Hi,

I know that it is basic math, but, I don't remember how to do it!

With what number I have to divide 29,7 to obtain a remainder equal to 0?

thanks!

## Find the number to divide with another one to obtain zero remainder.

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### Re: Find the number to divide with another one to obtain zero remainder.

Are you looking for an integer n for which n/29 is an integer and n/7 is an integer?

All such numbers are the integer multiples of the least common multiple of 7 and 29.

Or are you looking for an integer k for which 29/k is an integer and 7/k is an integer?

All such numbers are factors of the greatest common divisor of 7 and 29.

All such numbers are the integer multiples of the least common multiple of 7 and 29.

Or are you looking for an integer k for which 29/k is an integer and 7/k is an integer?

All such numbers are factors of the greatest common divisor of 7 and 29.

wee free kings

### Re: Find the number to divide with another one to obtain zero remainder.

Sorry, the number is 29.7!

So I looking for an integer k for which 29.7/k retrieve a zero remainder.

So I looking for an integer k for which 29.7/k retrieve a zero remainder.

- Sableagle
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### Re: Find the number to divide with another one to obtain zero remainder.

Cant be done. To have a zero remainder is to give an integer answer. The product of two integers is an integer. 29.7 is not an integer.

You could find factors of it, things by which it can be divided to give a zero remainder, but they won't be integers.

Let's go with 297 instead and we can get integers.

You can ignore 1, because of course it's divisible by that and 1 isn't a prime number anyway (don't ask me why not).

The next prime number is 2. 297 is an odd number, so not divisible by 2.

The next prime number is 3. To test whether something's divisible by 3, add the digits. 2 + 9 + 7 = 18. 1 + 8 = 9. 9 is divisible by 3.

297 = 3 x ___

Well, you could do long division on this but once you're used to multiples of 11 that one kind of stands out. It's also 3 less than 300, which is an even better shortcut.

297 = 3 x 99

Sticking with 3, is 99 divisible by 3? Well, yes. Obvs.

297 = 3 x 3 x 33

Is 33 divisible by 3? Yes.

297 = 3 x 3 x 3 x 11

Is 11 divisible by 3? No. By 5? No. By 7? No. By 11 ... well, er, yeah. So that's that. 297 is exactly divisible by 3 and 11.

29.7 is exactly divisible by 0.3 and 1.1, if that helps.

...

To find the smallest number that can be divided by 297 and 7 you'd do the same process to both numbers, getting:

3 x 3 x 3 x 11 and 7 (because it's a prime number). You can't cancel out anything there, so you get 3 x 3 x 3 x 7 x 11 = 2079, I hope.

That was a lousy illustration. Let's have some better numbers for that: 6 and 4, and 297 and 63.

6 is 2 x 3 and 4 is 2 x 2, so you can ignore one "2x" and get 2x2x3 = 12, divisible by both.

63 is 3 x 3 x 7, so we're looking for the lowest common multiple of 3 x 3 x 7 and 3 x 3 x 3 x 11. At least two cases of "3x" each side, so ignore two and get 7x3x3x3x11 = 2079 again.

...

To get the largest number by which each can be divided without remainder, you take only the prime factors found in both numbers. The largest number by which 6 and 4 can be divided without remainder is 2, and the largest number by which both 297 and 63 can be divided is 9.

...

... right? I've had one beer, so that probably worked but no promises.

You could find factors of it, things by which it can be divided to give a zero remainder, but they won't be integers.

Let's go with 297 instead and we can get integers.

You can ignore 1, because of course it's divisible by that and 1 isn't a prime number anyway (don't ask me why not).

The next prime number is 2. 297 is an odd number, so not divisible by 2.

The next prime number is 3. To test whether something's divisible by 3, add the digits. 2 + 9 + 7 = 18. 1 + 8 = 9. 9 is divisible by 3.

297 = 3 x ___

Well, you could do long division on this but once you're used to multiples of 11 that one kind of stands out. It's also 3 less than 300, which is an even better shortcut.

297 = 3 x 99

Sticking with 3, is 99 divisible by 3? Well, yes. Obvs.

297 = 3 x 3 x 33

Is 33 divisible by 3? Yes.

297 = 3 x 3 x 3 x 11

Is 11 divisible by 3? No. By 5? No. By 7? No. By 11 ... well, er, yeah. So that's that. 297 is exactly divisible by 3 and 11.

29.7 is exactly divisible by 0.3 and 1.1, if that helps.

...

To find the smallest number that can be divided by 297 and 7 you'd do the same process to both numbers, getting:

3 x 3 x 3 x 11 and 7 (because it's a prime number). You can't cancel out anything there, so you get 3 x 3 x 3 x 7 x 11 = 2079, I hope.

That was a lousy illustration. Let's have some better numbers for that: 6 and 4, and 297 and 63.

6 is 2 x 3 and 4 is 2 x 2, so you can ignore one "2x" and get 2x2x3 = 12, divisible by both.

63 is 3 x 3 x 7, so we're looking for the lowest common multiple of 3 x 3 x 7 and 3 x 3 x 3 x 11. At least two cases of "3x" each side, so ignore two and get 7x3x3x3x11 = 2079 again.

...

To get the largest number by which each can be divided without remainder, you take only the prime factors found in both numbers. The largest number by which 6 and 4 can be divided without remainder is 2, and the largest number by which both 297 and 63 can be divided is 9.

...

... right? I've had one beer, so that probably worked but no promises.

Zohar wrote:You don't know what you're talking about. Please spare me your quote sniping and general obliviousness.

CorruptUser wrote:Just admit that you were wrong ... and your entire life, cyberspace and meatspace both, would be orders of magnitude more enjoyable for you and others around you.

### Re: Find the number to divide with another one to obtain zero remainder.

Sableagle wrote:So that's that. 297 is exactly divisible by 3 and 11.

29.7 is exactly divisible by 0.3 and 1.1, if that helps.

297 is also divisible by 9, 27, 33 and 99

so 29.7 is divisible by 0.3, 0.9, 1.1, 2.7, 3.3 and 9.9.

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### Re: Find the number to divide with another one to obtain zero remainder.

None of those are integers.

- Sableagle
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**Posts:**2153**Joined:**Sat Jun 13, 2015 4:26 pm UTC**Location:**The wrong side of the mirror-
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### Re: Find the number to divide with another one to obtain zero remainder.

measure wrote:Sableagle wrote:So that's that. 297 is exactly divisible by 3 and 11.

29.7 is exactly divisible by 0.3 and 1.1, if that helps.

297 is also divisible by 9, 27, 33 and 99

so 29.7 is divisible by 0.3, 0.9, 1.1, 2.7, 3.3 and 9.9.

Oh, yeah. Sorry. Only had one beer. Forgot to include groups of factors. 297 is 3 x 3 x 3 x 11 and is therefore divisible by:

1

3

11

3 x 3

3 x 11

3 x 3 x 3

3 x 3 x 11

3 x 3 x 3 x 11

Sableagle wrote:Cant be done. To have a zero remainder is to give an integer answer. The product of two integers is an integer. 29.7 is not an integer.

You could find factors of it, things by which it can be divided to give a zero remainder, but they won't be integers.

gmalivuk wrote:None of those are integers.

Sableagle wrote:Cant be done. 29.7 is not an integer. ... they won't be integers.

gmalivuk wrote:None of those are integers.

Sableagle wrote:... they won't be integers.

gmalivuk wrote:None of those are integers.

Zohar wrote:You don't know what you're talking about. Please spare me your quote sniping and general obliviousness.

CorruptUser wrote:Just admit that you were wrong ... and your entire life, cyberspace and meatspace both, would be orders of magnitude more enjoyable for you and others around you.

### Re: Find the number to divide with another one to obtain zero remainder.

If we're going to admit non-integer solutions, why bother with factoring at all? We're looking for x such that 29.7/x = n, where n is an integer. Then, 29.7 = xn, so x = 29.7/n, where again n is an integer. This completely classifies the possible divisors: for each integer n, 29.7/n is one solution to the problem, and all solutions are generated in this way.

(∫|p|

Thanks, skeptical scientist, for knowing symbols and giving them to me.

^{2})(∫|q|^{2}) ≥ (∫|pq|)^{2}Thanks, skeptical scientist, for knowing symbols and giving them to me.

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