# The Patterns to Solve GMAT Questions with Reversed-Digit Numbers – Part II

, I wrote about the GMAT’s tendency to ask questions regarding the number properties of two two-digit numbers whose tens and units digits have been reversed.

The biggest takeaways from that post were:

1. Anytime we add two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 11.
2. Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.

For the hardest GMAT questions, we’re typically mixing and matching different types of number properties and strategies, so it can be instructive to see how the above axioms might be incorporated into such problems.

Take this challenging Data Sufficiency question, for instance:

When the digits of two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?

(1) The integer (M –N) has 12 unique factors.

(2) The integer (M –N) is a multiple of 9.

The average test-taker looks at Statement 1, sees that it will be very difficult to simply pick numbers that satisfy this condition, and concludes that this can’t possibly be enough information. Well, the average test-taker also scores in the mid-500’s, so that’s not how we want to think.

First, let’s concede that Statement 1 is a challenging one to evaluate and look at Statement 2 first. Notice that Statement 2 tells us something we already know – as we saw above, anytime you have two two-digit numbers whose tens and units digits are reversed, the difference will be a multiple of 9. If Statement 2 is useless, we can immediately prune our decision tree of possible correct answers. Either Statement 1 alone is sufficient, or the statements together are not sufficient, as Statement 2 will contribute nothing. So right off the bat, the only possible correct answers are A and E.

If we had to guess, and we recognize that the average test-taker would likely conclude that Statement 1 couldn’t be sufficient, we’d want to go in the opposite direction – this question is significantly more difficult (and interesting) if it turns out that Statement 1 gives us considerably more information than it initially seems.

In order to evaluate Statement 1, it’s helpful to understand the following shortcut for how to determine the total number of factors for a given number. Say, for example, that we wished to determine how many factors 1000 has. We could, if we were sufficiently masochistic, simply list them out (1 and 1000, 2 and 500, etc.). But you can see that this process would be very difficult and time-consuming.

Alternatively, we could do the following. First, take the prime factorization of 1000. 1000 = 10^3, so the prime factorization is 2^3 * 5^3. Next, we take the exponent of each prime base and add one to it. Last, we multiply the results. (3+1)*(3+1) = 16, so 1000 has 16 total factors. More abstractly, if your number is x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

Now let’s apply this process to Statement 1. Imagine that the difference of M and N comes out to some two-digit number that can be expressed as x^a * y^b. If we have a total of 12 factors, then we know that (a+1)(b+1) = 12. So, for example, it would work if a = 3 and b = 2, as a + 1 = 4 and b + 1 = 3, and 4*3 =12. But it would also work if, say, a = 5 and b = 1, as a + 1 = 6 and b + 1 = 2, and 6*2 = 12. So, let’s list out some numbers that have 12 factors:

1. 2^3 * 3^2 (3+1)(2+1) = 12
2. 2^5 * 3^1 (5+1)(1+1) = 12
3. 2^2 * 3^3 (2+1)(3+1) = 12

Now remember that M – N, by definition, is a multiple of 9, which will have at least 3^2 in its prime factorization. So the second option is no longer a candidate, as its prime factorization contains only one 3. Also recall that we’re talking about the difference of two two-digit numbers. 2^2 * 3^3 is 4*27 or 108. But the difference between two positive two-digit numbers can’t possibly be a three-digit number! So the third option is also out.

The only possibility is the first option. If we know that the difference of the two numbers is 2^3 * 3^2, or 8*9 = 72, then only 91 and 19 will work. So Statement 1 alone is sufficient to answer this question, and the answer is A.

Algebraically, if M = 10x + y, then N = 10y + x.

M – N = (10x + y) – (10y + x) = 9x – 9y = 9(x – y).

If 9(x – y) = 72, then x – y = 8. If the difference between the tens and units digits is 8, the numbers must be 91 and 19.

Takeaway: the hardest GMAT questions will require a balance of strategy and knowledge. In this case, we want to remember the following:

• Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.
• If one statement is easier to evaluate than the other, tackle the easier one first. If it’s the case that one statement gives you absolutely nothing, and the other is complex, there is a general tendency for the complex statement alone to be sufficient.
• For the number x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTubeGoogle+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

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

What 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!

# Quarter Wit, Quarter Wisdom: Try to Answer This GMAT Challenge Question!

Today, we will give you a GMAT challenge question. The challenge of reviewing this question is not that the question is hard to understand – it is that you will need to solve this official question within a minute using minimum calculations.

Let’s take a look at the question stem:

 Date of Transaction Type of Transaction June 11 Withdrawal of \$350 June 16 Withdrawal of \$500 June 21 Deposit of x dollars

For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was \$1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was \$1,000, what was the amount of the deposit on June 21?

(A) \$1,000
(B) \$1,150
(C) \$1,200
(D) \$1,450
(E) \$1,600

Think about how you might answer this question:

The average of daily balances = (Balance at the end of June 1 + Balance at the end of June 2 + … + Balance at the end of June 30) / 30 = 1000

Now we have been given the only three transactions that took place:

• A withdrawal of \$350 on June 11 – so on June 11, the account balance goes down to \$650.
• A withdrawal of \$500 on June 16 – so on June 16, the account balance goes down to \$150.
• A deposit of \$x on June 21 – So on June 21, the account balance goes up to 150 + x.

Now we can plug in these numbers to say the average of daily balances = [1000 + 1000 + …(for 10 days, from June 1 to June 10) + 650 + 650 + … (for 5 days, from June 11 to June 15) +  150 + … (for 5 days, from June 16 to June 20) + (150 + x) + (150 + x) + … (for 10 days, from June 21 to June 30)] / 30 = 1000

One might then end up doing this calculation to find the value of x:

[(1000 * 10) + (650 * 5) + (150 * 5) + ((150 + x) * 10)] / 30 = 1000
x = \$1,450
The answer is D.

But this calculation is rather tedious and time consuming. Can’t we use the deviation method ? After all, we are dealing with large values here! How?

Note that we are talking about the average of certain data values. Also, we know the deviations from those data values:

• The amount from June 11 to June 30 is 350 less.
• The amount from June 16 to June 30 is another 500 less.
• The amount from June 21 to June 30 is x in excess.

Through the deviation method, we can see the shortfall = the excess:

350 * 20 + 500 * 15 = x * 10
x = 1,450 (D)

This simplifies our calculation dramatically! Though saving only one minute on a question like this may not seem like a very big deal, saving a minute on every question by using a more efficient method could be the difference between a good Quant score and a great Quant score!

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!