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I thought of this as a combination problem with a duplicate choice 4! / 2 !

4! because I have to put the numbers in order A*B*C*D

2! because I have a duplicate choice (3)

Can anyone tell me if this will work for other situations? or did I just confuse myself and everyone else?

Alternatively, I write these questions out starting with the largest and smallest factor (I am using the .... to show that I write them a good distance apart on my scratch paper.) 1 .............................................................................................................. 90 next 1, 2,...........................................................................45, 90 1, 2, 3 30, 45, 90 1, 2, 3, 5, 18, 30, 45, 90 If find that doing things this way prevents me from having duplicates and quickly gets me to a situation where I know I have covered all the possibilities. This method is really only good for number less than 100. For example if you had 87452, it might take too much time to do this.

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Back to the original question:

How many distinct integers are factors of 90?

A. 6 B. 8 C. 9 D. 10 E. 12

\(90=2*3^2*5\), which means that the number of factors of 90 is: \((1+1)(2+1)(1+1)=12\).

To know the number of number of factors for a number , we need to split the number in the following way

Number = a^x * b^y * c^z .......so on

The number of factors would be (x+1)*(y+1)*(z+1)...so on

Applying the above to the current example

90 = 2^1 * 3^2 * 5^1

The number of factors = (1+1) * (2+1) * (1+1) = 2*3*2 = 12

Please send kudos if you like this

This is definitely a more methodical approach to the question. But it requires double the work. Since you anyways have to prime factor number N, I found it easier to factorize the non-prime factor further until it is represented as prime. For example, 90 = 45 x 2. Then 45 = 15 x 3. 15 = 5 x 3 and so on. At the end it is a simple math of counting the factors!
_________________

-DK --------------------------------------------------------- If you like what you read then give a Kudos! Diagnostic Test: 620 The past is a guidepost, not a hitching post. ---------------------------------------------------------

To know the number of number of factors for a number , we need to split the number in the following way

Number = a^x * b^y * c^z .......so on

The number of factors would be (x+1)*(y+1)*(z+1)...so on

Applying the above to the current example

90 = 2^1 * 3^2 * 5^1

The number of factors = (1+1) * (2+1) * (1+1) = 2*3*2 = 12

Please send kudos if you like this

This is very useful if you have to determine the number of factors of very large numbers (kudos!). Finding each factor by trial-and-error would cost way to much time. Try 2496: prime factorization is much easier (2^5)*3*13 -> (5+1)*(1+1)*(1+1)=24

I got the correct answer, (E): 12 however I do have one question, do negative integers not qualify as "distinct" factors? Meaning the answer would be 12 positive distinct factors + 12 negative distinct factors = 24 distinct factors in total?