How to Calculate Divisors of a Number with Recurring Factors

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SUMMARY

The discussion focuses on calculating the total number of divisors for the number 378, which factors into 2, 3, and 7. The prime factorization is expressed as 378 = 21 x 33 x 71. The total number of divisors is determined using the formula (e1 + 1)(e2 + 1)(e3 + 1), where e1, e2, and e3 are the exponents of the prime factors. For 378, this results in (1 + 1)(3 + 1)(1 + 1) = 2 * 4 * 2 = 16 divisors.

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MarekS
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Hello!

There's a combination exercise that has been bewildering me for some time now: how many divisors does the number 378 have?

I know it can be done like this: 378=2x3x3x3x7.
Divisors are as follows: 1,2,3,6,7,9,12,14,18,21,42,54,63,126,189,378. 16 all together.

But, in essence, it has to be a combinations task, the only catch is that the element 3 is recurrent.

Is there a formula that takes this aspect into account?

MarekS
 
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Lets say you have three "pools" amongst which you can draw your factors. Let one contain 2^0 and 2^1; one 3^0, 3^1, 3^2, and 3^3; and the last 7^0 and 7^1. Each distinct divisor is made by choosing one number from each pool. So it can easily be seen that the number of total divisors is 2 * 4 * 2 = 16.
 

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