salsal wrote:
If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?
(a) 10
(b) 11
(c) 12
(d) 13
(e) 14
My approach:
Let's take the four primes: 2, 3, 5 and 7 --> each of the 4 primes will be a factor of the product
All the possible combinations of the four primes will be factors (10 possible combinations):
2x3
2x5
2x7
3x5
3x7
5x7
2x3x5
2x3x7
2x5x7
3x5x7
Thus, in total there are 10 + 4 = 14 factors.
Is my reasoning correct? Is there a faster way to figure out the number of possible combinations than listing them?
No that;s not correct.
Finding the Number of Factors of an IntegerFirst 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 QUESTION:We are told that
x=abcd, where a, b, c, and d are four distinct prime numbers. According to the above the # of factors of x including 1 and x is (1+1)(1+1)(1+1)(1+1)=16. Excluding 1 and x the # of factors is 16-2=14.
Answer: E.
Hope it's clear.
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