How to Calculate Divisors of a Number with Recurring Factors

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To calculate the divisors of the number 378, which factors into 2^1, 3^3, and 7^1, the method involves considering the distinct powers of each prime factor. Each prime factor's exponent is increased by one and then multiplied together: (1+1) for 2, (3+1) for 3, and (1+1) for 7. This results in the calculation 2 * 4 * 2, yielding a total of 16 divisors. The recurring factor of 3 is accounted for by its exponent in the formula. Understanding this method allows for accurate divisor calculations for numbers with recurring factors.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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