem ==
 
== Problem ==
Find the number of ordered triple <math>(a,b,c)</math> where <math>a</math>, <math>b</math>, and <math>c</math> are positive [[integer]]s, <math>a</math> is a [[factor]] of <math>b</math>, <math>a</math> is a factor of <math>c</math>, and <math>a+b+c=100</math>.
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Find the number of ordered triples <math>(a,b,c)</math> where <math>a</math>, <math>b</math>, and <math>c</math> are positive [[integer]]s, <math>a</math> is a [[factor]] of <math>b</math>, <math>a</math> is a factor of <math>c</math>, and <math>a+b+c=100</math>.
  
 
== Solution ==
 
== Solution ==

Revision as of 20:16, 30 July 2009

Problem

Find the number of ordered triples $(a,b,c)$ 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 (ProblemsAnswer KeyResources)
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