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## Comment on

Introduction to Divisibility## Hi Brent Could you explain

## You bet.

You bet.

"x is divisible by y" is the SAME AS "x = ky for some integer k"

For example, since 20 is divisible by 2, we can also say that 20 = 2k for some integer k (in this case k = 10)

Since 33 is divisible by 11, we can also say that 33 = 11k for some integer k (in this case k = 3)

Since 42 is divisible by 6, we can also say that 42 = 6k for some integer k (in this case k = 7)

Does that help?

Cheers,

Brent

## https://in.edugain.com/forum

Hi Brent!

In this question, when I use 101 as the value of y, I get x as 133. However, Barron's answer guide says that its 132. I was wondering whether my logic was correct or no?

4y-3x=5;

4(101)-3x=5

3x=399

x=133

This is how I solved the equation.

Thanks for your help.

## Question link: https://in

Question link: https://in.edugain.com/forum/Question/3155/if-4y-3x-5-what-is-the-smalle...

Just because y > 100, we can't say that y = 101

It could be the case that y = 100.1 or 100.2 or....

Here's my solution:

Given: 4y - 3x = 5

Add 3x to both sides: 4y = 3x + 5

Solve: y = (3x + 5)/4

We want y to be GREATER THAN 100

So, let's first see what it takes for y to EQUAL 100

Set y = 100 to get: 100 = (3x + 5)/4

Multiply both side by 4 to get: 400 = 3x + 5

So, 3x = 395

Solve: x = 395/3

Rewrite as: x = 131 2/3

So, when x = 131 2/3, y EQUALS 100

So, when x is greater than 131 2/3, y will be GREATER THAN 100

132 is the smallest INTEGER that's greater than 131 2/3, so the correct answer is 132

Cheers,

Brent

## Is there any divisibility

## There is a divisibility rule

There is a divisibility rule for 11, but it is not tested on the GRE.

On test day, you need to know the divisibility rules for 2, 3, 4, 5, 6, 9, and 10

That said, we can combine some of the rules to make up other rules.

For example, if a number is divisible by 15, that it must be divisible by 3 and by 5.

So, for example, we know that 10035 must be divisible by 15, because we can confirm that 10035 is divisible by both 3 and 5.

Likewise, if a number is divisible by 12, that it must be divisible by 3 and by 4.

So, for example, we know that 11232 must be divisible by 12, because we can confirm that it's divisible by both 3 and 4.

## Hi Brent, could you please

My reasoning was - 15 * 80 = 3 * 5^2 * 2^4. So any answer choice that has an equal or greater number of 3s, 5s and 2s would be an acceptable answer.

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/which-of-the-following-has-15-80-as-a-f...

The issue with your solution is that 15^80 represents the product (15)(15)(15)(15)(15)(15).....(15)(15)(15) not (15)(80)

Does that help?

## https://gre.myprepclub.com

Hey Brent can u please comment on my approach.

What i did was take a nice number which was divisible by both 3 and 7 like 21, and found 7 to be 2-bedroom apartments, and 1 in front apartment.

So narrowed it to two choice 42 and 56 but took 42 as we have 7 rooms. as if i added 1 + 7 = 8 than i would be double counting.

however i also had another approach which didn't make much sense like i subtracted 1 - 1/3 to get other rooms and 1 - 6/7 to get other sides but i couldnt figure out how to combine it via fractions to find rooms.....is there a fractional approach here?

## Question link: https://gre

Question link: https://gre.myprepclub.com/forum/in-a-certain-apartment-building-exactly...

Testing 21 apartments was a good idea.

In fact, if 21 were among the answer choices, then that answer would have been correct.

Upon seeing that 21 is not among the answer choices, we should test 42 (which is twice 21).

If there are 42 apartments, then 6 of those apartments are two-bedroom apartments.

And if there are 6 two-bedroom apartments, then 2 of those two-bedroom apartments are in front.

So everything works perfectly when there are 42 apartments.

Note: There's no double counting in the above solution

Among the 42 apartments, there are 6 of those apartments are two-bedroom apartments.

Among 6 two-bedroom apartments, 2 are in the front, and 4 are in the back.

Here's why answer choice D (56) is incorrect.

If there are 56 apartments, then 8 of those apartments are two-bedroom apartments.

And if there are 8 two-bedroom apartments, then 8/3 of those two-bedroom apartments are in front.

Stop!

Since 8/3 is not an integer value, we know that answer choice D (56) is incorrect.