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## Comment on

Introduction to Remainders## how come W can be 3, 3 is not

## You're referring to the

You're referring to the practice question that starts at 4:45 in the video.

Think of it this way:

An integer (K) is EITHER divisible by integer N, OR not divisible by integer N. One of those two things must be true.

If K IS divisible by N, then there is no remainder when we divide K by N. For example, 12 is divisible by 4, and when we divide 12 by 4, we get 3 with remainder 0.

On the other hand, if K is NOT divisible by N, then there is a remainder when we divide K by N. For example, 13 is not divisible by 5, and when we divide 13 by 5, we get 2 with remainder 3.

In the practice question, I say that 3 divided by 8 leaves a remainder of 3. So, let's first ask, "Is 3 divisible by 8?" Since it's clear that 3 is not divisible by 8, then there must be a remainder when we take 3 and divide it by 8. What is that remainder?

To answer this question, let's see what happens when we divide OTHER values by 8.

84 divided by 8 equals 10 with remainder 4. In other words, 8 divides INTO 84 ten times, with remainder 4.

17 divided by 8 equals 2 with remainder 1. In other words, 8 divides INTO 17 two times, with remainder 1.

10 divided by 8 equals 1 with remainder 2. In other words, 8 divides INTO 10 one time, with remainder 2.

And finally, 3 divided by 8 equals 0 with remainder 2. In other words, 8 divides INTO 3 zero times, with remainder 3.

Does that help?

## If x is divisible by 78,

Indicate all possible values.

A. x/78

B. x

C. x+78

D. 78/x

E. 78-x

F. 78x

G. 78x+78

## If x is divisible by 78, then

If x is divisible by 78, then x = 78k for some integer k.

Take each answer choice and replace x with 78k. We get:

A. 78k/78 = k. k need not be divisible by 78k (aka x)

B. 78k is definitely divisible by 78k (KEEP)

C. 78k + 78 = 78(k+1). This need not be divisible by 78k

D. 78/78k = 1/k. This need not be divisible by 78k

E. 78-78k = 78(1 - k). This need not be divisible by 78k

F. 78(78k). This is definitely divisible by 78k (KEEP)

G. 78(78k) + 78 = 78(78k + 1). This need not be divisible by 78k

Answer: B & F

## I am confused about this one

Then, if option A which is (x/78), i.e. (0/78) is divisible by x, that would mean 0/0 which is undefined.

For option B, that too would be 0/0. How can B be correct here?

Help would be appreciated.

## You make a great point!!

You make a great point!!

Most (probably all actually) official GRE remainder questions have some kind of proviso that limits the numbers to POSITIVE integers. The question above would benefit from such a proviso.

## Hello Sir ,I have a following

what is the smallest integer when divided by 34 and 58 leaves remainder 28 and 52 ?

A) 968

B) 974

C) 980

D) 988

Could you please give the explanation too.

## we have a nice rule that says

we have a nice rule that says:

If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.

For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Since all 4 answer choices are between 950 and 1000, let's find some possible values that yield the necessary remainders.

First condition: when divided by 34 the remainder is 28

(25)(34) = 850

So, 28 + (25)(34)= 878

One possible value is 878

Now keep adding 34's to find other possible values.

So, more possible values are: 878, 912, 946, 980, 1014...

STOP!!!

Only one of the answer choices (C) meets the first condition.

So, C MUST be correct.

ASIDE: If we test C, we'll see that it meets the second condition (the number, when divided by 58 leaves remainder 52)

## how do we know that we should

## Hi bimla,

Hi bimla,

I should have made that step clearer.

We're told that when the number is divided by 34 the remainder is 28. So, the number is 28 more than some multiple of 34.

There are infinitely many numbers that meet this condition. However, since the answer choices range from 968 to 988, we must focus on values in this range.

So, I ask myself, "What multiple of 34 is in this range?"

How about (20)(34)? This equals 680, which isn't near the 968 to 988 range.

How about (25)(34)? This equals 850 . . . getting closer.

How about (27)(34)? This equals 918. Getting closer.

How about (28)(34)? This equals 952. REMEMBER that the number is 28 MORE than some multiple of 34. So, one possible value = 952 + 28 = 980 (IN RANGE!)

So, 980 meets the FIRST condition (when the number is divided by 34 the remainder is 28).

Does 980 also meet the SECOND condition (when the number is divided by 58 the remainder is 52)? Yes it does, since 980 = (16)(58) + 52

So, the number in question must be 980

Does that help?

Cheers,

Brent

## Thank you Brent. It's clear

## https://gre.myprepclub.com/forum

How do I identify the value i.e option C x=6 and y=3?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-x-and-y-are-different-positive-integ...

Just before we chose those values, we learned that x = 2y

So, one possible solution to that equation is y = 3 and x = 6

We could have also chosen y = 4 and x = 8

Or y = 5 and x = 10

etc.

Each one of those pairs of values will also show that statement III could be true.

Does that help?

Cheers,

Brent

## Hello Brent,

I have solved this question as follows

https://gre.myprepclub.com/forum/the-remainder-when-120-is-divided-by-single-digit-integer-m-5969.html

From our formula

120/m = R(P)---equation 1

120/n = R(P)---equation 2

from 1 and 2

120/m = 120/n , further cross multiplying and equating

0 = m-n , hence answer is 0.

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/the-remainder-when-120-is-divided-by-si...

Unfortunately, it's just a coincidence that you reached the correct answer.

In your calculations, you have:

120/m = R(P)---equation 1

120/n = R(P)---equation 2

We cannot conclude that R(P) is the same value for each equation.

Yes, dividing by m and n will yield a quotient and a remainder, but we can't say the quotient and remainder will be exactly the same in each case.

Also, we can't conclude that 0 = m - n, since that would imply that m = n, and the question tells us that m > n.

Cheers,

Brent

## Hi Brent,

can you provide the solution of this question?

https://gre.myprepclub.com/forum/n-is-a-positive-integer-that-is-divisible-by-1836.html

## Here's my full solution:

Here's my full solution: https://gre.myprepclub.com/forum/n-is-a-positive-integer-that-is-divisib...

## Hi Brent,

for this question.

https://gre.myprepclub.com/forum/gre-math-challenge-8-a-b-and-c-are-multiple-of-175.html

Can I plug in the number by 15,30 and 45?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/gre-math-challenge-8-a-b-and-c-are-mult...

Yes, you can plug in those values.

When you do so, you'll get:

QUANTITY A: 15

QUANTITY B: 0

Quantity A > Quantity B

So, the correct answer is either A or D

From this point, you can either test more values of a, b and c, or you can apply some Remainder concepts.

Cheers,

Brent

## https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/when-a-number-a-is-divided-by-6-the-remainder-is-3-and-when-17785.html

For these questions I notice that you take an algebraic approach. What I have done instead is just pick numbers and go with that instead and I still get the right answer. Would it be better to approach remainder problems in an algebraic way? I really find the number approach much faster to deal with.

## Question #1: https:/

Question #1: https://gre.myprepclub.com/forum/x-y-and-z-are-three-consecutive-multipl...

Question #2: https://gre.myprepclub.com/forum/when-a-number-a-is-divided-by-6-the-rem...

Choosing numbers that satisfy the given information is a great idea for multiple choice questions in which we select one answer choice.

For example:

When positive integer N is divided by 12, the remainder is 5. What is the remainder when N is divided by 6?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Here, we can choose a number that satisfies the given information here because exactly ONE of the five answer choices is correct.

So for example, it could be the case that N = 17

When we divide 17 by 6, the remainder is 5.

Since we can't have more than one correct answer, we can assume that all possible values of N will have a remainder of 5 when divided by 6.

Now consider this question:

When positive integer N is divided by 5, the remainder is 3.

QUANTITY A: The remainder when N is divided by 10

QUANTITY B: 3

In this case, the correct answer isn't hiding among 5 answer choices. In fact, in Quantitative Comparison questions, a quantity can have many possible values.

For example, it could be the case that N = 13. When we divide 13 by 10, the remainder is 3. In this case, the two quantities are equal.

It could also be the case that N = 18. When we divide 18 by 10, the remainder is 8. In this case, Quantity A is greater.

So, for multiple choice questions with exactly one correct answer, testing numbers is often a great strategy.

We can also use this strategy for Numeric Entry questions

However, the strategy doesn't work with Quantitative Comparison questions.

## https://gre.myprepclub.com/forum

I did the same logic that you followed but ended up with bot I and III. This is the logic I followed:Now check the answer choices..

i) x = 9 and y = 3

We get: 18k + 9 = 6j + 3 (it this possible given that k and j are positive integers?)

Rearrange to get: 6j - 18k = 9 - 3

Simplify to get: 6(j - 3k) = 6

Divide both sides by 6 to get: j - 3k = 1.

So for the last logic x = 13 y = 7

18k +13 = 6j + 7

18k-6j = -6

6(j-3k) = -6

j-3k = -1... if j = 2 and k = 1 that would satisfy the equation above

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/when-positive-integer-n-is-divided-by-1...

You made a mistake when you went from 18k-6j=-6 to 6(j-3k)=-6

When we factor 18k-6j, we get 6(3k-j)

## https://gre.myprepclub.com/forum

for this question I ended up with zy +q = x. Just from a mathematical perspective why is this wrong: (zy)/z +q = x/z -> therefore x/z = y +q. therefore landing on D. Why is this a wrong algebraic way of solving?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-when-x-is-divided-by-z-the-result-is...

Given: when x is divided by z, the result is y remainder q

So, for example, the values x = 17, z = 5, y = 3 and q = 2 satisfy the given information.

That is, when 17 is divided by 5, the result (aka quotient) is 3 and the remainder is 2.

When we plug these values into zy + q = x, we get: (5)(3) + 2 = 17 WORKS

When we plug these values into your expression, (zy)/z +q = x/z, we get: (5)(3)/(5) + 2 = 17/5

When we simplify we get: 5 = 3.4, which isn't true.

## Hey Brent,

I left a comment about your question.

Can you please explain why the answer B is an impossible scenario?

https://gre.myprepclub.com/forum/if-x-and-y-are-different-positive-integers-which-of-the-11483.html#p103284

## Question link: https://gre

Question link: https://gre.myprepclub.com/forum/if-x-and-y-are-different-positive-integ...

Given: If x and y are different positive integers

B) When 2x is divided by y, the remainder is x

From B, we can see that possible values of 2x are: x, x + y, x + 2y, x + 3y, . . . etc.

Now we'll demonstrate that each of these possible values of 2x are impossible.

Examine the first option: 2x = x.

Solve to get x = 0, but we're told x is POSITIVE No good.

Now examine the second option: 2x = x + y.

Solve to get x = y.

We are told that x and y are DIFFERENT positive integers. No good.

Before we check the next set of values, I would like to remind you of an important rule regarding remainders:

When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D

For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0

Similarly, since option B tells us "When 2x is divided by y, the remainder is x," our above property tells us that 0 ≤ x < y

Now examine the third option: 2x = x + 2y.

Solve for x to get: x = 2y

Now take our inequality 0 ≤ x < y and replace x with 2y to get: 0 ≤ 2y < y

As you can see, the inequality does not work.

Now examine the fourth option: 2x = x + 3y.

Solve for x to get: x = 3y

Now take our inequality 0 ≤ x < y and replace x with 3y to get: 0 ≤ 3y < y

As you can see, the inequality does not work.

As you can see, if we keep testing possible values of 2x, we'll keep getting an inequality that doesn't work.

Does that help?

## Thank you for the explanation

I think that I know what was my mistake.

I thought that we can reduce when working with reminders.

For example, if I divide 6 by 10, what is the reminder? Is it 6 or 3?

I thought about the 3 because if we reduce 6/10 we get 3/5 (6/10=3/5).

But after your explanation, I think we can't reduce the numbers when working with reminders.

Can you please confirm it?

And if we can't reduce the numbers when looking for reminders, why can't we reduce that?

## That's correct; when working

That's correct; when working with remainders, we can't reduce values.