HCF of (2^3 × 3^2 × 5), (2^2 × 3^3 × 5^2 ) and (2^4 × 3 × 5^3 × 7) is (a) 30 (b) 48 (c) 60 (d) 105
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Answer:
(c) 60
Step-by-step explanation:
The power of 2 dividing the HCF:
- can't be more then 2^3 since it must divide the first number;
- can't be more than 2^2 since it must divide the second number;
- can't be more than 2^4 since it must divide the third number.
All together, the largest it can be is 2^2.
Likewise, the power of 3 dividing the HCF must be 3^1 since needing to divide the third number means it can't be any larger than this.
Likewise, the power of 5 dividing the HCF must be 5^1 (see the first number).
No other primes (7, 11, 13, etc.) can divide the HCF since, for instance, the HCF must divide the first number and it has no other prime factors.
Therefore the HCF is
2^2 x 3 x 5 = 60
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