Question

By what integer should 540 be divided, so that remainder would make 75% of quotient.

Answer 1

I had found a way to solve this by writing an equation 180/N = y + x/N where x - remainder, y - quotient and N - the divisor that we need. However, after simplifying it (taking x to be 0.75y and adding the fractions) it turns out that I would need to put quite a lot different values into y and partially "guess" the answer.
Is there a solution that doesn't require any random putting of 20 or so numbers and checking for each case?

Answer 2

Let "a" be the required integer. You have to solve :

```
540 = a*q + r
4r = 3q
```

Answer 3

Multiply first equation by 4 in order to eliminate r :

```
540*4 = 4aq + 4r
```

Answer 4

```
540*4 = 4aq + 3q
```

Answer 5

```
540*4 = q*(4a + 3)
```

Answer 6

Here is the question :
What do you notice about an integer of form `4a+3` ? Is it divisible by 4 ?

Answer 7

Thank you, @ganeshie8 I thought about going that way, however, when I got to the 540*4 = q*(4a + 3) I didn't know what else to do....

4a + 3 isn't divisible by 4. The remainder is 3 or -1. So, with that information we could notice that q is divisible by 4:
540 * 4 congruent to q(4a + 3) (mod 4)
0 congruent to 3q (mod 4), so q must be a multiple of 4.

However, what to do now?

Answer 8

@Zyberg that's right, but more importantly notice that \(4a+3\) is "not" divisible by \(4\).

Answer 9

@ganeshie8 hm, but what would that allow me to do?

Answer 10

```
540*4 = q*(4a + 3)
```

Answer 11

factor the left hand side and kick out powers of 4

Answer 12

http://www.wolframalpha.com/input/?i=factor+540*4

Answer 13

Can we now say that \(4a+3\) must be a divisor of \(\large 3^3*5\) ?

Answer 14

No, we can't. Or, we shouldn't be able to as far as I can see.

Answer 15

Why not ?

Answer 16

Okay, then we can. But I don't see why we can.

Answer 17

Let me give you a simple example

Answer 18

If \( x*y = a*b\) and if you know that \(x\) and \(a\) are coprime, is below true ?

\(a\) must be a divisor of \(y\).

Answer 19

example :

\(6*35 = 7*30 \)

Clearly \(7\) is a divisor of \(35\)

Answer 20

Ohh, I understand it now. Made up a few questions and found answers to them ;)

Answer 21

Good

```
4^2*(3^3*5)= q*(4a + 3)
```

Since 4^2 and 4a+3 are coprime, we can say that 4a+3 is a divisor of 3^3*5

Answer 22

Thanks!
So, what should we do now? Is there any more simplification left or should we try putting some values?

Answer 23

We can simplify further

Answer 24

use that same fact again :
when a number of form 4a+3 is divided by 4, we must get a remainder 3

Answer 25

Look at 3^3*5

Answer 26

Same thing. Remainder 3.

Answer 27

When 5 is divided by 4, the remainder is 1.
When 3 is divided by 4, the remainder is 3.
When 3^2 is divided by 4, the remainder is 1.
When 3^3 is divided by 4, the remainder is 3.

Answer 28

So 4a+3 has to be 3 or 3*5 or 3^3*5

(because only these combinations leave the remainder 3)

Answer 29

3 cases, shouldn't be hard ?

Answer 30

Yeah, not hard at all :) Thank you very much!

Answer 31

Np :)

Answer 32

Looks I have missed one case, we get 4 cases right ?

```
So 4a+3 has to be 3 or 3^3 or 3*5 or 3^3*5

(because only these combinations leave the remainder 3)
```

Answer 33

Well, 3^3 isn't a, so it didn't do much damage to miss it. I found that only a = 33 (from 4a + 3 = 3^3 * 5) solves the problem.
Thanks again :)

Answer 34

looks good..