Page 163 - New Medley 5 Semester 1

∴ HCF of 24 and 40 = 2 × 2 × 2 = 8

To use this method, first, find the prime factorisation of all

the given numbers and then take the primes which appear
the least number of times in any prime factorisation for

finding HCF.

Here, 2 appears 3 times in each case, so take three 2s.

3 appears 1 time in first and 0 times in second, so leave it.

5 appears 0 times in first and 1 time in second, so leave it.

Finally, HCF of 24 and 40 = 2 × 2 × 2 = 8

Method III

24 40 1 Divide the bigger number by the smaller

– 24 number. If the remainder is left, divide
16 24 1 the first divisor by it (i.e., remainder).

– 16 If the remainder is left, divide the last
8 16 2 divisor by it. Continue this process till

– 16 the remainder 0 is left. The final divisor
0 is the HCF of the given numbers.
HCF
∴ HCF of 24 and 40 is 8.

Note that if three numbers are given, perform the process
with any two numbers first and the HCF of first two with

the third number. The final HCF (divisor) is the required
HCF.

Example 4: Find the HCF of 60, 90 and 110.

Solution: 260 290
230 345
315 315

5 5 5 5

1 1

∴ 60 = 2 × 2 × 3 × 5 ∴ 90 = 2 × 3 × 3 × 5

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