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 $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$ . Condition $\text{(c)}$ states that $p\mid B-D-a$ , $p|B-a+d$ , and $p\mid B+D-a-d$ . We subtract the first two to get $p\mid-d-D$ , and we do the same for the last two to get $p\mid 2d-D$ . We subtract these two to get $p\mid 3d$ . So $p\mid 3$ or $p\mid d$ . The second case is clearly impossible, because that would make $c=a+d>p$ , violating condition $\text{(b)}$ . So we have $p\mid 3$ , meaning $p=3$ . Condition $\text{(b)}$ implies that $(b,a,c)=(0,1,2)$ or $(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3}). Now we return to condition$ \text{(c)} $, which now implies that$ (A,B,C)\equiv(-2,0,2)\pmod{3} $. Now, we set$ B=3k $for increasing positive integer values of$ k $.$ B=0 $yields no solutions.$ B=3 $gives$ (A,B,C)=(1,3,5) $, giving us$ 1 $solution. If$ B=6 $, we get$ 2 $solutions,$ (4,6,8) $and$ (1,6,11) $. Proceeding in the manner, we see that if$ B=48 $, we get 16 solutions. However,$ B=51 $still gives$ 16 $solutions because$ C_\text{max}=2B-1=101>100 $. Likewise,$ 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}=16\cdot 17=\boxed{272}$.

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2013 AIME I Problems/Problem 15

Problem 15

Solution

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