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

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==Problem 15==
 
==Problem 15==
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Let <math>N</math> be the number of ordered triple <math>(A,B,C)</math> of integers satisfying the conditions (a) <math>0\le A<B<C\le99</math>, (b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>, (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and (d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</math>.
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Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions (a) <math>0\le A<B<C\le99</math>, (b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>, (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and (d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</math>.
  
 
==Solution==
 
==Solution==
 
From condition (d), we have <math>(A,B,C)=(B-\Delta,B,B+\Delta)</math> and <math>(b,a,c)=(a-\delta,a,a+\delta)</math>. Condition (c) states that <math>p|B-\Delta-a</math>, <math>p|B-a+\delta</math>, and <math>p|B+\Delta-a-\delta</math>. We subtract the first two to get <math>p|-\delta-\Delta</math>, and we do the same for the last two to get <math>p|2\delta-\Delta</math>. We subtract these two to get <math>p|3\delta</math>. So <math>p|3</math> or <math>p|\delta</math>. The second case is clearly impossible, because that would make <math>c=a+\delta>p</math>, violating condition (b). So we have <math>p|3</math>, meaning <math>p=3</math>. Condition (b) implies that <math>(b,a,c)=(0,1,2)</math>. Now we return to condition (c), which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us one solution. If <math>B=6</math>, we get two solutions. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives 16 solutions. <math>B=54</math> gives 15 solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=\boxed{272}</math>.
 
From condition (d), we have <math>(A,B,C)=(B-\Delta,B,B+\Delta)</math> and <math>(b,a,c)=(a-\delta,a,a+\delta)</math>. Condition (c) states that <math>p|B-\Delta-a</math>, <math>p|B-a+\delta</math>, and <math>p|B+\Delta-a-\delta</math>. We subtract the first two to get <math>p|-\delta-\Delta</math>, and we do the same for the last two to get <math>p|2\delta-\Delta</math>. We subtract these two to get <math>p|3\delta</math>. So <math>p|3</math> or <math>p|\delta</math>. The second case is clearly impossible, because that would make <math>c=a+\delta>p</math>, violating condition (b). So we have <math>p|3</math>, meaning <math>p=3</math>. Condition (b) implies that <math>(b,a,c)=(0,1,2)</math>. Now we return to condition (c), which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us one solution. If <math>B=6</math>, we get two solutions. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives 16 solutions. <math>B=54</math> gives 15 solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=\boxed{272}</math>.

Revision as of 21:21, 15 March 2013

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

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