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xud | 21:19 Thu 16th Feb 2017 | Education
15 Answers
364 is a multiple of 7 but not 3
384 is a multiple of 3 but not 7
Which number between 364 and 384 is a multiple of both?
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378
If it has to be a multiple of 3 and 7 then it has a Lowest Commoner Denominator of (3x7) ie 21
So the number your are looking for has to be a multiple of 21
378
Don't you think it would be better to know how to find the answer than what the answer is? this is not a hard question.
Didn't Captian answer that?
Not there when I was posting Zac no and that's not the easy answer anyway. It has nothing to with LCD. It's multiples of 7 starting from 364 that are also divisible by 3.
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Cheers guys.
Prudie, we are not all experts in maths. Your comments have added no value to this thread whatsoever.
Thank you
-- answer removed --
If I was doing this with one of my children I would get them to divide all the numbers between 364 and 384 by 7 and then repeat by dividing by 3. Not saying this is the correct way to do it but it'll find the answer.
if i was answering this q i would add on 7 to 364, and try to divide the result by 3, then if it wouldnt, i'd add on anther 7 and repeat
I like bednobs' method. Counting on from 364 the answer has to be either 371 or 378 (since 385 would be too high so there are at most only 2 numbers to check for divisibility by 3. You can use a calculator or short division but a simple test for divisibility by 3 is whether the sum of the digits is a multiple of 3 (3+7+8 =18 which is divisible by 3)
I think Captain just mean LCM not LCD. The answer does have to be a multiple of 21. There are often several ways to solve these problems.
>Prudie, we are not all experts in maths. Your comments have added no value to this thread whatsoever

Prudie made a very point in my opinion. She was the first to point out that you can simply start at 364 and count up in 7s until you reach the answer a few seconds later.

Hopefully the person who needed the answer didn't just copy down the answer- they also thought through how they should approach similar problems in future in a logical manner.

I would use ff's method. Gives the answer in seconds.

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