kiran882 wrote:
Find the number of factors of a three digit even number xyz where x, y and z are distinct prime numbers (2, 3 and 5)
A. 8
B. 10
C. 12
D. 16
E. 24
Since xyz is an even number then it's \(532=2^2*7*19\) or \(352=2^5*11\).In either case the # of factors is 12:
For \(532=2^2*7*19\) the # of factors will be \((2+1)(1+1)(1+1)=12\);
For \(352=2^5*11\) the # of factors will be \((5+1)(1+1)=12\).
Answer: C.
In case one doesn't know.
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.
For more on number properties check:
math-number-theory-88376.htmlHope it helps.