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2007-04-27 17:13:54 · 10 answers · asked by stantx5 1 in Science & Mathematics Mathematics

10 answers

The answer is 120...

2007-04-29 03:32:02 · answer #1 · answered by Anonymous · 0 0

The Least Common Multiple is the Number for which a Given pair (or more) of Numbers are Factors

In your example you are being asked to find the lowest number into which 8 , 15 and 20 are all divisble with NO REMAINDER

This can be done by Multiplying 8 x 15 x 20 = 2400

Then by Inspection Decide if this can be Reduced and still be True

2400 / 20 = 120 ( which is 8 x 15 and 15 x 8 and 20 x 3)

Therefore 120 IS the Least Common Multiple of 8 ,15 & 20.

2007-04-28 02:52:39 · answer #2 · answered by Rod Mac 5 · 0 0

It is 120.
And you mean Lowest Common Multiple. (Or at least that's what they call it in the UK). This is the smallest number that has factors of 8, 15 and 20.
As 8 and 15 are coprime, 8 x 15 =120. 20 shares factors with 8 and 15 you don't need to count it.

2007-04-28 02:51:10 · answer #3 · answered by ArchieBabes 2 · 0 0

20 40 60 80 100 120 140 etc
15 30 45 60 75 90 105 120 135 150 etc
8 16 32 48 56 64 72 80 88 96 104 112 120 etc
From above, L.C.M. = 120

2007-04-27 20:15:53 · answer #4 · answered by Como 7 · 0 0

120. Since 8 and 15 have no common factor, the LCM must be some multiple of their product, which is 120. Since 120 is a multiple of 20, the answer is 120.

2007-04-27 17:15:49 · answer #5 · answered by Anonymous · 3 0

8, 15, 20 div by 2
4, 15, 10 div by 2
2, 15, 5 div by 2
1, 15, 5 div by 3
1, 5, 5 div by 5
1, 1, 1

Hence 2x2x2x3x5 = 120

So the Lowest Common Multiple is 120.

2007-04-28 16:23:41 · answer #6 · answered by Kemmy 6 · 0 0

15

2007-04-27 17:16:22 · answer #7 · answered by Faisal R 3 · 0 2

factorise:
2 x 2 x 2
3 x 5
2 x 2 x 5

then cancel out:
2 x 2 x 2 x 3 x 5 = 120

2007-04-27 19:10:15 · answer #8 · answered by Tropic-of-Cancer 5 · 0 0

120

2007-04-27 17:16:28 · answer #9 · answered by Anonymous · 0 0

http://www.dreamwork.dk/whocares2.jpg

2007-04-27 17:15:52 · answer #10 · answered by tbear 5 · 0 0

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