A hard (but very interesting!) math puzzle... help!?

Question

Find the largest integer which is both an n-digit number and an nth power.

If you find the answer, let me know how you got it as well as what it is! 10 points for the correct answer with the clearest explanation.

Thank you.

Answer 1

Start with one digit numbers. The largest is 9^1. 10 would be too much.
Next you need a two digit number and a 2nd power. The largest here would be 9² = 81, because 10² will be too much.
Continuing with three digits, you need a cube that is less than 1000. 9^3 = 729. Again 10^3 is too much.
With four digits you need a 4th power that is less than 10,000. Again you have 9^4 = 6561.
Continuing the pattern...

1 --> 9
2 --> 81
3 ---> 729
4 ---> 6,561
5 ---> 59,049
6 ---> 531,441
7 ---> 4,782,969
8 ---> 43,046,721
9 ---> 387,420,489
10 ---> 3,486,784,401
11 ---> 31,381,059,609
12 ---> 282,429,536,481
13 ---> 2,541,865,828,329
14 ---> 22,876,792,454,961
15 ---> 205,891,132,094,649
16 ---> 1,853,020,188,851,841
17 ---> 16,677,181,699,666,569
18 ---> 150,094,635,296,999,121
19 ---> 1,350,851,717,672,992,089 (19 digits)
20 ---> 12,157,665,459,056,928,801 (20 digits)
21 ---> 109,418,989,131,512,359,209 (21 digits)
22 ---> 984,770,902,183,611,232,881 (21 digits)

Notice that you get to 9^21 which is 21 digits. After that you have 9^22 but it is only 21 digits so it won't work.

Therefore answer is 9^21 which equals 109,418,989,131,512,359,209

You can also solve this using logarithms.
We know the following:
9^1 > 10^0
9^2 > 10^1
...
In general, we are looking for largest value of n where:
9^n > 10^(n-1)

If you take the log of both sides you get:
log(9^n) > log(10^(n-1))

One of the rules of logs is that:
log(a^b) = b log a
So you have:
n log(9) > (n-1) log(10)

And log(10) is just 1:
n log(9) > n - 1

Now subtract n from both sides. Addition and subtraction don't affect the direction of the inequality.
n log(9) - n > -1

Now factor out n.
n(log(9) - 1) > -1

Finally divide both sides by (log(9) - 1). This is a negative value so it will switch the direction of the inequality.
n < -1 / (log(9) - 1)

Equivalently you could write this as:
n < 1 / (1 - log(9))

If you plug this in your calculator you get:
n < 21.85434533...

So n must be the largest integer less than 21.8543... That would be n = 21.

In other words, you'll have a 21 digit number when you have 9^21 (109,418,989,131,512,359,209), but fewer digits with any exponent higher than 21.

Answer 2

Since you want the largest integer that ch is both an n-digit number and an nth power, you have to pick 9 as the base of the exponential. The largest number is 9^21.

Answer 3

10^n is an n+1 digit number, so the number must be a power of some number less than ten. Choose 9 because you can multiply it by itself the most times before you have n-1 digits.

9^21 is the larger power of nine that is both 9^n and an n-digit number.

109,418,989,131,512,359,209 is the 21-digit number 9^21.

Answer 4

10^n is 1 with n zeroes after it. i.e. it's an n + 1 digit number. So it must be 9 to a power.

The highest number is one which begins with less than 1111.... as anything higher than this will multiply by 9 to give more than 10, which will add another digit.

Answer 5

When you say nth power, do you mean that it is the nth power of an integer?

so you are looking for a two digit number that is also a square of some other integer?