Please, does someone can take a look at this problem?

Thanks.

didieravoaka wrote:Please, does someone can take a look at this problem?

Thanks.

If a and b are integers and (ab)^5 = 96y, y could be:

(A) 5
(B) 9
(C) 27
(D) 81
(E) 125

The prime factorization of 96 = 2^5 * 3. We can rewrite the original equation as

a^5 * b^5 = 2^5 * 3 * y.

If a and b are integers, we know that each prime base should be raised to a multiple of 5.

If y were 81, or 3^4, we'd have a^5 * b^5 = 2^5 * 3 * 3^4 or a^5 * b^5 = 2^5 * 3 ^5. Looks good to me. Answer is D.

Thanks David for your answer.

Please tell me, how to know that we should consider 81 as y?

didieravoaka wrote:Thanks David for your answer.

Please tell me, how to know that we should consider 81 as y?

Once we've gotten to this point: a^5 * b^5 = 2^5 * 3 * y, we know that each base must be raised to a multiple of 5. We already have 2^5, so the issue is 3. In order to get from 3 to 3^5, we need to multiply by 3^4. We can take the prime factorization of each answer choice to determine which one will donate those additional four 3's.

A) 5
B) 3^2
C) 3^3
D) 3^4
E) 5^3

Only D will donate the four 3's we need.

Another approach here:

(ab)� = 96y

a�b� = 2�*3*y

Since the left side has two fifth powers, the right side must also have two fifth powers. If 3y were a fifth power, we'd be set, so

3y = 3�

or

y = 81

is a valid solution. (There are others, but this is the simplest, and the answer is there, so we're set!)