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CHAPTER 2 L.C.M. AND H.C.F.
1 Common Factor A common factor of two or more numbers is a number that divides each of them exactly. Ex. Let the numbers be 27 and 81. In this 3 is common factor. 2. Common Multiple A common multiple of two or more numbers is a number which is exactly divisible by each of them. Ex. 12 is common multiple of numbers 4 and 6. 3. L.C.M. L.C.M. stands for least common multiple. L.C.M. of two or more given numbers is the least number which is exactly divisible by each one of them. Thus, 20 is a common multiple of 4 and 5. 40 is a common multiple of 4 and 5. 60 is a common multiple of 4 and 5. But 20 is the least common multiple (L.C.M.) of 4 and 5 2 1, 2, 6 3

 L.C.M of 8, 24 is 2 × 2 × 2 × 3 × 1 = 24. Ex. 3. Find the L.C.M of 12, 15, 90, 108, 135 and 150. Sol. 2 3 3 5 12, 15, 90, 108, 135, 150 45, 54, 135, 75 18, 45, 25 6, 15, 25 6, 3, 5 • • • • ... (ii) ... (iii) ... (iv) ... (v) ... (i)

in row (i) 12 and 15 are factors of 108 and 90 respectively. Hence we struck off 12 and 15. in row (ii) 45 is factor of 135, therefore 45 is struck off. in row (v) 3 is factor of 6, therefore 3 is struck off. The required L.C.M. = 2 × 3 × 3 × 5 × 6 × 5 = 2700.

Methods of finding L.C.M Method-I. By Prime factorization
Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of the highest powers of all the factors. Ex. 1. Find the L.C.M. of 72, 108 and 2100. Sol. 72= 2 × 2 × 2 × 3 × 3 = 23 × 32 108= 3 × 3 × 3 × 2 × 2 = 33 × 22 2100= 2 × 2 × 5 × 5 × 3 × 7 = 22 × 52 × 3 × 7 L.C.M. of 72, 108 and 2100 = Product of highest powers of 2, 3, 5 and 7.  L.C.M. of 72, 108 and 2100 = 23 × 33 × 52 × 7 = 37800.

4.H.C.F. H.C.F. stands for Highest Common Factor or Greatest Common Multliple, i.e., G.C.M. H.C.F. of two or more numbers is the greatest number that divides each of them exactly. Thus, 7 is the H.C.F. of 21 and 28. Because there is no number greater than 7 that divides both 21 and 28. ... (1) Methods of Finding H.C.F. Method I-By ... (2) factorization Prime Express each one of the given numbers as the product of ... (3) prime factors. The product of least powers of common prime factors gives the H.C.F. Ex. 4. Find the H.C.F of 48 and 66 Sol. 48 = 2 × 2 × 2 × 2 × 3 = 24 × 3 66 = 2 × 3 × 11

Method-II. Short Cut method.
Write down the given numbers in a line separating each of them by commas. Divide by any one of the prime numbers 2, 3, 5, 7, etc which will exactly divides at least any two of the given numbers. Set down the quotients and the undivided numbers in a line below the first. Repeat the process until you get a line of numbers which are prime to one another. The product of all divisors and the numbers in the last line will be the required L.C.M. Ex. 2. Find the L.C.M of 8, 24. Sol. 2 8, 24 2 4, 12

H.C.F. = 2 × 3 = 6 Ex. 5. Find the H.C.F. of 70 and 42. Sol. 70 = 2 × 5 × 7 42 = 2 × 3 × 7 H.C.F. = 2 × 7 or H.C.F = 14. Ex. 6. Find the H.C.F. of 16, 18, 20 and 36. Sol. 16 = 2 × 2 × 2 × 2 18 = 2 × 3 × 3 20 = 2 × 2 × 5

1

36 = 2 × 2 × 3 × 3 In above eqns. (i), (ii), (iii) and (iv) only one 2 is common in all four sets of prime factors of the given numbers. Hence H.C.F. is 2.

or

... (iv) b = 240 × 24 24 × [ a = 24] or b = = 240 Second number is 240.

SOLVED EXAMPLES Method-II. By division method
In this method divide the greater number by the smaller numbers divide the divisior by the remainder, divide the remainder by the next remainder and so on, until no remainder is left. The last divsior is the required H.C.F. Ex. 7. Find the H.C.F. of 513, 1134 and 1215. Sol. 1134 )1215 ( 11134 81 ) 1134 ( 14 810 324 324 ××× H.C.F. of 1134 and 1215 is 81. Required H.C.F. = H.C.F. of 513 and 81 81 ) 513 ( 6 486 27 ) 81 ( 3 81 ×× H.C.F of the given numbers = 27. 5. Special formulae on H.C.F and L.C.M (1) First number × Second number = L.C.M. × H.C.F. (2) H.C.F. and L.C.M. of fractions (i) L.C.M. = The greatest length will be the H.C.F of  Sol. exact number...
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