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EXAMPLE 7: In a shop, there are three clocks which chime at intervals of 15, 20 and 30 respectively. EXAMPLE 11: Find the HCF and the LCM of 1152 and 1664.
They all chime together at 10 a.m. At what time will they all chime together again? SOLUTION: We first find the HCF of the given numbers.
SOLUTION: Required time = LCM of 15, 20, 30 minutes.
1152 1664 1
5 15, 20, 30
3 3, 4, 6 - 1152
2 1, 4, 2 512 1152 2
1, 2, 1 - 1024
128 512 4
..LCM of 15, 20, 30 = (5 × 3 × 2 × 2) = 60. - 512
So, all the clocks will chime together again after 60 minutes, ×
i.e., after 1 hour, i.e., at 11 a.m.
HCF =128.
PROPERTIES OF HCF AND LCM OF GIVEN NUMBERS And, LCM = Prpoduct of the numbers 1152 × 1664 = 14976.
thier HCF 128
(i) The HCF of a group of numbers ls not greater than any of the given numbers. HCF = 128 LCM = 14976.
(ii) The HCF of two co-primes is 1.
(iii) The LCM of a group of numbers ts not less than any of the given numbers.
(iv) The LCM of two co-primes ts equal to their product.
(v) The HCF of a group of numbers ts always a factor of their LCM. EXAMPLE 12: The HCF of two numbers is 16 and their product ts 3072. Find their LCM.

EXAMPLE 8: Consider the numbers 12, 16, 36, 40. 2 12, 16, 36, 40 SOLUTION: We know that
Clearly, the HCF of the given numbers = 4. And. their LCM
= 2 × 2 × 3 × 2 × 2 × 3 × 5 = 720. 2 6, 8, 18, 20 (i) LCM = product of the given two numbers = 3072 = 192.
Clearly, 4 is a factor of 720. 3 3, 4, 9, 10 their HCF 16
(vi) If a and bare two given numbers such that a is a factor of b 2 1, 4, 3, 10
then their HCF = a and LCM = b 1, 2, 3, 5 EXAMPLE 13: HCF of two numbers is 23 and their LCM is 1449. lf one of the numbers is 161. find the other.
SOLUTION: We know that
EXAMPLE 9: We know that 8 is a factor of 32. Then, clearly HCF of 8 and 32 is 8. And, LCM of 8 and 32 (one number) x (the other number)= (HCF × LCM).
is 32.
(vii) AN IMPORTANT PROPERTY if two numbers are given then the product of the two numbers = Hence, the required number = = ( 23 ×1449
the product of their HCF and LCM. 161 )= 207
EXAMPLE 14: Can two numbers have 16 as their HCF and 204 as their LCM? Give reason.
EXAMPLE 10: Consider the the numbers product 48 of and the 60.

SOLUTION: We know that the HCF of two or more numbers must divide their LCM exactly. But,16 does not
SOLUTION: We have. 48 = 2 × 2 × 2 × 2 ×3 = 2 x 3. divide 204 exactly. So. there can be no two numbers with 16 as their HCF and 204 as their LCM.
4
And, 60 = 2 × 2 × 3 × 5 = 2 × 3 × 5.
2
So, the HCF of 48 and 60 is 2 × 3 = 12. JUST TRY 1: Find the LCM of 36, 72, 96 by the prime factorisation method.
2
And, the LCM of 48 and 60 is 2 × 3 × 5 = 240. JUST TRY 2: Find the LCM of 25, 75 and 100 by the prime factorisation method.
4
Now, the product of the given numbers = 48 × 60 = 2880. JUST TRY 3: Find the LCM of 72, 90 and 108 by the common division method.
Product of their HCF and LCM= (12 × 240) = 2880. JUST TRY 4: Find the LCM of 36, 72, 96 by the prime factorisation method.
:. product of two numbers = (their HCF) × (their LCM). JUST TRY 5: Find the LCM of 25, 75 and 100 by the prime factorisation method.
REMARKS: Thus, for any two given numbers, we have: JUST TRY 6: Find the LCM of 72, 90 and 108 by the common division method.
(one number) × (the other number) JUST TRY 7: Find the LCM of 220, 440 and 660 by the long division method.
(i) LCM =
their HCF JUST TRY 8: Find the smallest number which when divided by 20, 25 and 40 leaves a remainder 5 in each
(one number) × (the other number) case.
(ii) HCF =
their LCM
The Relationship Between Two Numbers and their HCF and LCM
First number. second number = HCF . LCM