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by [email protected] » Sat Nov 04, 2017 11:33 am
Hi LUANDATO,

You should notice that all of the 'bases' are PRIME numbers, so this question is all about 'prime factorization.' To start, we have to prime factor 270,000,000 down into its 'pieces'...

(270)(1,000)(1,000) =

270 = (3)(3)(3)(2)(5)
1,000 = (2)(2)(2)(5)(5)(5)

Thus 270,000,000 = (2^7)(3^3)(5^7)

The product of those three exponents is (7)(3)(7) = 147

Final Answer: B

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by EconomistGMATTutor » Sat Nov 04, 2017 2:17 pm
What is the product of integers a, b, and c if
$$2^a\ \cdot\ 3^b\ \cdot\ 5^c\ =\ 270,000,000$$

A. 141
B. 147
C. 162
D. 235
E. 270

The OA is B.

Please, can any expert assist me with this PS question? I need help to solve it. Thanks.
Hi LUANDATO,
Lets take a look at your question.

$$2^a\ \cdot\ 3^b\ \cdot\ 5^c\ =\ 270,000,000$$
$$2^a.3^b.5^c=27\times10,000,000$$ $$2^a.3^b.5^c=\left(3\times3\times3\right)\times\left(2\times5\right)\times\left(2\times5\right)\times\left(2\times5\right)\times\left(2\times5\right)\times\left(2\times5\right)\times\left(2\times5\right)\times\left(2\times5\right)$$
$$2^a.3^b.5^c=3^3\times2^7\times5^7$$
$$a=3,\ b=7,\ c=7$$
$$abc=3\times7\times7$$
$$abc=147$$

Therefore, option B is correct.

I am available if you'd like any follow up.
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by Jeff@TargetTestPrep » Tue Nov 14, 2017 6:06 am
LUANDATO wrote:What is the product of integers a, b, and c if
$$2^a\ \cdot\ 3^b\ \cdot\ 5^c\ =\ 270,000,000$$

A. 141
B. 147
C. 162
D. 235
E. 270
Let's break 270,000,000 into its prime factors.

270,000,000 = 27 x 10,000,000 = 3^3 x 10^7 = 2^7 x 3^3 x 5^7, so a x b x c = 7 x 3 x 7 = 147.

Answer: B

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