Thanks to help.
integers
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didieravoaka wrote:Please, does someone can take a look at this problem?
Thanks.
The prime factorization of 96 = 2^5 * 3. We can rewrite the original equation asIf a and b are integers and (ab)^5 = 96y, y could be:
(A) 5
(B) 9
(C) 27
(D) 81
(E) 125
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.
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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.didieravoaka wrote:Thanks David for your answer.
Please tell me, how to know that we should consider 81 as y?
A) 5
B) 3^2
C) 3^3
D) 3^4
E) 5^3
Only D will donate the four 3's we need.
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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!)
(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!)