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

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== Solution ==
 
== Solution ==
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Denote <math>x = \frac{b}{a}</math> and <math>y = \frac{c}{a}</math>. The last condition reduces to <math>\displaystyle a(1 + x + y) = 100</math>. Therefore, <math>\displaystyle 1 + x + y</math> is equal to one of the 9 factors of <math>\displaystyle 100 = 2^25^2</math>.
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Denote <math>x = \frac{b}{a}</math> and <math>y = \frac{c}{a}</math>. The last condition reduces to <math>a(1 + x + y) = 100</math>. Therefore, <math>1 + x + y</math> is equal to one of the 9 factors of <math>100 = 2^25^2</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>.
+
Subtracting the one, we see that <math>x + y = \{0,1,3,4,9,19,24,49,99\}</math>. There are exactly <math>n - 1</math> ways to find pairs of <math>(x,y)</math> if <math>x + y = n</math>. Thus, there are <math>0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}</math> solutions of <math>(a,b,c)</math>.
  
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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>.
+
Alternatively, note that the sum of the divisors of <math>100</math> is <math>(1 + 2 + 2^2)(1 + 5 + 5^2)</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>1 \Rightarrow -1</math>, so we have to add one to account for that. We get <math>\boxed{200}</math>.
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 +
== Video Solution by OmegaLearn ==
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https://youtu.be/LqrXinQbk1Q?t=73
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 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 07:41, 4 November 2022

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 $a(1 + x + y) = 100$. Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$.

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

Alternatively, note that the sum of the divisors of $100$ is $(1 + 2 + 2^2)(1 + 5 + 5^2)$ (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 $1 \Rightarrow -1$, so we have to add one to account for that. We get $\boxed{200}$.

Video Solution by OmegaLearn

https://youtu.be/LqrXinQbk1Q?t=73

~ pi_is_3.14

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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