Lesson 14ª

 

 

 

 

 

   

HOW DO WE CALCULATE THE L.C.M. FOR MORE THAN THREE NUMBERS?

1st.  Calculate the l.c.m. for the first two numbers.
2nd.  Calculate the l.c.m. for the third number and the l.c.m. for the number found in the previous step.
3rd. Take the fourth number and the l.c.m. from the previous step, and so on successively.

3.77   Calculate the l.c.m. for (11,111,1111,11111)

1st.   We calculate the l.c.m. for 11 and 111

11=11
111=3x37
l.c.m. (11,111) = 3x11x37 = 1221

2nd. We calculate the l.c.m. for 1111 and 1221
1111=11x101
1221=3x11x37
l.c.m. (1111,1221) = 123321

3rd. We calculate the l.c.m. for 11111 and 123321
11111=41x271
123321=3x11x37x101
l.c.m. (11111,123321) = 1370219631

As we can see, the lowest number that can be divided by 11, 111, 1111 and 11111 and rendering exact quotients is a very high number. In these divisions, the remainder is zero.

3.78  Calculate the l.c.m. for (12,42,126)
Answer: the l.c.m. for (12,42) = 84; the l.c.m. for (84,126) =

h

WHAT IS THE USE OF KNOWING HOW TO CALCULATE THE LOWEST COMMON MULTIPLIER?

They help us solve some problems:

3.79   Four friends live in a town and study in different places. The first one comes home every 6 days; the second one, every 8 days; the third friend, every 10 days; and the fourth and final friend, every 12 days. If today is March 7th, and the four friends are in town, when will they all be in town at the same time again?

Solution:
We have to find a number, the smallest one, which is a multiple of 6, 8, 10 and 12. As we can see, we have to calculate the l.c.m. for (6,8,10,12).

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They shall see each other again 120 days after. because this number can be divided exactly by 6, 8, 10 and 12.

Since we are using March 7th as today's day:
We have 31 – 7 = 24 days left in March
                              April…………..30
                              May…………...31
                              June…………...30
Total of days up to June 30th: 24+30+31+30 = 115 days

We still have left: 120 – 115 = 5 days. Then, we count 5 days after June 30th. They shall meet again on July 5th.

3.80  Could we load a truck with machines weighting 625 kilos each or with iron sheets of 500 kilos each or with iron beams of 80 kilos each?

Answer: Yes

Solution:

We calculate a multiple, the smallest one, that contains 625, 500 and 80 at the same time. This means, the lowest common multiple for 625, 500 and 80:

We calculate first the l.c.m. for (625,500):

g

j

With a weight of 10000 kilos, we can transport:

k

3.81 Three copper cables measuring 110, 90 and 75 metres each divided into pieces of equal length without any left over. What would be the length of each piece?

Answer: each piece should be 5 metres long.

Solution:
We have to calculate the highest common denominator for 110, 90 and 75.

The h.c.d. for (110,90,75) = 5.

If we divide 110, 90 and 75 in 5 metre pieces, we see that the length of all cables is used. We have no remainder. t