my brother needs help with this problem
2007-11-12 09:38:36 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The answer is 84.
For small numbers, the easiest way to do it is to take the largest number and adding its various multiples and then testing it against the other numbers:
7 isn't a multiple of 3 or 4
14 isn't a multiple of 3 or 4
21 is a multiple of 3 but not 4
28 is a multiple of 4 but not 3
35 isn't a multiple of 3 or 4
42 isn't a multiple of 3 or 4
and so on...
The other way is called prime factorization:
Break down each number into primes.
3 = 1 * 3
4 = 2 * 2
7 = 1 * 7
Now, take the largest number of times each multiple appears and multiply them together to get the LCM.
3 * 2 * 2 * 7 = 84
Good luck!
2007-11-12 09:41:42 · answer #1 · answered by Eagle1 Fox2 7 · 2⤊ 3⤋
Lowest Common Multiple Of 4 And 14
2016-12-17 06:29:43 · answer #2 · answered by sittloh 4 · 0⤊ 0⤋
Hi,
There seems to be quite a lot of disagreement among responders. Part of the problem seems to be that some appear to think that only one factor,2, of 4 is used in the LCM. Actually you must use the factor with the highest exponent that appears in any one of the numbers.
So,
4 = 2²
3=3*1
7=7*1
LCM = 3*2² *7 = 84
Check it out 42/4=10.5, so 42 is NOT the LCM.
FE.
2007-11-12 10:14:17 · answer #3 · answered by formeng 6 · 1⤊ 0⤋
1
2007-11-12 09:41:32 · answer #4 · answered by Anonymous · 0⤊ 5⤋
Lcm Of 4 And 7
2016-10-07 01:06:51 · answer #5 · answered by ? 4 · 0⤊ 0⤋
7/10 and 3/4
2014-01-25 05:06:59 · answer #6 · answered by Connie 1 · 0⤊ 0⤋
84
Do a prime factorization of each:
3
2 x 2
7
If you can't cross off primes (say 2 exists in each one, then you could cross out the 2), then LCM = 3 x 2 x 2 x 7 =84
2007-11-12 09:43:55 · answer #7 · answered by McDudette 3 · 1⤊ 1⤋
84
2007-11-12 09:43:30 · answer #8 · answered by ironduke8159 7 · 2⤊ 1⤋
42
2007-11-12 09:42:14 · answer #9 · answered by luisfernu@sbcglobal.net 2 · 1⤊ 2⤋
84 is the right answer (3*4*7 since none of them have common factos)
2007-11-12 09:44:56 · answer #10 · answered by someone1254 2 · 1⤊ 1⤋