# Quantitative Section - GMAT Problem Solving

## GMAT Problem Solving Questions

Problem solving (PS) questions may not be new to you. You must have seen these types of questions in your high school or college days; they are essentially multiple-choice questions. The format is as follows: There is a question stem and is followed by five options, out of which, only one option is correct or is the best option that answers the question correctly.

PS questions measure your skill to solve numerical problems, interpret graphical data, and assess information. These questions present to you five options, and no option is phrased as "None of these". Mostly the numeric options, unlike algebraic expressions, are presented in ascending order from option A through E, occasionally in descending order until there is a specific purpose not to do so.

A typical Problem Solving question look like the following:

If $a12a$ is a four-digit number with the digit $a$ in both the thousands and units places, what is the value of $a$ if the number is divisible by $6$?

- 2
- 3
- 4
- 6
- 8

As stated earlier, this question format is generally comfortable since most of us have attempted many similar multiple-choice questions in our secondary school and university experience.

There can be more than one approach to solve a PS question. Though for many, these types of problems are approached with traditional methods, it is not necessary that you would do the same for every the GMAT PS question. Traditional approaches are certainly one possible option to solving any GMAT Problem Solving question. The key is to apply optimum approach to a question.

Note that the GMAT is not a test on your mathematical proficiency; rather, it tests your ability to efficiently handle mathematical data to make decisions. The GMAT does not test mathematical concepts beyond the high-school level.

Let's solve this question with three approaches. Note that none of the approaches may be the best one; it depends on you which of them clicks for you on the test day.

**Question**

If $a12a$ is a four-digit number with the digit $a$ in both the thousands and units places, what is the value of $a$ if the number is divisible by $6$?

- 2
- 3
- 4
- 6
- 8

*Approach 1:*

A number is divisible by '6' implies that it is divisible by both of its two prime factors: '2' and '3'.

Since $a12a$ is divisible by '2', the digit in the units place, i.e., $a$ must be even. Thus, the possible values of a are: 2, 4, 6 or 8 ($a$ cannot be '0' since the digit in the thousands place of a four-digit number cannot be '0').

Again, since the number is divisible by '3,' too, the sum of its digits must be divisible by '3,' too.

Sum of the digits of the number $a12a=a+1+2+a=2a+3$.

Thus, $(2a+3)$ must be divisible by '3'. Since the constant 3 in $(2a+3)$ is divisible by 3, $2a$ must also be divisible by 3, i.e., $a$ is divisible by 3.

Thus, checking with the possible values of $a$, i.e., 2, 4, 6 or 8, we see than only $a=6$ satisfies the above condition.

Hence, we have $a=6$.

The correct answer is Option D.

In this approach, we applied the concept of divisibility rules of '2' and '3' for the given number.

Let's see another approach.

*Approach 2:*

Note that if a number is divisible by an even number, here 6, the number must be even. Thus, $(2a+3)$ must be even; however, this does not help us much as we are able to eliminate only one choice: Option B = 3, an odd number.

Let's apply the plug-in value approach.

We start with Option A: $a=2$

Thus, $a12a=2122$. Diving 2,122 by 6 leaves a remainder of 4. Thus, 2,122 + 4 = 2,126 is divisible by 6; however, here the thousands digit and the units digit are not same. So, let's add 4 to our original answer, 2, and make the thousands digit equal to 6. We can then see whether 6,126 is divisible by 6. We see that it is divisible by 6; thus, $a=6$.

The correct answer is Option D.

In this approach, we applied the traditional method of division and at the second stage, we made a logical conclusion.

*Approach 3:*

We can rewrite $a12a$ as $a\times 1000+1\times 100+2\times 10+a\times 1=a000+100+20+a$

Let's divide $a000+100+20+a$ by $6$. We get $\frac{\left(a000\right)}{6}+\frac{100}{6}+\frac{20}{6}+\frac{a}{6}$.

Looking at $\frac{a}{6}$, we see that if $a=6$, it is divisible, so is $\frac{\left(a000\right)}{6}$; however, $\frac{100}{6}$ and $\frac{20}{6}$ are not. With a little thinking, you would observe that if we add 100 and 20, we get 120, which is divisible by $6$. So, $(a000+120+a)$ is divisible by $6$ if $a=6$.

The correct answer is Option D.

In this approach, we applied the concept of place value and at the second stage, we made a logical conclusion.