It is known that n = 10^100 - 1 has 100 digit 9. How many digit 9 are there in n^3?

Answer 1

Ian is almost right... at least his method works well.

n^3 = (10^100 - 1)^3
= 10^300 - 3*10^200 + 3*10^100 - 1

You can break this down into:
(10^300 - 1) - 3*10^200 + 3*10^100.

10^300 would be a one followed by 300 zeroes. But we subtract one making it a number with 300 nines.

Subtracting 3*10^200 would leave the first 200 digits untouched (counting from the right) but change the 201st nine (from the right) into a 6.

Adding 3*10^100 would leave the first 100 digits untouched but change the 101st nine (from the right) into a 2, and would carry a 1.

Here's where Ian made the mistake. The 1 would carry forward to 102nd digit, which would be come 0 and carry to the 103rd digit, etc. This would carry all the way to the 201st digit. At this point the 6 would become a 7 and stop the carrying.

The number looks like:
{9999.... 9999}7{0000.... 0000}2{9999.... 9999}

Or stated another way:
{ninety-nine 9s}7{ninety-nine 0s}2{one hundred 9s}

There are 99 nines in the first set and 100 nines in the last set. So all together 199 nines.

The other answers are also a good way to see this. Rather than breaking it the way Ian did, do it like this:

(10^100 - 3)*10^200 + (3*10^100 - 1)

The first number in parentheses would be a 1 with one-hundred 0s, but we subtract 3. So it becomes 99999....9997 (ninety-nine 9s followed by a 7). This is shifted over 100 places to the left.

The other part would be a 3 followed by 100 zeros, but again we subtract 1. This makes it 299999....9999 (2 followed by one-hundred 9s).

Again that looks like:
{ninety-nine 9s}7{ninety-nine 0s}2{one hundred 9s}

And we again see 99 + 100 = 199

Answer 2

n^3
= 10^300 -- 3*10^200 + 3*10^100 -- 1
= (10^300 -- 1) -- 3*10^200 + 3*10^100
now (10^300 -- 1) has 300 digits 9 but one digit 9 will turn to 6 by -- 3*10^200 and that 6 will turn to 7 and all 9s to the right into 0 upto 101th from the right.
so total digits 9 in the number remaining
= 99 + 100, rest one is 7, one is 2 and 99 are 0.
so only 199 nines are there.

Answer 3

n^3 = (10^100 - 1)^3
= 10^300 - 3*10^200 + 3*10^100 - 1
So this has 300 digits.
The 200th digit has 3 subtracted from it so it is not 9.
The 100th has 3 added to it and it carries over to the 101st which carries over to the 102nd which carried over and voer until the 200th digit.
So the number 9 appears from the first digit to the 99th digit then from the 201st digit to the 300th digit.
So the number 9 appears 199 times.

This can also be seen from the pattern:
(10^1 - 1)^3 = 9^3 = 729 ... one nine
(10^2 - 1)^3 = 99^3 = 970299 ... three nines
(10^3 - 1)^3 = 999^3 = 997002999 ... five nines
(10^4 - 1)^3 = 9999^3 = 999700029999 ... seven nines
Looking at the pattern there are 2n-1 nines.

Answer 4

n = 10^100 - 1

n^3
= (10^100 - 1)^3
= 10^300 - 3*10^200 + 3*10^100 - 1
= 10^200 (10^100 - 3) + 3*10^100 - 1

answer
= 99 + 100
= 199

Answer 5

199