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, 2010

Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

For what minimum integer number $N$ one can change "/" to "+" and "-" in the following expression: $1 \ast 2 \ast 3 \ast \dots \ast N$ in order to get:
a) $2010$ ; b) $2011$ ?

Solution

a) If we have all $+$ , then: $1 + 2 + 3 + \dots + 62 = 1953 < 2010$ , $1 + 2 + 3 + \dots + 63 = 2016 > 2011$ . Hence, $N \ge 63$ . From the other hand, $N = 63$ will do: $1 + 2 - 3 + \dots + 62 + 63 = 2016 - 6 = 2010$ .

b) $N = 63, 64$ are not the solutions, because change of sign before any number does not change the parity of the expression itself and $1+2+3+\dots+63 = 2016$ and $1+2+3+\dots+64$ are even. Hence, $N \ge 65$ . Now it's easy to locate signs. Since $1+2+3+\dots+65 = 2145$ , then $-2 \cdot (2+65)+1+2+3+\dots+65 = 2145-134=2011$ .

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