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Question 1199569: How many positive integers n < 2022 are there for which the sum of the odd positive divisors of n is 24?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
If the sum of the odd positive divisors is 24, then the number of those divisors must be even.
The odd positive integer 1 is a divisor of every integer, so the number of odd positive divisors other than 1 must be odd.
If there are only 2 odd divisors and the sum of the odd divisors is 24, then those odd divisors must be 1 and 23.
If there are 4 odd divisors, the largest of them must be the product of the other two. There is only one set of odd divisors that satisfies that condition and has a sum of 24 -- 1, 3, 5, and 15.
Obviously, the only other prime factors must be even; that means the only other prime factors are 2.
With the two odd prime factors 1 and 23, any number less than 2022 of the form 23(2^n) satisfies the conditions of the problem. 2022/23 is about 87.9, which is between 2^6=64 and 2^7=128, so the number of factors of 2 is between 0 and 6, inclusive. That gives 7 numbers that satisfy the conditions.
With the four odd prime factors 1, 3, 5, and 15, any number less than 2022 of the form 15(2^n) satisfies the conditions of the problem. 2022/15 is about 134.8, which is between 2^7=128 and 2^8=256, so the number of factors of 2 is between 0 and 7, inclusive. That gives 8 other numbers that satisfy the conditions.
ANSWER: the number of positive integers less than 2022 for which the sum of the odd positive divisors is 24 is 7+8 = 15.
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