Difference between revisions of "2007 AIME II Problems/Problem 2"
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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>. | 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>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 = 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>. |
− | 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>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>. |
== See also == | == See also == |
Latest revision as of 12:15, 29 July 2017
Problem
Find the number of ordered triples where , , and are positive integers, is a factor of , is a factor of , and .
Solution
Denote and . The last condition reduces to . Therefore, is equal to one of the 9 factors of .
Subtracting the one, we see that . There are exactly ways to find pairs of if . Thus, there are solutions of .
Alternatively, note that the sum of the divisors of is (notice that after distributing, every divisor is accounted for). This evaluates to . Subtract for reasons noted above to get . Finally, this changes , so we have to add one to account for that. We get .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.