Difference between revisions of "2013 AIME I Problems/Problem 15"
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− | From condition (d), we have <math>(A,B,C)=(B- | + | 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}$. |
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2013|n=I|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:32, 12 February 2016
Problem 15
Let be the number of ordered triples
of integers satisfying the conditions (a)
, (b) there exist integers
,
, and
, and prime
where
, (c)
divides
,
, and
, and (d) each ordered triple
and each ordered triple
form arithmetic sequences. Find
.
Solution
From condition (d), we have and
. Condition
states that
,
, and
. We subtract the first two to get
, and we do the same for the last two to get
. We subtract these two to get
. So
or
. The second case is clearly impossible, because that would make
, violating condition
. So we have
, meaning
. Condition
implies that
or
\text{(c)}
(A,B,C)\equiv(-2,0,2)\pmod{3}
B=3k
k
B=0
B=3
(A,B,C)=(1,3,5)
1
B=6
2
(4,6,8)
(1,6,11)
B=48
B=51
16
C_\text{max}=2B-1=101>100
B=54
15
B=96
B=99
N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}$.
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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