2000 AMC 10 Problems/Problem 14

Revision as of 17:54, 7 January 2009 by BOGTRO (talk | contribs) (New page: 71, 76, 80, 82, 91. The sum of the first 2 must be even, so we must choose 2 evens or the 2 odds. Let us look at the numbers (mod 3). 2,1,2,1,1. If we choose the two odds, the next num...)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

71, 76, 80, 82, 91.

The sum of the first 2 must be even, so we must choose 2 evens or the 2 odds.

Let us look at the numbers (mod 3).

2,1,2,1,1.

If we choose the two odds, the next number must be a multiple of 3, of which there is none.

Similarly, if we choose 76,80 or 80,82, the next number must be a multiple of 3, of which there is none.

So we choose 76,82 first.

The next number must be 1 (mod 3), of which only 91 remains.

The sum of these is 249. This is equal to 1 (mod 4).

Thus, we need to choose one number that is 3 (mod 4). 71 is the only one that works.

Thus, 80 is the last score entered.

C.

Invalid username
Login to AoPS