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# help number theory

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Compute the sum of all positive integers k for which the number k*1984*2 has exactly 21 positive divisors.

Mar 12, 2022

#1
+23198
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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.

Mar 12, 2022
#3
+2442
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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. )

BuilderBoi  Mar 12, 2022
edited by BuilderBoi  Mar 12, 2022
#2
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ສສສວງ

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Mar 12, 2022