Difference between revisions of "2013 AIME I Problems/Problem 15"
(→Solution 1) |
Hashtagmath (talk | contribs) |
||
Line 1: | Line 1: | ||
− | ==Problem | + | ==Problem == |
Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions | Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions | ||
(a) <math>0\le A<B<C\le99</math>, | (a) <math>0\le A<B<C\le99</math>, |
Latest revision as of 19:14, 24 January 2021
Contents
Problem
Let be the number of ordered triples of integers satisfying the conditions (a) , (b) there exist integers , , and , and prime where , (c) divides , , and , and (d) each ordered triple and each ordered triple form arithmetic sequences. Find .
Solution 1
From condition (d), we have and . Condition states that , , and . We subtract the first two to get , and we do the same for the last two to get . We subtract these two to get . So or . The second case is clearly impossible, because that would make , violating condition . So we have , meaning . Condition implies that or . Now we return to condition , which now implies that . Now, we set for increasing positive integer values of . yields no solutions. gives , giving us solution. If , we get solutions, and . Proceeding in the manner, we see that if , we get 16 solutions. However, still gives solutions because . Likewise, gives solutions. This continues until gives one solution. gives no solution. Thus, .
Solution 2
Condition c gives us that , etc. Condition d then tells us that C and c can be expressed as and , respectively. However, plugging what we got from condition c into this, we find that . From there, we branch off into two cases; either , or . Realize then that the second case leads to a contradiction, due to condition b. Then, means that must be . The bash from there is pretty similar to what was done in Solution 1. We get . - Spacesam
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.