Lesson: Introduction to Remainders

Comment on Introduction to Remainders

how come W can be 3, 3 is not divisible hence there's shouldnt be R in possible Values?
greenlight-admin's picture

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, which of the following must be divisible by x?
Indicate all possible values.

A. x/78
B. x
C. x+78
D. 78/x
E. 78-x
F. 78x
G. 78x+78
greenlight-admin's picture

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 if x is equal to zero, since 0 is divisible by 78.

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.
greenlight-admin's picture

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 question :

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.
greenlight-admin's picture

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 take 25 times the remainder 34 to get 850 and then add 28 to it. THIS is regarding shubham4596's question.
greenlight-admin's picture

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 now.

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

How do I identify the value i.e option C x=6 and y=3?
greenlight-admin's picture

Question link: https://greprepclub.com/forum/if-x-and-y-are-different-positive-integers...

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://greprepclub.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.
greenlight-admin's picture

Question link: https://greprepclub.com/forum/the-remainder-when-120-is-divided-by-singl...

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

Hi Brent,

for this question.
https://greprepclub.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?
greenlight-admin's picture

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

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

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