How many even divisors of 1600 are not multiples of 16?
We can use two approaches to this problem. The first one is less time consuming but requires close familiarity with number properties. The second one is a manual method and thus more time consuming but still brings us to the correct answer. Memorizing the second is easier and thus helpful when we forget the first
FIRST APPROACH: We need to prime factorize 1600: \(1600 = 16*100 = 2^6*5^2\)
Overall number of factors 1600 have is the product of powers increased by 1: \((6+1)(2+1)=21\)
The number of Odd factors is found by removing \(2^6\) becuase it will make factors even. So we will have \(5^2\) which has \((2+1) = 3\) odd factors.
The number of Even factors \(= All factors - Odd factors = 21 - 3 = 18\). So 1600 have 18 even and 3 odd factors. Next, how many of 18 even factors are NOT divisible by \(16\) or \(2^4\)?
For a number NOT to be divisible by \(2^4\), it must have at most \(2^3\) as a prime factor. Hence, \(2^3*5^2\) will give us the number of factors not divisible by \(2^4\).
So
All factors not devisible \(= (3+1)*(2+1)=12\). We need to remove 3 odd factors: \(12-3=9\). Thus 9 even factors of 1600 are not divisible by 16.
SECOND APPROACH: For this manual method we again need to prime factorize 1600: \(1600 = 16*100 = 2^6*5^2\)
So even factors not devisibe by 16 would be the product of 2's and 5's we have when the highest possible power of 2 is 3.
Let's manually find those factors:
\(2\)
\(2^2\)
\(2^3\)
\(2*5\)
\(2^2*5\)
\(2^3*5\)
\(2*5^2\)
\(2^2*5^2\)
\(2^3*5^2\)
Overall 9.
Hence
C