Bunuel
If y is a positive integer, is (y^4+7)^2 divisible by 4?
(1) √y has exactly three Prime Factors
(2) All the prime Factors of y^7 are greater than 3
If a number is divisible by 4 then it will be an even number.(1) √y has exactly three Prime Factors\(\sqrt y\) is square of a prime, as only square of primes have total 3 factors.
If \(\sqrt y =4 \) then \(y = 16 \)
In this case \((y^4+7)^2\) turns out be odd and is
NOT divisble by \(4 \)
If \(\sqrt y =9 \) then \(y = 81 \)
\((81^4+7)^2 = (3^{16}+7)^2\) . Little calculation shows that this is divisble by \(4 \)
INSUFF.(2) All the prime Factors of \(y^7 \) are greater than \(3\)All prime numbers greater than \(3\) are odd,So \(y\) will be odd and the expression \((y^4+7)^2 \) will be of the form:
\((3^4+7)^2 =\hspace{2mm}\)Divisible by \(4 \)
\((5^4+7)^2 =\hspace{2mm}\)Divisible by \(4\)
\((7^4 +7)^2=\hspace{2mm}\)Divisible by \(4\)
\((11^4 +7)^2=\)Divisible by \(4\)
etc.
Note: One needs to only calculate the last two digits of the above expression to determine whether it's divisble by \(4.\)
Hence we can definitely answer yes to the question
SUFF.Ans B
Hope it's clear