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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 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>.
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Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions  
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(a) <math>0\le A<B<C\le99</math>,  
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(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>,  
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(c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and  
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(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 1==
 
==Solution 1==

Revision as of 20:06, 25 October 2020

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 1

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

Condition c gives us that $A \equiv a \pmod p$, etc. Condition d then tells us that C and c can be expressed as $2B - A$ and $2a - b$, respectively. However, plugging what we got from condition c into this, we find that $3a \equiv 3b \pmod p$. From there, we branch off into two cases; either $p = 3$, or $a \equiv b \pmod p$. Realize then that the second case leads to a contradiction, due to condition b. Then, $p = 3$ means that $(b,a,c)$ must be $(0, 1, 2)$. The bash from there is pretty similar to what was done in Solution 1. We get $\boxed{272}$. - Spacesam

See also

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