9514 1404 393
Answer:
55
Step-by-step explanation:
Here's one way to count them.
12 consecutive integers will have 3 each of the integers whose value is 0, 1, 2, or 3 mod 4. To make a total that is 0 mod 4, any of these combinations of mod 4 values may be used:
000 -- 1 combination
013 -- 27 combinations, 3 each of mod 0, mod 1, and mod 3
022 -- 9 combinations, 3 of mod 0, and 3 ways to choose 2 of mod 2
112 -- 9 combinations, 3 of mod 2, and 3 ways to choose 2 of mod 1
233 -- 9 combinations, 3 of mod 2, and 3 ways to choose 2 of mod 3
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Consider the set of 12 consecutive integers 1 .. 12. Then here are the 9 "233" combinations. Mod 2 integers are 2, 6, 10, and mod 3 integers are 3, 7, 11.
(2, 3, 7), (2, 3, 11), (2, 7, 11),
(3, 6, 7), (3, 6, 11), (6, 7, 11),
(3, 7, 10), (3, 10, 11), (7, 10, 11)
The total number of combinations that will be divisible by 4 is ...
1 + 27 + 9 + 9 + 9 = 55 . . . . combinations with a sum divisible by 4
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<em>Additional comment</em>
The number of ways that 3 can be chosen from 12 is 220, so this is 1/4 of that total. It is perhaps not surprising that the remainders from division by 4 are uniformly distributed among the possibilities.
A computer program to list the 220 subsets and the mod 4 remainders of their totals confirms this is the correct number.