What is the mean of all three-digit positive integers whose digits are in the set {2, 0, 1, 9}?

Solutions

List all the numbers

A straightfoward way is to list all possible 3-dgit numbers that have digits 2, 0, 1, and 9 only, and then calcuate the mean of all the numbers.

Since we consider 3-digit numbers only, the hundreds digit cannot be 0. All the numbers we list should start with 2, 1, or 9 (having 2, 1, or 9 at the hundreds place).

Let us start with numbers that have 1 as the hundreds digit. Note that a digit can appear more than once. For example, 100 should be included because all the digits in it, 1, 0, and 0, are in the set {2, 0, 1, 9}.

Hundreds Tens Units
1 0 0
1 0 1
1 0 2
1 0 9
1 1 0
1 1 1
1 1 2
1 1 9
1 2 0
1 2 1
1 2 2
1 2 9
1 9 0
1 9 1
1 9 2
1 9 9

There are a total of 16 such kind of numbers. Why? Because there are four choices at the tens place and four choices at the units place.

You may continue to list numbers that have 2 or 9 in the hundreds place. Each gives you 16 numbers. The pattern in the tens and units places are the same. In total you will have 48 3-digit numbers that all their digits are in the set {2, 0, 1, 9}. You can then compute the mean.

However, it will take a while to add up 48 numbers, even if you use a calculator. Is there a faster way?

Find the mean

A faster way to find the mean of the 48 numbers is to find the digits in the mean one by one.

If you look at the numbers in the table carefully, you can see that 0, 1, 2, and 9 appear the same number of times, four times, in the tens and units place. If you count the nubmers that start with 2 and 9, each of 0, 1, 2, and 9 appears 12 times. Therefore, the units digit of the mean is the mean of 0, 1, 2, and 9. It can be calculated as:

\[\frac{(0 + 1 + 2 + 9) \times 12}{4 \times 12} = \frac{0 + 1 + 2 + 9}{4} = 3.\]

The tens digit of the mean can be calculated similarly because 0, 1, 2, and 9 appear the same number of times at the tens place. It is 3.

The hundreds digit of the mean is a little different becuase 0 is not there. Each of 1, 2, and 9 appears 16 times. So the hundreds of digit of the mean is

\[\frac{(1 + 2 + 9) \times 16}{3 \times 16} = \frac{1 + 2 + 9}{3} = 4.\]

Now, we have figured out all the digits in the mean. The answer is 433.

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