Difference between revisions of "2013 AIME I Problems/Problem 15"
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− | ==Problem | + | ==Problem == |
− | 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>. | + | 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-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>. | + | 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>. |
− | == Solution 2 == | + | |
− | + | ==Solution 2== | |
+ | Let <math>(A, B, C)</math> = <math>(B-x, B, B+x)</math> and <math>(b, a, c) = (a-y, a, a+y)</math>. Now the 3 differences would be | ||
+ | <cmath>\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}</cmath> | ||
+ | |||
+ | Adding equations <math>(1)</math> and <math>(3)</math> would give <math>2B - 2a - y</math>. Then doubling equation <math>(2)</math> would give <math>2B - 2a + 2y</math>. The difference between them would be <math>3y</math>. Since <math>p|\{(1), (2), (3)\}</math>, then <math>p|3y</math>. Since <math>p</math> is prime, <math>p|3</math> or <math>p|y</math>. However, since <math>p > y</math>, we must have <math>p|3</math>, which means <math>p=3</math>. | ||
+ | |||
+ | |||
+ | If <math>p=3</math>, the only possible values of <math>(b, a, c)</math> are <math>(0, 1, 2)</math>. Plugging this into our differences, we get | ||
+ | <cmath>\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*}</cmath> | ||
+ | The difference between <math>(4)</math> and <math>(5)</math> is <math>x+1</math>, which should be divisible by 3. So <math>x \equiv 2 \mod 3</math>. Also note that since <math>3|(5)</math>, <math>3|B</math>. Now we can try different values of <math>x</math> and <math>B</math>: | ||
+ | |||
+ | When <math>x=2</math>, <math>B=3, 6, ..., 96 \Rightarrow 17</math> triples. | ||
+ | |||
+ | When <math>x=5</math>, <math>B=6, 9, ..., 93\Rightarrow 15</math> triples.. | ||
+ | |||
+ | ... and so on until | ||
+ | |||
+ | When <math>x=44</math>, <math>B=45\Rightarrow 1</math> triple. | ||
+ | |||
+ | So the answer is <math>17 + 15 + \cdots + 1 = \boxed{272}</math> | ||
+ | |||
+ | ~SoilMilk | ||
== 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}} |
Latest revision as of 22:14, 26 May 2023
Contents
Problem
Let be the number of ordered triples of integers satisfying the conditions (a) , (b) there exist integers , , and , and prime where , (c) divides , , and , and (d) each ordered triple and each ordered triple form arithmetic sequences. Find .
Solution
From condition (d), we have and . Condition states that , , and . We subtract the first two to get , and we do the same for the last two to get . We subtract these two to get . So or . The second case is clearly impossible, because that would make , violating condition . So we have , meaning . Condition implies that or . Now we return to condition , which now implies that . Now, we set for increasing positive integer values of . yields no solutions. gives , giving us solution. If , we get solutions, and . Proceeding in the manner, we see that if , we get 16 solutions. However, still gives solutions because . Likewise, gives solutions. This continues until gives one solution. gives no solution. Thus, .
Solution 2
Let = and . Now the 3 differences would be
Adding equations and would give . Then doubling equation would give . The difference between them would be . Since , then . Since is prime, or . However, since , we must have , which means .
If , the only possible values of are . Plugging this into our differences, we get
The difference between and is , which should be divisible by 3. So . Also note that since , . Now we can try different values of and :
When , triples.
When , triples..
... and so on until
When , triple.
So the answer is
~SoilMilk
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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