How many 2-digit numbers are there that contain at least one 2?
How many 2-digit numbers are there that contain at least no 2 at all?
How many 3-digit numbers are there that contain at least one 3?
How many 5-digit numbers are there that contain at least one 5?
This problem is about learning how to count without counting, because the possibilities are too numerous. To solve the problem, students will need to apply a systematic approach, logic and reasoning about our number system and its patterns, and algebra. In so doing they will discover more about number, and recognize the efficiency of using an algebraic approach.
How many r-digit even numbers are there?
Is there another way to do this though? After all, if we were asked to find the number of 7-digit numbers that contained 2, we would have to produce a very long list.
There are 9 numbers that start with a 1; there are 9 numbers that start with a 3; there are …; there are 9 numbers that start with a 9. So we have 9 times 8 numbers altogether. The 8 comes from the fact that there are 8 numbers in the sequence 1, 3, 4, 5, …, 9. That means that there are 72 numbers that don’t have a 2 in them.
Consider: What have we found so far? There are 18 2-digit numbers that have a 2 and there are 72 2-digit numbers that don’t. 18 + 72 = 90. is that familiar? Surely there are 90 2-digit numbers altogether? So we were wasting our time when we started listing and then counting, the 2-digit numbers without a 2. Subtracting 18 from 90 is more straight-forward.
First of all count all 3-digit numbers, then count all 3-digit numbers with no threes, then subtract the second number from the first. First the 3-digit numbers: There are 9 possible digits for the first place (you can’t use 0), 10 for the second place (you can use anything from 0 to 9) and 10 for the third. That makes 9 x 10 x 10 = 900.
For the 3-digit numbers with no threes: Their first (hundreds) digit can be chosen in just 8 ways (no 0 and no 3), their second digit in just 9 (no 3 remember), and their third digit in 9 ways. So there are 8 x 9 x 9 of these. That’s 648 altogether.
So the number of 3-digit numbers with at least one three is 900 - 648 = 252. (That would have been a long list!)
We couldn’t possibly write down a list here. Though, if you are stuck at this point, then writing down the list for r = 2 and r = 3 might give you some inspiration.
When is a number even? It is even if it ends in 0, 2, 4, 6, or 8. This means that we have a choice of 5 numbers for the last digit. Once again we have a choice of 9 for the first digit, then 10 for the next digit, then 10 for the next, … , and then 5 for the last digit. So we have to multiply together one 9, one 5 and r – 2 10s. This gives 5 x 9 x 10r-2. That’s 45 with r – 2 zeros.
Printed from https://meaningfulmaths.nt.edu.au/mmws/nz/resource/how-many-numbers at 8:59pm on the 26th February 2024