You know what makes a number large: it’s more number of digits and greater digits in each place value. Based on that, if you want to have the largest two-digit number for example, you’d simply take a two-digit number and fill both the place values with 9’s because the digit 9 has the highest value among all the digits in the decimal number system.
In the same way, if you were looking for the smallest three-digit number, you’d simply take three digits. The 1’s digit, you could simply fill it with a zero and do the same thing for the 10’s place value. The last place value, 100’s, could not be filled with a 0 otherwise you’d end up with a 000 which is practically a 0 (zero).
So the greatest two-digit number for instance becomes 99, the greatest three-digit number, 999, and so on. The smallest two-digit number is 10, the smallest three-digit number, 100, and so on.
The interesting thing here is that if you,
- 9 (largest one-digit number) +1 = 10 (smallest two-digit number) = 10 * 1 = 10 * 10^0
- 99 (largest two-digit number) + 1 = 100 (smallest three-digit number) = 10 * 10 = 10 * 10^1
- 999 (largest three-digit number) + 1 = 1000 (smallest four-digit number) = 10 * 100 = 10 * 10^2
- 9999 (largest four-digit number) + 1 = 10000 (smallest five-digit number) = 10 * 1000 = 10 * 10^3
- and so on …
- largest n-digit number + 1 = smallest (n+1)-digit number = 10 * 10^(n-1)
What that means is that, if you add 1 to the largest n-digit number, you’ll always get the smallest (n+1)-digit number. It also means that if you add 1 to the largest n-digit number, that can be expressed as [10 * 10^(n-1)].
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