If `n=2^3xx3^4xx5^4xx7`, where n is a natural number, then find the number of consecutive zeros in n

If n=2^(3)xx3^(4)xx5^(4)xx7, then the number of consecutive zeros in n, where n is a natural number,is ( a) 2( b ) 3 (c) 4 (d) 7

If the prime factorization of a natural number n is 2^(3)*3^(2)*5^(2)*6, write the number of consecutive zeros in n .

If n(A)=6 and n(B)=3, then find the number of subsets of A xx B .

If N=12^(3)xx3^(4)xx5^(2) , then the total number of even factors of N is

Assertion (A): 6^(n) ends with the digit zero where n is a natural number. Reason (R):Any number ends with digit zero, if its prime factor is of the form 2^(m)xx5^(n) , where m,n are natural numbers.

The set (0,2,6,12,20) in the set-builder form is (1) {x:x=n^(2)-3n+2," where "n'" is a natural number "&1 (2) {x:x=n^(2)-3n+2," where "'n'" is a natural number "&1 (3) {x:x=n^(2)-3n+4," where "'n'" is a natural number "&1 (4) {x:x=n^(2)+5n-6," where 'n' is a natural number "&1<=n<=5}

If M=2^(2)xx3^(5),N=2^(3)xx3^(4) , then the number of factors of N that are common with factors of M is

A 2n digit number starts with 2 and all its digits are prime,then the probability that the sum of all 2 consecutive digits of the number is prime is a.4xx2^(3n) b.4xx2^(-3n) c.2^(3n) d. none of these

If four xx the number of permutations of n things 3 together is equal to five xx the number of permutations of n-1things 3 together,find n.