Problem 2 · AMC 8 Stretch
Core
Logic & Word Problems
Counting & ProbabilityNumber Theory
account-for-all-possibilitiesreduce-and-expandcounting-principle
How many \(2\)-digit whole numbers are there? Then generalize: how many \(n\)-digit whole numbers are there, where \(n\) is a whole number bigger than \(1\)? (Hint to start small: first try counting using only the digits \(0\) and \(1\), then using \(0,1,2,3\).)
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Answer: 90 two-digit numbers; in general 9 × 10^(n−1)
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Hint 1 of 4
There are two nice ways to do this. You can count a range of numbers, or you can count digit by digit using the multiplication (counting) principle.
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Hint 2 of 4
Range way: the smallest \(2\)-digit number is \(10\) and the biggest is \(99\). How many whole numbers are there from \(10\) to \(99\), counting both ends?
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Hint 3 of 4
Counting way: the first digit can't be \(0\) (or it wouldn't be a \(2\)-digit number), so how many choices does it have? The second digit can be anything \(0\) through \(9\). Multiply the two counts.
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Approach: Count a range, or use the multiplication counting principle
- Way 1 (count a range): the whole numbers from \(a\) to \(b\) number \(b-a+1\). The 2-digit numbers go from 10 to 99, so there are \(99-10+1=90\).
- Way 2 (counting principle): the first digit can be \(1,\dots,9\) (9 choices, no leading 0) and the second digit can be \(0,\dots,9\) (10 choices), giving \(9\times10=90\).
- Generalizing to \(n\) digits: 9 choices for the leading digit and 10 for each of the other \(n-1\) digits, so \(9\times10^{\,n-1}\).
- Check: \(n=2\) gives \(9\times10=90\); \(n=3\) gives \(9\times100=900\), matching the three-digit numbers from 100 to 999.
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