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# If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is GMAT Explanation, Video Solution, and More Practice!

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

A. 6
B. 12
C. 24
D. 36
E. 48

We get this one a lot in GMAT tutoring sessions. Most people fail to understand what the questions is asking and then just start working without a plan. It ends up being that work itself isn't totally off but the student doesn't really know where to take it in order to narrow down the answer choices.

What also tends to confuse is that the question asks for the largest positive integer but we end up choosing the smallest possible value of n.

## Define the question

the largest positive integer that must divide n

There it is. But let's not leave it like that. Always try to do something with the questions. In this case because we're looking for something that divides evenly into n let's set up a fraction:

n/z = integer

And we want to maximize z because we're looking for the largest integer that must divide n.

The largest integer that divides any integer is itself. So really we're looking for n. Not that important to make that inference but just wanted to point it out.

## Setup

Now let's gather the information from the question and get things set up so we can make some inferences. We know that n is a positive integer and then n squared is divisible by 72. We can write out an equation with that second piece of information.

n^2/72 = integer

We're really looking to solve for n so let's go ahead and simplify this equation.

n^2 = integer*72

Take the square root of both sides.

n = √72√integer

Now let's pull out perfect squares from 72.

n = √9√4√2√integer

n = 6√2√integer

Now we can use the first piece of information that n is a positive integer. So 6√2√integer is a positive integer. Somehow the radicals have to disappear. So √2•√integer must be an integer. What's an easy way to do that? Make integer equal 2 so you have √2•√2 = 2.

n = 6*2 = 12

What's the largest positive integer that must divide n? 12

Now, you might be thinking: is 12 the only possibility for n? Or put another way, is 2 the only possible value for the integer? Good question. No. There are an infinite number of possible values for the integer and consequently for n. Any number that cancels out the radicals will work.

√2√8 = √2√2√4 = 4*6 = 24

√2√18 = √2√2√9 = 6*6 = 36

√2√32 = √2√2√16 = 8*6 = 48

Any number that has √2•perfect square will work.

So why is the integer 2 and the correct n 12? This is coming back to what tends to confuse GMAT tutoring students (looking for the largest integer that must divide but then the answer is actually the smallest possible value of n).

Because the question is a MUST. So you need the most basic building block of n. Look at all of the values we came up with for n: 12, 24, 36, 48. What's the biggest integer they have in common (least common multiple)? 12

Regardless of which multiple of n you come up with it will always be divisible by 12.

## Additional GMAT Divisibility Practice Question

Here's another divisibility question from GMAT Official Guide: If x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200, then x must be a multiple of which of the following?

Here's a tricky exponents divisibility puzzle from GMAT question of the day

And another from the GMAT official guide that's not the same but has a similar puzzle/exponents/divisibility vibe with factorials in the mix: If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?

# If n = (33)^43 + (43)^33 what is the units digit of n? GMAT Explanation, Video Solution, and More Practice!

If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0
B. 2
C. 4
D. 6
E. 8

Units digits questions always come up in tutoring. They're great because they're very formulaic (unlike a lot of the GMAT). I remember really struggling for time on an official GMAT I was taking and then question 31 was: What's the units digit of 8^52? Blasted through that one is 20 seconds and: 770! I was psyched! Anyways, rest assured that GMAT units digits questions are easy.

Every units digit has a multiplication pattern with the longest patterns being only 4 variations long. So let's look at some units digit patterns and then we'll apply what we learn to the question at hand. With each of these we're just raising the power to get to the next number in the pattern:

1: Always ends in 1

2: 2, 4, 8, 6 (2^1, 2^2, 2^3, 2^4). Then you can see that 2^5 goes back to "2".

3: 3, 9, 7, 1

4: 4, 6

5: Always ends in 5

6: Always ends in 6

7: 7, 9, 3, 1

8: 8, 4, 2, 6

9: 9, 1

0: Always ends in 0

You don't have to memorize these! They are easy enough to derive in the moment. An important note when creating units digit patterns: you don't have to multiply the entire number to yield the units. You just need to multiply the units digits. So 37*37, no need to multiple the whole thing. Just take 7*7 and you know that number ends in 9.

Ok, easy right? Let's get back to: If n = (33)^43 + (43)^33 what is the units digit of n?

Let's figure out the units for each one and then, guess what, you just add those units digits. Adding and subtracting units digits is also do-able without having to deal with the entire number.

You might be wondering how we're going to get to the 43rd number in the pattern for (33)^43. Well, that's just a question of divisibility. The units is 3 so we have a pattern of 4 numbers. How many 4's fit into 43? 10 with 3 leftover. The leftover is the important part.

3

9

7

1

That 3 means we end in 7. There are 40 complete patterns of 4 and then 3, 9, 7 gets you to 43. Put another way, 7 is the 43rd number in the pattern.

Same thing for 33. There are 32 complete patterns of 4 with 1 leftover. So that leaves a units digit of 3.

You can test this all out by just writing out 3, 9, 7, 1 a bunch and counting to the 33rd and 43rd terms. You will find 1 and 7 respectively. But again, easy way to figure out where you are is to divide the exponent by the number of numbers in the pattern.

Then just add the units to yield the units of the sum of the numbers: 7 + 3 = 10. So the units digit of n is 0.

## Additional GMAT Units Digit Practice Questions

Here is a very similar units digit question from GMAT Question of the Day but it has an important twist. Make sure you follow through with all of you basic math rules.

This GMAT question of the day puzzle goes beyond digits into powers of 10 but is a great one to be familiar with as I've seen this style on a multiple official GMATs.

Here's a digits Data Sufficiency Question from GMAT Question of the Day It's a bit more involved but good practice.

# The product of all the prime numbers less than 20 is closest to which of the following powers of 10? GMAT Explanation, Video Solution, and More Practice!

The product of all the prime numbers less than 20 is closest to which of the following powers of 10?

(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 10^5

I really like this question. The type of thinking required for success here is exactly what you need to succeed overall on the GMAT and what we try to reinforce each and every tutoring session.

Prime numbers less than 20

Let's list out those numbers. Take your time. Double check. This isn't a race. Or at least it isn't a sprint. You'll end up saving time by doing work in a careful and considered way.

2, 3, 5, 7, 11, 13, 17, 19

Just a quick note on primes. 2 is the only even prime and the smallest prime. Important thing to have memorize for the GMAT.

Ok, so we're looking to match a power of 10 with the product of these numbers. A power of 10 is 10*10*10... some number of times. So let's figure out a way to translate the product of the those primes to powers of 10. Start easy. You don't have to figure out everything all at once.

2*5 = 10. Great, we've got one 10.

3*7 = 21 which is basically 2*10

11 is basically 10

13 is a bit of a stretch but let's call it 10 for now and judge whether it's OK once we've got the first round done.

17 let's call 20 and maybe it balances out 13.

19 let's call 20 and it balances out 11.

So we've got 10*10*10*10*10*10*2*2*2*2. So 10^6*8. That's closest to 10^7.

Now, I get that you might be skeptical of the approximation BUT we did a reasonable approximation and 10^6*8 is way closer to 10^7 than 10^6. It's not close. In GMAT land I think that's good enough and I'd pick C and move on.

There is an underlying principle that proves this without a doubt that's worth knowing.

Consider:

2*100 = 200

2*101 = 202

3*100 = 300

When we added 1 to 100 we moved from 200 to 202.

When we added 1 to 2 we moved from 200 to 300.

So adding a fixed value to smaller number creates a bigger change. That makes sense since the fixed value increases the smaller by a greater percent than it increases the bigger number.

Back to our question!

2*5 = 10

3*7

11 (-1) vs 19 (+1) Given the above idea which move has more influence? The smaller one, 11.

13 (-3) vs 17 (+3) Same thing here. 13 has more influence.

So with those two we 100% underestimated. And then with 7*3 we also went down to 20. So it's clear that our approximation was even smaller than the actual number guaranteeing that the product of all the primes less than 20 is closest to 10^7.

Just a quick note beyond this question. So we showed above that when adding a fixed value to a number the magnitude of the number you're adding to changes the affect of the number you add (adding a fixed value to 2 will have more impact on 2 than on a larger number say 100000).

Multiplication is different. Multiplying something by 2 doubles it's value regardless of the original magnitude (2*2 is 4, 2*100 is 200). Of course you know that. But it's important on the GMAT to put this all together in a meaningful way. We call it "absolute vs relative" and it comes up a decent amount especially on the GMAT Data Sufficiency. It tends to be that relative info (multiplication) which stays constant regardless of the magnitude of the underlying numbers is usually more helpful than absolute info, addition/subtraction.

## Additional GMAT Puzzle Question Practice

Not the same but requires similar GMAT-puzzle-thinking. Here's a GMAT Question of the Day puzzle dealing with exponents and primes

Another similar but different puzzle question. This GMAT Question of the Day Exponents Puzzle deals more with constraints but again still tests the same type of organization skills that lead to success in the above primes question.

# (0.99999999/1.0001) − (0.99999991)/1.0003 GMAT Explanation, Video Solution, and Additional Practice!

(0.99999999/1.0001) − (0.99999991)/(1.0003) =

(A) 10^(−8)
(B) 3∗10^(−8)
(C) 3∗10^(−4)
(D) 2∗10^(−4)
(E) 10^(−4)

Give it a shot on your own and then you can reference the explanation and video solution. Good luck!

This is a tough GMAT exponents puzzle from the GMAT Official Guide. I love seeing these because:

1. The question type is easy to see. This .99999999 or 1.0000001... sticks out.
2. Once you know the question you're half done! And the follow through on these is somewhat formulaic.

If you follow our explanations or are a tutoring student then you know that we're always striving to start by defining the question. In this case though there's nothing to define. Also, there's not much of a distinction between setting the question up and solving it.

That said, we're going to call the initial inference the "setup" and then the number crunching/simplifying the equation the "solve".

## Setup

First off, let's change the format of these numbers. It's going to be a challenge working with .99999999 as is. You're not going to be able to divide .99999999/1.00001 without a calculator and the GMAT isn't going to expect you to do any brutal arithmetic. Again, let's change the format. The row of 9's should make you think about 0's. Why? Because a long string of 0's becomes a long string of 9's once you subtract from those zeroes (because of borrowing). GMAT loves to use this property. Now let's consider what we need to subtract from the nearest integer to create this .99999999. The nearest integer is 1. So we need to think 1 - ? = .99999999

Well, you guessed it, to create a 9 at the end you need to subtract a 1 at the end. You can always gives yourself a simple example to test. Think 1000 - ? = 999. 1000 - 1 = 999. So again we just need a 1 at the end. That's it. Easy.  Now translate that to our situation. How do we get a 1 at the end? 1.00000000. We need a one 8 places back so all of those zeroes turn to 9's. So 1*10^-8. The -8 sends the 1 8 places back. Be comfortable with powers of 10, including negative powers, as you'll be using them all over the test.

Next up, 1.0001. GMAT tutoring students usually find this one easier. 1 + 10^-4.

The second fraction we've got .99999991. Same thing but with a 1 at the end. So think simple: 1000 - ? = 991. Yep, 9. So instead of 1 8 places back we need a 9. 1 - 9^10-8. And, 1.0003 = 1 + 3*10^-3.

## Solve

Finally all together:

(1 - 10^-8)/(1 + 10^-4) - (1 - 9*10^-3)/(1 + 3*10^-4)

What's next? You might be thinking that this still looks complicated. Let's look for clues!

We've got exponents and subtraction. We also have perfect squares.

1 is a tricky perfect square everyone forgets. Even exponents always create perfect squares. That should get you thinking about difference of squares. Foggy on that concept? Dig in. You'll need to know difference of squares backwards and forwards to have a good chance at succeeding on your GMAT.

So 1 - 10^-8 = (1 - 10^-4)(1 + 10^-4)/(1 + 10^-4)

The entire first fraction looks like this: (1 - 10^-4)(1 + 10^-4)/(1 + 10^-4). We can cross out (1 + 10^-4) to leave (1 - 10^-4).

Let's do the same thing with the second fraction remembering that we also take the root of the coefficients, in this case 9.

(1 - 9*10^-3) = (1 + 3*10^-4)(1 - 3*10^-4).

The entire thing: (1 + 3*10^-4)(1 - 3*10^-4)/(1 + 3*10^-4). We can cross out (1 + 3*10^-4). That leaves (1 - 3*10^-4).

Putting things together: (1 - 10^-4) - (1 - 3*10^-4)

Remember to distribute the negative sign.

1 - 10^-4 - 1 + 3*10^-4

Cross out the ones and re-arrange

3*10^-4 - 10^-4

You could factor out 10^-4 but GMAT workflow-wise you're almost always better off recognizing like terms and adding or subtracting them. In this case the like term is 10^-4. We have 3 of them minus 1 of them so that leaves 2*10^-4. If that's confusing just sub in the variable x for 10^-4. 3x - x = 2x. Then sub 10^-4 back in.

Again, these are great to get on your GMAT. Most other people will get this one wrong but learn the above and you'll have a solid shot at blasting through a tough GMAT question. Here's a diagram with the complete solution. This was done on the online GMAT whiteboard to give you sense for how that works. You can see me work through it on the whiteboard in the video solution and here's a link to our suggestions for using the GMAT Online Whiteboard and whether to take the Online GMAT.

## Additional GMAT Exponents Puzzle Example Questions

Similar but different exponents puzzle from GMAT Question of the Day You need to think about the digits in the same way.

Here is more difference of squares practice from the Official GMAT Practice Tests: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of , where b is an integer?

# If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p? GMAT Explanation

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?

A. 10

B. 12

C. 14

D. 16

E. 18

Here's a GMAT factorials puzzle from the GMAT Official Guide. I always tell GMAT tutoring students that this is a fantastic questions to get on a test, even though, at least initially, most people get it wrong. Why is it a good one? Once you know how to solve this puppy:

1. It's pretty easy to recognize it again
2. It's relatively quick to solve

Let's break this down.

p is the product of the integers from 1 to 30. This just means 30!. What's 30!? It's a the product of the integers from 1 to 30. Oh. Yeah. 30*29*28*27......*3*2*1

3^k is a factor of p. This means that 30! must be divisible by 3 raised to the k power. So 30!/3^k = integer.

We want to find the GREATEST integer k. Clearly you could make k = 1 and that would work as 30!/3 is an integer. But, again, we want the greatest k. So the question boils down to, how many 3's are in 30!?

There are two ways to do this. Let's start with the slow but practical way.

List out the components of 30! that are divisible by 3 and then count up all of your 3's. 30, 27, 24, 21, 18, 15, 12, 9, 6, 3. Keep in mind that 27 has 3x3's, 18 has 2x3's, and 9 has 2x3's. So you end up with 14 3's. Not bad.

Here's the faster/cleaner way to approach these factorial questions. Take the factorial and divide it by the number you're testing, in this case 3.

30/3 = 10. That means that there are 10 numbers from 1-30 that are divisible by 3. Done! Nope. Not yet because there are some numbers from 1-30 that have more than one 3 (see above list).

30/9 = 3. Next, up the power of 3 so that we're dividing by 9. This tells us that there are three numbers from 1-30 that have two 3's. Great - that was easy! Well, not quite there yet because there are numbers from 1-30 that have three 3's.

30/27 = 1. Up the power once more so that we're dividing by 27. That counts the number of 27's from 1-30 or the number of numbers with three 3's. There's just one, 27.

30/81 = doesn't fit. 81 doesn't fit into 30 so that tells us that there are no numbers from 1-30 that have four 3's. That's where you stop.

Then just add those, 10 + 3 + 1 = 14.

The explanation for this second method is much longer BUT the method itself tends to be much quicker than the manual calculation.

## Additional GMAT Factorial Puzzle Practice Questions

Not the same factorial setup but a solid GMAT Question of the Day puzzle example to get you thinking in a GMAT kind of way

Here's another GMAT Question of the Day to sharpen your GMAT puzzle skills.

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