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

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==Solution==
 
==Solution==
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From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p|B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3}). Now we return to condition </math>\text{(c)}<math>, which now implies that </math>(A,B,C)\equiv(-2,0,2)\pmod{3}<math>. Now, we set </math>B=3k<math> for increasing positive 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 </math>1<math> solution. If </math>B=6<math>, we get </math>2<math> solutions, </math>(4,6,8)<math> and </math>(1,6,11)<math>. Proceeding in the manner, we see that if </math>B=48<math>, we get 16 solutions. However, </math>B=51<math> still gives </math>16<math> solutions because </math>C_\text{max}=2B-1=101>100<math>. Likewise, </math>B=54<math> gives </math>15<math> 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}=16\cdot 17=\boxed{272}$.
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From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p|B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive 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 <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> 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}=16\cdot 17=\boxed{272}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2013|n=I|num-b=14|after=Last Problem}}
 
{{AIME box|year=2013|n=I|num-b=14|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:33, 12 February 2016

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

See also

2013 AIME I (Problems • Answer Key • Resources)
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