Lowest common multiple (LCM) of two or more numbers is the smallest number which is exactly divisible by all of them e.g.,
LCM of 5, 7, 10 = 70
LCM of 2, 4, 5 = 20
LCM of 11, 10, 3 =330
Factorization Method to find LCM
To find LCM of the given numbers, first resolve all the numbers into their prime factors and then the LCM is the product of highest powers of all the prime factors.
Example: Find the LCM of 40, 120 and 380
Solution: 40 = 4 x 10 = 2 x 2 x 2 x 5 = 23 x 51
120 = 4 x 30 =2 x 2 x 2 x 5 x 3 = 23 x 51 x 31
380 = 2 x 190 = 2 x 2 x 95 = 2 x 2 x 5 x 19 = 22 x 51 x 191
∴ Required LCM = 23 x 51 x 31 x 191 = 2280.
Division method to find LCM
Write the given numbers separately. Then divide by 2 and write the result below the numbers divisible by 2. If it is not divisible by 2 then try with 3, 5, 7, .... etc. Leave the others (those not divisible) untouched. Do the same for all steps till you get 1 as the remainder in each column.
Example: Find the LCM of 6, 10, 15, 24 and 39.
Solution:
Note: If we have to find the least number which is exactly divisible by p, q and r then the required number = LCM of p, q and r.
Example: Find the least number that is exactly divisible by 6, 5 and 7.
Solution: Required number = LCM of 6, 5 and 7 =210.
If we are to find the least number which when divided by p, q and r leaves the remainders a, b and c respectively, then if it is observed that, (p -a) = (q -b) = (r -c) = k (say), then
The required number = LCM of (p, q and r) - (k)
Example: Find the least number which when divided by 6, 7 and 9 leaves the remainders 1, 2 and 4 respectively.
Solution: Here, (6 -1) = (7 - 2) = (9 - 4) = 5
Required number = (LCM of 6, 7 and 9) - 5 = 126 - 5 = 121.
If we have to find the least number which when divided by p, q and r leaves the same remainder 'a' each case, then required number = (LCM of p, q and r) + a.
Example: Find the LCM of 25 and 35 if their HCF is 5.
Solution:
LCM =
Product of Two Numbers
HCF
25 x 35
5
LCM = 175.
Example: By using the rule that LCM = Product of two numbers ÷ HCF.
Solution: 442 = 2 x 17 x 13 ⇒ HCF = 26 && LCM = (26 x 442) / 26 = 442
Example: Find the LCM of 6, 10, 15, 24 and 39.
Solution:
2 6 10 15 24 39
2 3 5 15 12 39
2 3 5 15 6 39
3 3 5 15 3 39
5 1 5 5 1 13
13 1 1 1 1 13
1 1 1 1 1
LCM = 2 x 2 x 2 x 3 x 5 x 13 = 1560.
Note: If we have to find the least number which is exactly divisible by p, q and r then the required number = LCM of p, q and r.
Example: Find the least number that is exactly divisible by 6, 5 and 7.
Solution: Required number = LCM of 6, 5 and 7 =210.
If we are to find the least number which when divided by p, q and r leaves the remainders a, b and c respectively, then if it is observed that, (p -a) = (q -b) = (r -c) = k (say), then
The required number = LCM of (p, q and r) - (k)
Example: Find the least number which when divided by 6, 7 and 9 leaves the remainders 1, 2 and 4 respectively.
Solution: Here, (6 -1) = (7 - 2) = (9 - 4) = 5
Required number = (LCM of 6, 7 and 9) - 5 = 126 - 5 = 121.
If we have to find the least number which when divided by p, q and r leaves the same remainder 'a' each case, then required number = (LCM of p, q and r) + a.
Example: Find the LCM of 25 and 35 if their HCF is 5.
Solution:
LCM =
Product of Two Numbers
HCF
25 x 35
5
LCM = 175.
Example: By using the rule that LCM = Product of two numbers ÷ HCF.
Solution: 442 = 2 x 17 x 13 ⇒ HCF = 26 && LCM = (26 x 442) / 26 = 442
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