H.C.F. AND L.C.M. OF NUMBERS
IMPORTANT FACTS AND FORMULAE
I. Factors and Multiples : If a number a divides another number b
exactly, we say that a is a factor of b. In this case, b is called a multiple
of a.
II. Highest
Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common
Divisor (G.C.D.): The H.C.F.
of two or more than two numbers is the greatest number that divides each of
them exactly.
There are two methods of finding the
H.C.F. of a given set of numbers :
1. Factorization Method : Express each one of the given numbers as the
product of prime factors.The product of least powers of common prime factors
gives H.C.F.
2. Division Method: Suppose we have to find the H.C.F. of two
given numbers. Divide the larger number
by the smaller one. Now, divide the divisor by the remainder. Repeat the
process of dividing the preceding number by the remainder last obtained till
zero is obtained as remainder. The last divisor is the required H.C.F.
Finding
the H.C.F. of more than two numbers : Suppose we have to find the H.C.F. of three numbers. Then, H.C.F. of
[(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given
numbers.
Similarly, the
H.C.F. of more than three numbers may be obtained.
III. Least Common Multiple (L.C.M.) : The least number which is exactly divisible
by each one of the given numbers is called their L.C.M.
1. Factorization Method of Finding L.C.M.: Resolve each one of the given numbers into a
product of prime factors. Then, L.C.M. is the product of highest powers of all
the factors,
2. Common Division Method {Short-cut Method) of
Finding L.C.M.: Arrange the
given numbers in a row in any order. Divide by a number which divides exactly
at least two of the given numbers and carry forward the numbers which are not
divisible. Repeat the above process till no two of the numbers are divisible by
the same number except 1. The product of the divisors and the undivided numbers
is the required L.C.M. of the given numbers,
IV. Product of
two numbers =Product of their H.C.F. and L.C.M.
V. Co-primes: Two numbers are said to be co-primes if
their H.C.F. is 1.
VI. H.C.F. and
L.C.M. of Fractions:
1.H
C F= H.C.F. of Numerators 2.L C M
= L.C.M of Numerators__
L.C.M. of Denominators H.C.F. of Denominators
VII. H.C.F. and
L.C.M. of Decimal Fractions:
In given numbers, make the same number of decimal places by annexing zeros in
some numbers, if necessary. Considering these numbers without decimal point,
find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many
decimal places as are there in each of the given numbers.
VIII.
Comparison of Fractions: Find
the L.C.M. of the denominators of the given fractions. Convert each of the
fractions into an equivalent fraction with L.C.M. as the denominator, by
multiplying both the numerator and denominator by the same number. The
resultant fraction with the greatest numerator is the greatest.
SOLVED EXAMPLES
Ex. 1. Find the
H.C.F. of 23 X 32 X 5 X 74, 22 X 35 X 52 X 73,23 X 53 X 72
Sol.
The prime numbers common to given numbers are 2,5 and 7.
H.C.F. = 22
x 5 x72 = 980.
Ex. 2. Find the
H.C.F. of 108, 288 and 360.
Sol. 108
= 22 x 33, 288 = 25 x 32 and 360 =
23 x 5 x 32.
H.C.F. = 22 x
32 = 36.
Ex. 3. Find the
H.C.F. of 513, 1134 and 1215.
Sol.
1134
81 ) 1134 ( 14 81
324
324
x
\H.C.F. of 1134 and 1215 is 81.
So, Required
H.C.F. = H.C.F. of 513 and 81.
______
81 ) 513 ( 6
__486____
27) 81 ( 3
81
0
H.C.F. of given numbers = 27.
Ex. 4. Reduce 391 to lowest terms .
667
to lowest terms.
Sol.
H.C.F. of 391 and 667 is 23.
On dividing the
numerator and denominator by 23, we get :
391 = 391 ¸ 23 = 17
Ex.5.Find the
L.C.M. of 22 x 33 x 5 x 72 , 23 x 32
x 52 x 74 ,
2 x 3 x 53 x 7 x 11.
Sol. L.C.M. = Product of highest powers of 2, 3,
5, 7 and 11 = 23 x 33 x 53 x 74 x
11
Ex.6. Find the
L.C.M. of 72, 108 and 2100.
Sol. 72 = 23 x 32, 108 = 33
x 22, 2100 = 22 x 52 x 3 x 7.
L.C.M. = 23 x 33 x 52
x 7 = 37800.
Ex.7.Find the L.C.M. of 16, 24, 36 and 54.
Sol.
2
|
16
|
- 24
|
- 36
|
- 54
|
2
|
8
|
- 12
|
- 18
|
- 27
|
2
|
4
|
- 6
|
- 9
|
- 27
|
3
|
2
|
- 3
|
- 9
|
- 27
|
3
|
2
|
- 1
|
- 3
|
- 9
|
2
|
- 1
|
- 1
|
- 3
|
|
\
L.C.M. = 2 x 2 x 2 x 3 x 3 x 2 x 3 = 432.
Ex. 8. Find the
H.C.F. and L.C.M. of 2 , 8 , 16 and 10.
L.C.M of given fractions = L.C.M.
of 2,8,16,10 = 80_
Ex. 9. Find the H.C.F. and L.C.M. of 0.63, 1.05
and 2.1.
Sol.
Making the same number of decimal places, the given numbers are 0.63,
1.05 and 2.10.
Without decimal
places, these numbers are 63, 105 and 210.
Now, H.C.F. of 63,
105 and 210 is 21.
H.C.F. of 0.63,
1.05 and 2.1 is 0.21.
L.C.M. of 63, 105
and 210 is 630.
L.C.M. of 0.63,
1.05 and 2.1 is 6.30.
Ex. 10. Two numbers are in the ratio of 15:11. If
their H.C.F. is 13, find the numbers.
Sol. Let
the required numbers be 15.x and llx.
Then, their H.C.F.
is x. So, x = 13.
The numbers are
(15 x 13 and 11 x 13) i.e., 195 and 143.
Ex. 11. TheH.C.F. of two numbers is 11 and their
L.C.M. is 693. If one of the numbers is 77,find the other.
Sol.
Other number = 11 X 693 =
99
Ex. 12. Find the greatest possible length which
can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.
Sol. Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.
495 = 32 x 5 x 11, 900 = 22
x 32 x 52, 1665 = 32 x 5 x 37.
\H.C.F. = 32 x 5 = 45.
Hence, required length = 45 cm.
Ex. 13. Find the greatest number which on
dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Sol. Required number = H.C.F. of (1657 - 6) and
(2037 - 5) = H.C.F. of 1651 and 2032
_______
1651_______
381 )
1651 ( 4
1524_________
127 ) 381 ( 3
381
0
Required number =
127.
Ex. 14. Find the largest number which divides 62,
132 and 237 to leave the same remainder in each case.
Sol .
Required number = H.C.F. of (132
- 62), (237 - 132) and (237 - 62)
= H.C.F. of 70, 105 and 175 = 35.
Ex.15.Find the
least number exactly divisible by 12,15,20,27.
Sol.
3
|
12
|
- 15
|
- 20
|
- 27
|
4
|
4
|
- 5
|
- 20
|
- 9
|
5
|
1
|
- 5
|
- 5
|
- 9
|
1
|
- 1
|
- 1
|
- 9
|
Ex.16.Find the least number which when divided by 6,7,8,9,
and 12 leave the same remainder 1 each case
Sol. Required number = (L.C.M OF 6,7,8,9,12) + 1
3
|
6
|
- 7
|
- 8
|
- 9
- 12
|
4
|
2
|
- 7
|
- 8
|
- 3
- 4
|
5
|
1
|
- 7
|
- 4
|
- 3
- 2
|
1
|
- 7
|
- 2
|
- 3
- 1
|
\L.C.M
= 3 X 2 X 2 X 7 X 2 X 3 = 504.
Hence required number = (504 +1) = 505.
Ex.17. Find the largest number of four digits exactly
divisible by 12,15,18 and 27.
Sol. The Largest number of four digits is 9999.
Required number
must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999
by 540,we get 279 as remainder .
\Required
number = (9999-279) = 9720.
Ex.18.Find the smallest number of five digits exactly
divisible by 16,24,36 and 54.
Sol. Smallest number of five digits is 10000.
Required number
must be divisible by L.C.M. of 16,24,36,54 i.e 432,
On dividing
10000 by 432,we get 64 as remainder.
\Required
number = 10000 +( 432 – 64 ) = 10368.
Ex.19.Find the least number which when divided by 20,25,35
and 40 leaves remainders 14,19,29 and 34 respectively.
Sol. Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and
(40-34)=6.
\Required
number = (L.C.M. of 20,25,35,40) – 6 =1394.
Ex.20.Find the least number which when divided by 5,6,7,
and 8 leaves a remainder 3, but when divided by 9 leaves no remainder .
Sol. L.C.M. of 5,6,7,8 = 840.
\
Required number is of the form 840k + 3
Least value of k
for which (840k + 3) is divisible by 9 is k = 2.
\Required
number = (840 X 2 + 3)=1683
Ex.21.The traffic lights at three different road crossings
change after every 48 sec., 72 sec and 108 sec.respectively .If they all change
simultaneously at 8:20:00 hours,then at what time they again change
simultaneously .
Sol. Interval of change = (L.C.M of
48,72,108)sec.=432sec.
So, the lights
will agin change simultaneously after every 432 seconds i.e,7 min.12sec
Hence , next
simultaneous change will take place at 8:27:12 hrs.
Ex.22.Arrange the fractions 17 , 31, 43, 59 in the ascending order.
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