AIME 2013 I · 第 15 题

Solution

From condition (d), we have $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$.

Now, we substitute these values in. 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, $N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}$.

Solution 2

Let $(A, B, C)$ = $(B-x, B, B+x)$ and $(b, a, c) = (a-y, a, a+y)$. Now the 3 differences would be

$$\begin{aligned} \label{1} &A-a = B-x-a \\ \label{2} &B - b = B-a+y \\ \label{3} &C - c = B+x-a-y \end{aligned}$$ Adding equations $(1)$ and $(3)$ would give $2B - 2a - y$. Then doubling equation $(2)$ would give $2B - 2a + 2y$. The difference between them would be $3y$. Since $p|\{(1), (2), (3)\}$, then $p|3y$. Since $p$ is prime, $p|3$ or $p|y$. However, since $p > y$, we must have $p|3$, which means $p=3$.

If $p=3$, the only possible values of $(b, a, c)$ are $(0, 1, 2)$. Plugging this into our differences, we get

$$\begin{aligned} &A-a = B-x-1 \hspace{4cm}(4)\\ &B - b = B \hspace{5.35cm}(5)\\ &C - c = B+x-2 \hspace{4cm}(6) \end{aligned}$$ The difference between $(4)$ and $(5)$ is $x+1$, which should be divisible by 3. So $x \equiv 2 \mod 3$. Also note that since $3|(5)$, $3|B$. Now we can try different values of $x$ and $B$:

When $x=2$, $B=3, 6, ..., 96 \Rightarrow 32$ triples.

When $x=5$, $B=6, 9, ..., 93\Rightarrow 30$ triples..

... and so on until

When $x=47$, $B=48, 51\Rightarrow 2$ triple.

So the answer is $32 + 30 + \cdots + 2 = \boxed{272}$

~SoilMilk