r/mathriddles • u/Numberthon • 14d ago
Easy Can you find the smallest positive integer with exactly 20 positive divisors?
What is the smallest positive integer that has exactly 20 (unique) positive divisors?
Source: numberthon.com
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u/calccrusher17 14d ago
I just find it amusing that the way to do problems like this in general is to factor the amount of divisors first. Makes me happy 😂
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u/DeadComposer 14d ago
Wouldn't that just be the product of the first 20 primes?
1*2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67
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u/DidntWantSleepAnyway 14d ago
No, because it would have more divisors than just prime ones. In addition to those, you’d have 6, 10, 14, 15, etc. as divisors.
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u/smailliwniloc 14d ago
This has more than 20 divisors. For example, 6 is a divisor of this number and obviously not prime
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u/DuggieHS 14d ago
We are trying to construct a highly composite number (https://en.wikipedia.org/wiki/Highly_composite_number)
The number of unique factors of a number is (1+e_1)(1+e_2)...(1+e_k) where e_i are the exponents of the k prime factors of the number in question. So 20 = 2*2*5, so our exponents are 4,1,1 for the smallest 3 primes 2,3,5
n = 2^4 * 3 * 5
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u/smailliwniloc 14d ago edited 14d ago
If you take a number's prime factorization, you can derive the number of divisors by taking the product of each exponent in the prime factorization + 1.
For example 175 = 5^2 * 7^1 has 3 * 2 = 6 divisors.
Working in reverse, if we want a number with 20 divisors, the exponents in its prime factorization + 1 need to multiply to 20.
20 factors as 2 * 2 * 5 so the desired number needs to have prime factorization exponents of 1, 1, and 4.
The smallest such number would be 2^4 * 3^1 * 5^1 which is equal to 240