Compute the sum of all positive integers k for which the number k*1984*2 has exactly 21 positive divisors.
+23252 k·1984·2 can be written as k·3968 = k·27·31
27·31 has these 16 factors:
1, 2, 22, 23, 24, 25, 26, 27,
31, 31·2, 31·22, 31·23, 31·24, 31·25, 31·26, and 31·27
If k is a number such as 3, another 16 factors will be created (3 times each of the previous 16 factors).
If k is 2, two more factors are created, 28 and 31·28, for a total of 18 factors.
If k is 4 = 22, another two factors are created: 29 and 31·29, for a total of 20.
If k is 8 = 23, another two factors are created: 210 and 31·210, for a total of 22.
If you don't exclude 1 as a factor, I don't know how to get exactly 21.
+2668 Here's another way to prove it's impossible:
\(k \times 1984 \times 2 = k \times 2^7 \times 31\)
Using this logic, you can create the factors: 1, 2, 4, 8, 16, 31, 32, 62, 64, 124, 128, 248, 496, 992, 1984, and 3968.
Because we want 21 factors, we need the number to be a perfect square. The smallest way to do this is by multiplying \(3968 \times 31 \times 2\). This yields 27 factors, way too big. Any bigger and you get more factors.
Thus, this problem is impossible, like geno said (Note: If this problem was just \(k \times 1984\), we could multiply by 31, and get 21 factors. )
