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# What is the value of product abc?

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## Global Stats

Difficult

What is the value of product abc?

(1) 2^a * 3^b * 5^c = 1728
(2) a, b, and c are nonnegative integers

OA C

Source: Veritas Prep

### GMAT/MBA Expert

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BTGmoderatorDC wrote:
What is the value of product abc?

(1) 2^a * 3^b * 5^c = 1728
(2) a, b, and c are nonnegative integers

OA C

Source: Veritas Prep
We have to get the value of abc.

Let's take each statement one by one.

(1) 2^a * 3^b * 5^c = 1728.

Multiple solutions are possible. Note that we do not know that a, b and c are integers. They can be real numbers.

For example, at a = b = 0, we have 5^c = 1728 => the value of c is unique, of course it will be a real number; thus abc = 0. In the same way, at a = b = 1, we have 2*3*5^c = 1728 => 5^c = 288 => the value of c is unique; thus abc is some positive value. No unqiue value of abc. Insufficient.

(2) a, b, and c are nonnegative integers.

Certainly insufficient since the relationship among a, b and c is not known,

(1) and (2) together

So, we have 2^a * 3^b * 5^c = 1728

2^a * 3^b * 5^c = 2^6 * 3^3 * 5^c

Since a, b and c are nonnegative integers, we must have c = 0, a = 6 and b = 3; thus, abc = 6*3*0 = 0. Sufficient.

Hope this helps!

-Jay
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### GMAT/MBA Expert

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BTGmoderatorDC wrote:
What is the value of product abc?

(1) 2^a * 3^b * 5^c = 1728
(2) a, b, and c are nonnegative integers

Source: Veritas Prep
$? = abc$
$\left( 1 \right)\,\,\,{2^a} \cdot {3^b} \cdot {5^c} = 1728\,\,\,\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {6,3,0} \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {1728 = {2^6} \cdot {3^3}} \right] \hfill \\ \,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {{x_{\text{p}}},1,1} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,? = {x_{\text{p}}} > 0\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} \right.$
$\left( * \right)\,\,\,{2^x} = \frac{{1728}}{{15}}\,\,\,\,\, \Rightarrow \,\,\,\,x = {x_p} > 0\,\,\,{\text{unique}}\,\,\,\,\,\,\left( {{\text{see}}\,\,{\text{image}}\,\,{\text{attached}}} \right)$

$\left( 2 \right)\,\,a,b,c\,\,\, \geqslant 0\,\,\,\,{\text{ints}}\,\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {0,0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0 \hfill \\ \,\,Take\,\,\left( {a,b,c} \right) = \left( {1,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1 \hfill \\ \end{gathered} \right.$

$\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered} {2^a} \cdot {3^b} \cdot {5^c} = 1728 = {2^6} \cdot {3^3} \cdot {5^0} \hfill \\ \,a,b,c\,\,\, \geqslant 0\,\,\,\,{\text{ints}} \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {a,b,c} \right) = \left( {6,3,0} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,? = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\,\,\,\,\,\,\,$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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