The total number of factors of `2^3xx6^7xx3 9^5` is equal to

Find the total number of prime factors in 2^(17) xx 6^(31) xx 7^(5) xx 10^(11) xx 11^(10) xx (323)^(23) .

The number of factors of a number N=2^(3)xx3^(2)xx5^(3) is

Look at the following pattern : (i) 28 = 2^(2) xx 7^1 , Total number of factors (2+1) (1+1)= 3 xx 2 =6 28 is divisible by 6 factors i.e., 1,2,4,7,14,28 (ii) 30=2^(1) xx 3^(1) xx 5^(1) , Total number of factors ( 1+1)(1+1)(1+1)=2 xx 2 xx 2 =8 30 is divisible by 8 factor i.e., 1,2,3,5,6,10,15,30 find pattern

Look at the following pattern : (i) 28 = 2^(2) xx 7^1 , Total number of factors (2+1) (1+1)= 3 xx 2 =6 28 is divisble by 6 factors i.e., 1,2,4,7,14,28 (ii) 30=2^(1) xx 3^(1) xx 5^(1) , Total number of factors ( 1+1)(1+1)(1+1)=2 xx 2 xx 2 =8 30 is divisble by 8 factor i.e., 1,2,3,5,6,10,15,30

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

If (1, 3), (2, 5) and (3, 3) are the three elements of A xx B and the total number of elements in A xx B is 6 then the remaining elements of A xx B are

If (1,3),(2,5) and (3,3) are the three elements of A xx B and the total number of elements in A xx B is 6 the remaining elements of A xx B are