2013 AIME I Problems/Problem 15

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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 $N$.


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, $N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=\boxed{272}$.

See also

2013 AIME I (ProblemsAnswer KeyResources)
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