SOLUTION: Hi I really need help with this math problem-----> find the LCM of 5(y-11)and 55(y-11)
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Question 427267: Hi I really need help with this math problem-----> find the LCM of 5(y-11)and 55(y-11)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i believe the answer is going to be 55*(y-11)
5*(y-11) * 11 = 55*(y-11)
55*(y-11) * 1 = 55*(Y-11)
look at the multiples of 5*(y-11)
they will be:
5*(y-11)
10*(y-11)
15*(y-11)
20*(y-11)
25*(y-11)
30*(y-11)
35*(y-11)
40*(y-11)
45*(y-11)
50*(y-11)
55*(y-11) *****
etc.
the multiples of 55*(y-11) are:
55*(y-11)
1108(y-11) *****
etc.
to get the multiples, you just keep multiplying the original number by 1, then 2, then 3, then 4, etc.
alternatively, you just keep adding the number to itself.
the smallest multiple that can possibly be common to both would be 55 * (y-11)
5 * (y-11) * 11 = 55 * (y-11)
the smallest multiple therefore has to be 55 * (y-11)
you could have simplified the numbers together to make sure you were not doing anything wrong.
you would have gotten 5 * (y-11) = 5y - 55
you would also have gotten 55 * (y-11) = 55y - 605
your multiples for 5y - 55 would have been:
5y-55
10y-110
15y-165
20y-220
25y-275
30y-330
35y-385
40y-440
45y-495
50y-550
55y-605 *****
etc.
the common multiples for 55y-605 would have been:
55y-605 *****
110y-1210
etc.
you would get the same answer.
i believe your answer has to be that the least common multiple of 5*(y-11) and 55*(y-11) has to be 55*(y-11).
fyi:
the least common multiple has to be exactly divisible by each of the numbers involved.
in your problem the least common multiple is 55*(y-11)
that is exactly divisible by 5*(y-11). it divides into it exactly 11 times.
that is also exactly divisible by 55*(y-11). it divides into it exactly 1 time.
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