# Question: Remainder when Divided by 7

## Comment on Remainder when Divided by 7

### I set the equations equal to

I set the equations equal to each other: 5Q+1=7K+1 and I got Q= 7K/5. I plugged that into Q/7 and got K/5. Using divisibility rules, 0 was the only one divisible by 5. Is that another way to do it or did I get lucky?!

### That approach is solid!

That approach is solid!

### I did not get the approach

I did not get the approach mentioned here in comment.

### Here's the longer version of

Here's the longer version of mtl1212's solution:

There's a nice rule that says:

If N divided by D equals Q with remainder R, then N = DQ + R

For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2

Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

From the first sentence, we can write: x = 5q + 1
From the second sentence, we can write: x = 7k + 1

Since both equations are set equal to x, we can write:
5q + 1 = 7k + 1
Subtract 1 from both sides to get: 5q = 7k
Divide both sides by 5 to get: q = 7k/5

IMPORTANT: Notice that, in order for q to be an integer (which it is), 7k must be divisible by 5.

Since 7 is not divisible by 5, it MUST be the case that k is divisible by 5. Finally, if k is divisible by 5, then k/5 is an INTEGER, which means q = 7k/5 = 7(some integer)

Clearly 7(some integer) is a multiple of 7, which means q is a multiple of 7

The question asks, "What is the remainder when q is divided by 7?"

Since q is a multiple of 7, the remainder will be 0

### If we give the value 10 to k

If we give the value 10 to k it's also divisible by 5

### That's correct.

That's correct.
In fact, k can be ANY number that's divisible by 5 (e.g., 10, 35, 615, etc)

Cheers,
Brent

### Thanks, Understood :) the

Thanks, Understood :) the remainder will be zero. I suppose you meant that in the last statement. Thanks Brent.

### Good catch. I changed the

Good catch. I changed the last line to "Since q is a multiple of 7, the remainder will be 0"

### At the very end of the video,

At the very end of the video, when you say that x = 1 is the fastest approach, you say that 1 divided by 5 = 0 (1), so q = 0. But wouldn't 1 divided by 5 be 0 (5)?

### Think of it this way.

Think of it this way.

When we divide 17 by 5, we see that 5 divides into 17 THREE times (to make 15).
From the original 17, we have 2 left (after we subtract 15).
So, the remainder is 2.

Likewise, when we divide 1 by 5, we see that 5 divides into 1 ZERO times (to make 0).
From the original 1, we have 1 left.
So, the remainder is 1.

Alternatively, we can apply the formula for rebuilding the dividend.
It says:
If N divided by D equals Q with remainder R, then N = DQ + R

For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

Likewise, since 1 divided by 5 equals 0 with remainder 1, then we can write 1 = (5)(0) + 1
Works!!

You are suggesting that 1 divided by 5 equals 0 with remainder 5.
Applying the formula, then we can write: 1 = (5)(0) + 5
Doesn't work.

Does that help?

Cheers,
Brent

### Brent, thank you for your

Brent, thank you for your explanation. This helps!

### Hi Brent,

Hi Brent,

Can you give the solution of this question?
https://gre.myprepclub.com/forum/s-be-the-set-of-all-positive-integers-n-such-that-n-1760.html

### Yet again, just pick numbers

Yet again, just pick numbers and solve. Use x = 36 and BINGO. That's all you need to do. I would actually rate this question as EASY, 130-149.

### QA

QA: The remainder when 3^900 is divided by 4
QB: The remainder when 2^600 is divided by 7

For both these quantities I want to know if my approach is correct.
For QA I found the last digit is 1 and divided that by 4 so I got remainder 1
For QB I did the same approach and divided 6/7 but that gets me remainder 6. But that's not right

### Unfortunately, knowing the

Unfortunately, knowing the unit's digit doesn't help us here.
For example, for QA, you determined that the unit's digit of 3^900 is 1.
However, knowing that the unit's digit is 1, doesn't help us determine the remainder when that number is divided by 4.
For example, 21 divided by 4 leaves remainder 1.
On the other hand, 31 divided by 4 leaves remainder 3.

The same applies to QB.

Unless I'm missing something basic, this question is beyond the scope of the GRE since it requires knowledge of modular arithmetic.