Is the 5-digit positive integer abc000 divisible by 24?

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[Math Revolution GMAT math practice question]

Is the 5-digit positive integer abc000 divisible by 24?

1) The 3-digit integer abc is divisible by 8.
2) The 3-digit integer abc is divisible by 3.

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by fskilnik@GMATH » Thu Nov 15, 2018 9:45 am

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Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is the 6-digit positive integer abc000 divisible by 24?

1) The 3-digit integer abc is divisible by 8.
2) The 3-digit integer abc is divisible by 3.
$$\frac{{\left\langle {abc000} \right\rangle }}{{3 \cdot 8}}\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\frac{{\,1000 \cdot \left\langle {abc} \right\rangle }}{{3 \cdot 8}}\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\frac{{125 \cdot \left\langle {abc} \right\rangle }}{3}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\boxed{\,\,\,\frac{{\left\langle {abc} \right\rangle }}{3}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,\,\,\,\left( {a \ne {\text{0}}} \right)\,\,}$$
$$\left( 1 \right)\,\,\,{{\left\langle {abc} \right\rangle } \over 8}\,\, = \,\,\,{\mathop{\rm int}} \,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left\langle {abc} \right\rangle = \left\langle {160} \right\rangle \,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\left[ {\sum\nolimits_{3\,\,{\rm{digits}}} { = 7\,\,{\rm{not}}\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,3} } \right] \hfill \cr
\,{\rm{Take}}\,\,\left\langle {abc} \right\rangle = \left\langle {168} \right\rangle \,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\,\left[ {\sum\nolimits_{3\,\,{\rm{digits}}} { = 15\,\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,3} } \right]\,\, \hfill \cr} \right.\,$$
$$\left( 2 \right)\,\,\,{{\left\langle {abc} \right\rangle } \over 3}\,\, = \,\,\,{\mathop{\rm int}} \,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

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

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Fabio.
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by Jay@ManhattanReview » Thu Nov 15, 2018 10:22 pm

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Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is the 6-digit positive integer abc000 divisible by 24?

1) The 3-digit integer abc is divisible by 8.
2) The 3-digit integer abc is divisible by 3.
We have to ascertain whether abc000 is divisible by 24.

Factors of 24 = 3*8 = 3*2^3

So, if abc000 is divisible by 24, it must be divisible by 3 as well as 2^3.

Let's first check the rule for the divisibility of a number by 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Thus, for abc000 to be divisible by 3, (a + b + c) must be divisible 3. This is what Statement 2 states. However, we still need to check whether abc000 is divisible by 2^3.

> A number is divisible by 2 if its last digit is divisible 2.
> A number is divisible by 2^2 if the number formed out of the last two digits is divisible 2^2.
> A number is divisible by 2^3 if the number formed out of the last three digits is divisible 2^3.

We are interested in the last one.

So, the number abc000 is divisible by 2^3 is the number formed out of the last three digits, i.e., 000 is divisible 2^3. We see that 000 is divisible by 2^3. Thus, there is no need of Statement 1.

The correct answer: B

Hope this helps!

-Jay
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by Max@Math Revolution » Sun Nov 18, 2018 5:05 pm

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Since the last three digits 000 is a multiple of 8 and abc000 is a multiple of 8, the question "is abc000 divisible by 24?" is equivalent to "is abc000 divisible by 3?" or "is a + b + c divisible by 3?".
Thus, condition 2) is sufficient.

Condition 1) is not sufficient as 8 is not divisible by 3.

Therefore, the correct answer is B.
Answer: B