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

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

The number of consecutive zeros in 2^3xx\ 3^4xx\ 5^4xx\ 7, is

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

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

The number of divisors of the natural number N where N=2^(5)x3^(5)x5^(2)x7^(3) of the form 4n+2,n in W is

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}

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

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

For any natural number n, (15xx5^(2n))+(2xx2^(3n)) is divisible by