Quarter Wit, Quarter Wisdom: A GMAT Quant Question That Troubles Many!

Quarter Wit, Quarter WisdomWhat determines whether or not a question can be considered a GMAT question? We know that GMAT questions that are based on seemingly basic concepts can be camouflaged such that they may “appear” to be very hard. Is it true that a question requiring a lot of intricate calculations will not be tested in GMAT? Yes, however it is certainly possible that a question may “appear” to involve a lot of calculations, but can actually be solved without any!

In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.

This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.

We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:

† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?

(A) 121
(B) 361
(C) 576
(D) 961
(E) 1,089

The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.

The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?

As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”.  Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.

For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:

Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?

Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:

58 = 50 + 8 = 10*5 + 8
27 = 20 + 7 = 10*2 + 7
…etc.

Along those same lines:

AB = 10A + B
BA = 10B + A

Going back to our original question:

(AB)^2 – (BA)^2
= (10A + B)^2 – (10B + A)^2
= (10A)^2 + B^2 + 2*10A*B – (10B)^2 – A^2 – 2*10B*A
= 99A^2 – 99B^2
= 9*11*(A^2 – B^2)

We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.

We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Number Properties on the GMAT – Part VI

Quarter Wit, Quarter WisdomMost people feel that the topic of number properties is hard or at least a little tricky. The reason is that no matter how much effort you put into it, you will still come across new concepts every time you sit with some 700+ level problems of this topic. There will be some concepts you don’t know and will need to “figure out” during the actual test. I came across one such question the other day. It brought forth a concept I hadn’t thought about before so I decided to share it today:

Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product?

Say N = 3
The numbers are 5, 6, 7 (any three consecutive numbers)
Their sum is 5 + 6 + 7 = 18
Their product is 5*6*7 = 210
Note that both the sum and the product are divisible by 3 (i.e. N).

Say N = 5
The numbers are 2, 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 2 + 3 + 4 + 5 + 6 = 20
Their product is 2*3*4*5*6 = 720
Again, note that both the sum and the product are divisible by 5 (i.e. N)

Say N = 4
The numbers are 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 3 + 4 + 5 + 6 = 18
Their product is 3*4*5*6 = 360
Now note that the sum is not divisible by 4, but the product is divisible by 4.

If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even.
Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form:

(Multiple of N),
(Multiple of N) +1,
(Multiple of N) + 2,
… ,
(Multiple of N) + (N-2),
(Multiple of N) + (N-1)

In our examples above, when N = 3, the numbers we picked were 5, 6, 7. They would be written in the form:

(Multiple of 3) + 2 = 5
(Multiple of 3)       = 6
(Multiple of 3) + 1 = 7

In our examples above, when N = 4, the numbers we picked were 3, 4, 5, 6. They would be written in the form:

(Multiple of 4) + 3 = 3
(Multiple of 4)        = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
etc.

What happens in case of odd integers? We have a multiple of N and an even number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will always add up in pairs to give the sum of N:

1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N

So when you add up all the integers, you will get a multiple of N.

What happens in case of even integers? You have a multiple of N and an odd number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will add up to give integers of N but one will be leftover:

1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N

The middle number will not have a pair to add up with to give N. So when you add up all the integers, the sum will not be a multiple of N.

For example, let’s reconsider the previous example in which we had four consecutive integers:

(Multiple of 4)      = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
(Multiple of 4) + 3 = 3

1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4.

Let’s now consider the product of N consecutive integers.

In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N.

Now take a quick look at the GMAT question that brought this concept into focus:

Which of the following must be true?
1) The sum of N consecutive integers is always divisible by N.
2) If N is even then the sum of N consecutive integers is divisible by N.
3) If N is odd then the sum of N consecutive integers is divisible by N.
4) The Product of K consecutive integers is divisible by K.
5) The product of K consecutive integers is divisible by K!

(A) 1, 4, 5
(B) 3, 4, 5
(C) 4 and 5
(D) 1, 2, 3, 4
(E) only 4

Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer.

This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true.

5) The product of K consecutive integers is divisible by K!

We will leave it to you to try to prove this!

(For more advanced number properties on the GMAT, check out Parts I, II, III, IV and V of this series.)

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!