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What is the smallest integer that has 21 positive factors?

 Dec 16, 2018
 #1
avatar+568 
+3

576, as for odd factors we need a square number.

I just tried doing it out.

 

You are very welcome!

:P

 Dec 16, 2018
edited by CoolStuffYT  Dec 16, 2018
 #2
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+4

In order for a number to have an ODD number of divisors, it must be a "Perfect Square", and the product of its exponents must be 21. So, the smallest such number must be: 2^6 x 3^2 =576. The product of its exponents is:

[6 + 1] x [2 + 1] =7 x 3 =21. So, the divisors of 576, which is a Perfect square are:

576 = 1 | 2 | 3 | 4 | 6 | 8 | 9 | 12 | 16 | 18 | 24 | 32 | 36 | 48 | 64 | 72 | 96 | 144 | 192 | 288 | 576 (21 divisors)

 Dec 16, 2018
 #3
avatar+3988 
+3

We know that 21 has factors of 1, 3, 7, 21. Remember, when you want to find the number of factors a number has, you just add one to each power, then multiply. Therefore, by picking 7 and 3, we should subtract one from each power, yielding us with 6 and 2. For it to be the smallest integer, we just plug in 2 with the 6 and 3 with the 2. Thus, the answer is \(2^6*3^2=64*9=\boxed{576}.\)

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 Dec 16, 2018

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