120 can be expressed as a product of its prime factors as
(a) `5 xx 8 xx 3`
(b) `15 xx 2^3`
(c) `10 xx 2^2 xx 3`
(d) `5 xx 2^3 xx 3`

252 can be expressed as a product of primes as 2xx2xx3xx3xx7( b) 2xx2xx2xx3xx7 (c) 3xx3xx3xx3xx7 (d) 2xx3xx3xx3xx7

Find the LCM of (2^3 xx 3 xx 5^2 xx 7) , (2^4 xx 3^2 xx 5 xx 72 xx 1) and (2 xx 3^3 xx 5^4) .

Find the factors of 330. (a) 2xx4xx5xx11 (b) 2xx3xx7xx13(c)2xx3xx5xx13(d)2xx3xx5xx11

The number of zeros at the end of the product of : 2^3 xx 3^4 xx 4^5 xx 5^6 + 3^5 xx 5^7 xx 7^9 xx 8^10 + 4^5 xx 5^6 xx 6^7 xx 7^8 - 10^2 xx 15^3 xx 20^4 is :

HCF of (2^(3)xx 3xx 5) , (2^(2) xx 3^(3) xx5^(2)) and (2^(4) xx3xx 5^(3) xx7) is