USA Mathematical Talent Search
Solutions to Problem 5/4/18
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5/4/18. A sequence of positive integers (x1 , x2 , . . . , x2007 ) satisfies the following two con-
ditions:
(1) xn 6= xn+1 for 1 n 2006, and
x1 + x2 + + xn
(2) An = is an integer for 1 n 2007.
n
Find the minimum possible value of A2007 .
Credit This problem was a former proposal for the Canadian Mathematical Olympiad.
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Comments Finding the optimal sequence is not difficut, but a high degree of rigor and
careful reasoning must be employed to show conclusively that you have the minimum value.
In particular, using a greedy algorithm is not sufficient. Both conditions (1) and (2) must
be used effectively. Solutions edited by Naoki Sato.
Solution 1 by: Gaku Liu (11/FL)
We claim that the minimum value of An is n+1
2
. This value is achieved for the sequence
n+1
for odd n,
2
xn =
3n
for even n.
2
Indeed, if n 2 is even, then xn1 = n/2 and xn+1 = (n + 2)/2, both of which are less
than xn = 3n/2. Hence, no two consecutive terms are equal, so condition (1) is satisfied.
For even n,
(x1 + x3 + + xn1 ) + (x2 + x4 + + xn )
An =
n
(1 + 2 + + n/2) + (3 + 6 + + n/2)
=
n
4(1 + 2 + + n/2)
=
n
4 n/2 (n + 2)/2
=
2n
n+2 n+1
= = ,
2 2
USA Mathematical Talent Search
Solutions to Problem 5/4/18
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and for odd n,
(x1 + x3 + + xn2 ) + (x2 + x4 + + xn1 ) + xn
An =
n
[1 + 2 + + (n 1)/2] + [3 + 6 + + 3(n 1)/2] + (n + 1)/2
=
n
4[1 + 2 + + (n 1)/2] + (n + 1)/2
=
n
4 1/2 (n 1)/2 (n + 1)/2 + (n + 1)/2
=
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n+1 n+1
= = .
2 2
We now prove this is the minimum through induction. It is true for n = 1, because the
minimum of A1 is 1 = 1+1 2
. For n = 2, if A2 = 1, then x1 + x2 = 2 x1 = x2 = 1, which
2+1
contradicts (1). Hence, the minimum of A2 is 2 = 2 .
Now, assume that A2m 2m+1
2
= m + 1 for some positive integer m. Let Sn =
x1 + x2 + + xn . In particular, Sn must be a multiple of n. We have S2m = 2mA2m
2m(m + 1) = 2m2 + 2m. Also,
2m2 + m < 2m2 + 2m < 2m2 + 3m + 1
m(2m + 1) < 2m2 + 2m < (m + 1)(2m + 1),
so the least multiple of 2m + 1 greater than 2m2 + 2m is (m + 1)(2m + 1). Since S2m+1 >
S2m 2m2 + 2m, we have S2m+1 (m + 1)(2m + 1), so
(2m + 1) + 1
A2m+1 m + 1 = .
2
Note that 2m2 + 2m = m(2m + 2) is a multiple of 2m + 2. The next greatest multiple of
2m + 2 is (m + 1)(2m + 2). Suppose that S2m+2 = (m + 1)(2m + 2) = 2m2 + 4m + 2. Then
2m2 + 3m + 1 < 2m2 + 4m + 2 < 2m2 + 5m + 2
(m + 1)(2m + 1) < 2m2 + 4m + 2 < (m + 2)(2m + 1),
so the greatest multiple of 2m + 1 less than 2m2 + 4m + 2 is (m + 1)(2m + 1). Since
S2m+1 < S2m+2 = 2m2 + 4m + 2, we have S2m+1 (m + 1)(2m + 1). But we have already
shown that S2m+1 (m + 1)(2m + 1), so S2m+1 = (m + 1)(2m + 1).
Also,
2m2 + 2m < 2m2 + 3m + 1 < 2m2 + 4m
(m + 1)2m < 2m2 + 3m + 1 < (m + 2)2m,
USA Mathematical Talent Search
Solutions to Problem 5/4/18
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so 2m2 + 2m is the greatest multiple of 2m less than S2m+1 . Since S2m < S2m+1 = (m +
1)(2m + 1), we have S2m (m + 1)2m. But S2m (m + 1)2m, so S2m = (m + 1)2m. Then
x2m+1 = S2m+1 S2m = (m+1)(2m+1)(m+1)2m = m+1, and x2m+2 = S2m+2 S2m+1 =
(m + 1)(2m + 2) (m + 1)(2m + 1) = m + 1, which contradicts (1).
Hence, S2m+2 (m + 2)(2m + 2), so
(2m + 2) + 1
A2m+1 m+2= ,
2
2007+1
completing the induction. Therefore, the minimum value of A2007 is
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2
= 1004.
