Don't expect me to do these calculations!

You all must have noticed that

$10^{1} =$ The first appearance of a 2 digit number

$10^{2} =$ The first appearance of a 3 digit number

and so on

$10^{x-1} =$ The first appearance of a $x$ digit number .

So , to find the number of digits in $21!^{1000}$ , we do the following calculations $21!^{1000} = 10^{x-1} \\ log_{10} 21!^{1000} = (x-1) \\ 1000\cdot log_{10} 21! = (x-1) \\ 1000\cdot 19.70834 = (x-1) \\ \Rightarrow x=19709$

NOTE : Here even if $x-1$ evaluated out to be $19.709.9$ , we wouldn't round it to $19710$ since we are talking about the number of digits , not it's value .

There are many ways to solve this question. Two of them are : 1) An efficient code of c++; :P // generic solution template <class T> int numDigits(T number) { int digits = 0; if (number < 0) digits = 1; // remove this line if '-' counts as a digit while (number) { number /= 10; digits++; } return digits; }

// partial specialization optimization for 32-bit numbers template<> int numDigits(int32 t x) { if (x == MIN INT) return 10 + 1; if (x < 0) return numDigits(-x) + 1;

if (x >= 10000) {
    if (x >= 10000000) {
        if (x >= 100000000) {
            if (x >= 1000000000)
                return 10;
            return 9;
        }
        return 8;
    }
    if (x >= 100000) {
        if (x >= 1000000)
            return 7;
        return 6;
    }
    return 5;
}
if (x >= 100) {
    if (x >= 1000)

2) If no. of digits = x. Then,

$10^{x-1}$ = $21!^{1000}$ Taking log_10 on both sides,

x-1 = 1000*log(21!) Using a calculator, log(21!) = 19.708344 :P Therefore, x-1 = 19708.344 x = 19709.344 Taking the truncate value (not the round value) x = 19709

3) Another way is use of MySql code, Use Select length(21!^1000) from dual; But first convert 21!^1000 to integer. Enjoy !!!!!!!