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Difference between revisions of "2013 AIME I Problems/Problem 15"

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So the answer is <math>17 + 15 + \cdots + 1 = \boxed{272}</math>
 
So the answer is <math>17 + 15 + \cdots + 1 = \boxed{272}</math>
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~SoilMilk
 
~SoilMilk
  

Latest revision as of 22:14, 26 May 2023

Problem

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


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{align} \label{1} &A-a = B-x-a \\ \label{2} &B - b = B-a+y \\ \label{3} &C - c = B+x-a-y \end{align}

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{align*} &A-a = B-x-1 \hspace{4cm}(4)\\ &B - b = B \hspace{5.35cm}(5)\\ &C - c = B+x-2 \hspace{4cm}(6) \end{align*} 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 $x=2$, $B=3, 6, ..., 96 \Rightarrow 17$ triples.

When $x=5$, $B=6, 9, ..., 93\Rightarrow 15$ triples..

... and so on until

When $x=44$, $B=45\Rightarrow 1$ triple.

So the answer is $17 + 15 + \cdots + 1 = \boxed{272}$

~SoilMilk

See also

2013 AIME I (Problems • Answer Key • Resources)
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Problem 14 		Followed by
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