The sum of 1006 different positive integers is 1019057. If none of them is
greater than 2012, what is the minimum number of these integers which must be
odd?
【Solution】
Suppose we take the first 1006 even numbers. Then their sum is 1006 (2 2012)
2
× +
=
1013042. Now 1019057 − 1013042 = 6015. So we have to trade some even numbers
for an equal amount of odd numbers. Clearly, trading only one number can raise the total by at most 2011−2=2009. Hence we have to trade more than one number.
However, if we trade exactly two numbers, the total will increase by an even amount,
which is not what we want. Hence we must trade at least three numbers. If we trade 2,
4 and 6 for 2007, 2009 and 2011, the total will increase by 2007+2009+2011−2−4−6
=3×(2009−4)=6015 exactly. Hence the minimum number of the 1006 integers which
must be odd is 3.
ANS: 3