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## Comment on

Introduction to Remainders## Solution link: https:/

Solution link: https://gmatclub.com/forum/if-p-and-q-are-positive-integers-and-the-rema...

Let's look at some examples of what happens when p < q

Example 1: p = 3 and q = 4

So, p divided by q becomes 3 divided by 4

3 divided by 4 equals 0 with remainder 3 (3 is value of p)

Example 2: p = 9 and q = 20

So, p divided by q becomes 9 divided by 20

9 divided by 20 equals 0 with remainder 9 (9 is value of p)

Likewise, 7 divided by 11 equals 0 with remainder 11 (11 is value of p)

And 15 divided by 18 equals 0 with remainder 15 (15 is value of p)

Does that help?

## thanks a ton sir.

I comoleted integer parts.now i am going for geo.

After completing this module with 280 reinforment activities, i made a conclusion.

Three concepts. I grabed all of three.just one thing pinched me..

1/ when i was solving those problems ;numerous time i faced a issue...

That is """" suppose i am doing one quant;start with one rule;after some steps i realized that this one Won'twork"""" move to another one.....

Is this normal?? Am i missing something? If this thing will happen in exam day,i am sure i will face timing issue.

What's the solution? Sir

## Changing strategies while

Changing strategies while solving a math question is perfectly normal. In fact it's an integral part of mathematics.

There's a saying that goes something like this: In order to be a good mathematician, you must be both brave and afraid at the same time.

You must be brave enough to set out on a particular approach. However, while you're working on that approach, you must also be aware that your solution may lead you to a dead end (or it may simply be a solution that will take way too long to complete within the time constraints of the GMAT). So, while you must be brave to start one approach, it's important to also be afraid of the potential that you've chosen the wrong approach. If you feel that you have, indeed, chosen the wrong approach, then you must, again, be brave enough to abandon that approach and start a new one.

## Thanks sir.

Hopefully i Won't face this type of situation.

## Hi Brent,

https://gmatclub.com/forum/x-and-y-are-positive-integers-when-16x-is-divided-by-y-234977.html

could you please explain the logic behind choosing possible values for x?

## Question link: https:/

Question link: https://gmatclub.com/forum/x-and-y-are-positive-integers-when-16x-is-div...

Good question! I didn't explain that very well in my solution.

Let's start at the point where we have: x(16 - y) = 4

The important part here is that x and (16 - y) are both POSITIVE INTEGER

In other words, we have: (some integer)(some other integer) = 4

There aren't many ways to get a product of 4.

In fact there are only three possible options:

(1)(4) = 4

(2)(2) = 4

(4)(1) = 4

So, there are 3 possible solutions to the the equation: x(16 - y) = 4

#1: x = 1 and (16 - y) = 4, in which case y = 12

#2: x = 2 and (16 - y) = 2, in which case y = 14

#3: x = 4 and (16 - y) = 1, in which case y = 15

Does that help?

Cheers,

Brent

## got it. thanks!

## hi Brent,

Just noticed this. The 3 possible solutions, #1, #2 and #3 should equate to 4 if I am not wrong.

## That's correct.

That's correct.

Solutions #1, #2 and #3 are all such that x(16 - y) = 4

## Hi Brent,

Thank you for your quick response. I was referring to your comment where #2 and #3 are not equated to 4, if I have understood correctly.

#2 should be x = 2 and (16 - y) = 4

#3 should be x = 4 and (16 - y) = 4

## We want the PRODUCT of x and

We want the PRODUCT of x and (16 - y) to be 4

#2) If x = 2 and (16 - y) = 2, then (x)(16 - y) = (2)(2) = 4 (perfect)

#3) If x = 4 and (16 - y) = 1, then (x)(16 - y) = (4)(1) = 4 (perfect)

If we use your solutions, we get...

#2) If x = 2 and (16 - y) = 4, then (x)(16 - y) = (2)(4) = 8 (No good. We need a product of 4)

#3) If x = 4 and (16 - y) = 4, then (x)(16 - y) = (4)(4) = 16 (No good. We need a product of 4)

Does that help?

## Thank you so much!

I was looking at it like an equation - x * (16-y) = 4 and this caused the confusion. I didn't realize that you meant that (16-y) alone equals 2. But when I did the calculations in my approach, I still got the same the results. Could please let me know how is this possible?

The following was my calculation:

#1, x = 1

1*(16 - y) = 4

16 - y = 4

y = 16 - 4

y = 12

#2, if x = 2,

2*(16 - y) = 4

32 -2y = 4

2y = 32 - 4

2y = 28

y = 14

#3, if x = 4,

4*(16-y) = 4

64 - 4y = 4

4y = 64 - 4

4y = 60

y = 15

So, 12 + 14 + 15 = 41.

I didn't choose x = 3 because it didn't result in an integer.

## Those steps are perfect.

Those steps are perfect.

My only suggestion is that, once you get here: 4*(16-y) = 4

Just divide both sides by 4 to get: (16 - y) = 1

So, y = 15

This will save you some time.

You can do the same thing with the other solutions to.

Cheers,

Brent

## Thank you!

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