Among the numbers from 1 to 1984, there are 992 even numbers. Each of them contributes at least one factor of 2 to the prime factorization of 1984!. Two factors of 2 will be contributed by numbers divisible by 4 (there are 496 such numbers). Further, 3, 4, 5, 6, 7, 8, 9, and 10 factors of 2 will be contributed by 248, 124, 62, 31, 15, 7, 3, and 1 numbers, respectively, which are divisible by 8, 16, 32, 64, 128, 256, 512, and 1024, respectively. By summing these results, we obtain the desired number of twos: $992+496+248+124+62+31+15+7+3+1=1979$.

## Answer

1979 twos.
How many twos will be in the prime factorization of the number 1984! ?

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