How many n-digit numbers?

Algebra Level 2

How many $n$ -digit positive numbers are there?

9 \times 10^{n-1}

9 \times 10^{n-1} - 1

9 \times 10^{n}

10^{n}

2 solutions

Chew-Seong Cheong
May 10, 2017

The largest natural number up to $n$ -digits $N_n = \underbrace {999...999}_{\text{\# of 9s = n}} = 10^n-1$ .

Likewise, the largest natural number up to $n-1$ -digits $N_{n-1} = 10^{n-1}-1$ .

Therefore the number of $n$ -digit numbers $N_{n\text{-digit}}=N_n-N_{n-1}=10^n-1-(10^{n-1}-1) =10^n-10^{n-1}=\boxed{9\times 10^{n-1}}$ .

M Zadeh
May 9, 2017

The largest n-digit number is 999..... (9 written n times), and the smallest is 1000....( zero writen (n-1) times. So 999.... -1000...= 8999... (9 written n-1 times). Then we need to add 1 So, the answer is 8999....+1=9000... (zero written n-1 times, which is 9X10^(n-1)

It might be best to specify that the numbers are positive, since for $n = 1$ , if we included $0$ as a 1-digit number there would be $10$ 1-digit numbers in total, instead of the $9 \times 10^{1-1} = 9$ that the formula would indicate.

Brian Charlesworth - 4 years, 1 month ago

How many n-digit numbers?

2 solutions

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