2013 AIME I Problems/Problem 15
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) | ||
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