2013 AIME I Problems/Problem 15

Problem 15

Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions (a) $0\le A<B<C\le99$ , (b) there exist integers $a$ , $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ , (c) $p$ divides $A-a$ , $B-b$ , and $C-c$ , and (d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$ .

Solution

From condition (d), we have $(A,B,C)=(B-\Delta,B,B+\Delta)$ and $(b,a,c)=(a-\delta,a,a+\delta)$ . Condition (c) states that $p|B-\Delta-a$ , $p|B-a+\delta$ , and $p|B+\Delta-a-\delta$ . We subtract the first two to get $p|-\delta-\Delta$ , and we do the same for the last two to get $p|2\delta-\Delta$ . We subtract these two to get $p|3\delta$ . So $p|3$ or $p|\delta$ . The second case is clearly impossible, because that would make $c=a+\delta>p$ , violating condition (b). So we have $p|3$ , meaning $p=3$ . Condition (b) implies that $(b,a,c)=(0,1,2)$ . Now we return to condition (c), which now implies that $(A,B,C)\equiv(-2,0,2)\pmod{3}$ . Now, we set $B=3k$ for increasing integer values of $k$ . $B=0$ yields no solutions. $B=3$ gives $(A,B,C)=(1,3,5)$ , giving us one solution. If $B=6$ , we get two solutions. Proceeding in the manner, we see that if $B=48$ , we get 16 solutions. However, $B=51$ still gives 16 solutions. $B=54$ gives 15 solutions. This continues until $B=96$ gives one solution. $B=99$ gives no solution. Thus, $N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=\boxed{272}$ .

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

Page

Toolbox

Search

2013 AIME I Problems/Problem 15

Problem 15

Solution

See also