What is the value of product abc?

1728=(2^6)X(3^3)X(5^0)
=> c=0 by equating the powers of like terms on both sides.

So what ever the value of a & b, the product with c will be zero.

The statement that neither of the nos a,b,c are non negative doesnt give any solution.

So i feel the answer is A

Please correct me if I am wrong.

I chose A, as 1728 is 2^6*3^3 .

Please explain the OA.

It should be C.
(1) does not give me enough information. There are different variations of a,b,c (integers, non-integers, lagorithmus, positive and negative values) that will satisfy the equations. insufficient

(2) alone insufficient.

Combining (1) and (2) we have now the information that supplement statement (1). If we know that a,b,c are non-negative integers, we could then say that the only possible values are 6,3 and 0 (1728=2^6*3^3*5^0) and the value of a*b*c=0
Hence C

1) 2^a * 3^b * 5^c = 1728 => 2^6 * 3^3 * 5^0 . Therefore, abc = 6*3*0 =0 .
So, 1) should be sufficient.

2) a, b, c are non negative . i.e they can be positive or 0. However, this could lead to multiple values for abc. So, 2) is not sufficient.

Answer is A, 1) ALONE is sufficient

A for me.
2^a * 3^b * 5^c = 1728 would mean that the factor of 5 does not contribute to the multiplication (mutiples of 5 end with 5 or 0). Hence the value of c = 0. By knowing this, I can say that the value of abc = 0.

From statement 1 we can factor out 1728=2^6 *3^3 *5^1

Hey everyone - sorry...yesterday's question was formatted as "competition mode" and I can't figure out how to undo it, so I don't think anyone has seen my response yet. Here it is:

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Thanks for the responses, everyone! I love this question because of the strategy it brings up, which we call:

Why Are You Here?

Regarding statement 2, it's nowhere close to being sufficient on its own. So there are two likely reasons that it's there:

1) To trick you into thinking that you need it, and therefore picking C instead of A
2) To add information that IS, in fact, necessary to go with statement 1, so that the correct answer is C and not A

The GMAT doesn't use "red herring" statements - those that are simply so far out of scope that they're not even relevant - very often at all; if they provide a statement in a Data Sufficiency problem there has to be a reason...it's either a trap or it's necessary information. The good news for you is that you can use either case the same way: look at that statement to determine whether you really need it.

Here, although statement 1 may seem sufficient on its own (c must be 0 in order to make the 5 term equal to 1, since 1728 has no multiples of 5), that only fits if we know that they're all integers. Statement 2, by providing us with that information explicitly (they're nonnegative integers), should make us pause to think about statement 1: Do they have to be integers?

We don't need to use logarithms on the GMAT (thankfully!), but we should know enough that there would conceivably exist a set of noninteger exponents that would solve this problem. Even if you just assume that a and b are 1 so that:

5^c = 288

There is some value for c that will get us 288, so we can prove that statement 1 is not on its own sufficient. We need statement 2's help to determine that they're integers, so the correct answer is C.


So, strategically, when you see a statement that on its own is clearly not sufficient, ask yourself "why are you here?". Is it providing essential information, or is it there to make you think you need it?

I picked the answer as A and now I figure out what was my mistake.

But still, its very difficult to think that in the expression 2^a * 3^b * 5^c = 1728, even with a,b and c taking non integer value we can get the product,1728.

hi brian

so basically we should never assume that the numbers r integers if it is not given... also i need some explanation for statement one.... how is a,b,c being integers matter?... if they were fractions... how would they give us that number 1728?

When a, b, c are integers, and we have first statement as well, we need to put c=0 because 1728 has no 5 in it, and we have to make 5^c as 1. Even if they are fractions there can be two cases,

Case 1: They can be rational numbers/fractions, in which case the information coupled with statement 1 will be sufficient.

Case 2: They can be irrational numbers/fractions. In this case it won't suffice.

Hope it makes sense!!!

Hi,

Find value of abc

Statement 1: \(2^a * 3^b * 5^c = 1728\)
\(1728 = 2^6 * 3^3 * 5^0\)

Therefore, \(2^a * 3^b * 5^c = 2^6 * 3^3 * 5^0\)

a, b and c can have values: 6, 3 and 0 respectively.
or
if we take values for a and b to be 5, 3, we get: \(5^c = 2\) =>\(c=\frac{log 2}{log5}\)
if we take values for a and b to be 6, 2, we get: \(5^c = 3\) =>\(c= \frac{log 3}{log5}\)

Multiple solutions exist. So not sufficient

Statement 2: a, b, and c are nonnegative integers

Not sufficient

Combining both Statements,

we get, a = 6, b= 3 and c= 0

Hence,
Answer is C

\(? = abc\)

\(\left( 1 \right)\,\,\,{2^a} \cdot {3^b} \cdot {5^c} = 1728\,\,\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {6,3,0} \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {1728 = {2^6} \cdot {3^3}} \right] \hfill \\\\
\,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {{x_{\text{p}}},1,1} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,? = {x_{\text{p}}} > 0\,\,\,\,\,\,\,\,\,\, \hfill \\ \\
\end{gathered} \right.\)

\(\left( * \right)\,\,\,{2^x} = \frac{{1728}}{{15}}\,\,\,\,\, \Rightarrow \,\,\,\,x = {x_p} > 0\,\,\,{\text{unique}}\,\,\,\,\,\,\left( {{\text{see}}\,\,{\text{image}}\,\,{\text{attached}}} \right)\)


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

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


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

Regards,
Fabio.

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