We get the multiple of 7 only when the input is from \(\sqrt{49}\) to \(\ \sqrt{63}\).

exponent of 7 = (63-49)+1 = 15

C

Input: x= 1^(1/2), 2(1/2), 3^(1/2).....
g(x)= Y = greatest integer less than equal to X

g(1^(1/2))= 1; g(2^(0.5)= 1; g(3^0.5)= 1
so product of outputs of x from 1^0.5 to 3^0.5= 1 * 1* 1= 1^3

Similarly for next set of numbers
g(2^0.5)= 2; g(5^0.5)= 2; g(6^0.5)=2......g(8^0.5)= 2
there for product = 2*2*2*2*2= 2^5

The output value will change to next integer value every time with reach a perfect square (4,9,16,25....100)
and the powers of the integers increases in odd number
==> 1 repeats 3 time= 1^3
2 repeats 5 times= 2^5
3 repeats 7 times => 3^7
.
.
.
So on,
Therefore the product of all outputs= 1^3 X 2^5 X 3^7 X 4^9 X 5^11....... 9 ^ 19 X 10
Simplifying with prime factors only: 2^88 X ..... (I didnt calculate all since there is only one option where 2 has power of 88, OPTION C