1. ## three digit numbers

Here is a challenge problem I read in a recent publication of the MAA for undergraduates. I found it interesting that the author’s way of solving is completely different from mine.

“How many three digit numbers (no leading zeros) are there if the digits are distinct and one of the three digits is the average of the other two”?

2. I tried and got 69...?

Numbers like 987 and 978 are distinct, right? (where 8 is the mean of 9 and 7)

EDIT: Wait! I forgot to remove the non-leading zeroes...

This done, I get 65

3. That is a huge undercount.

4. Ok, back to work then.

5. Originally Posted by Plato
That is a huge undercount.
Ok, I realised I multiplied my numbers by 3 instead of 3!

Now, I get 196.

Maybe that's it?

6. A huge overcount.

7. Oh my

Ok, getting to work again.

8. Ok, this time should be the one.

121.

I counted some intermediates which I shouldn't have earlier such as [(3!x6)+4] while I needed only the ones with odd numbers. >.<

9. You are getting close. But on ring yet.

10. Okay, some limitation to the formula I made then... =/

I'm out of ideas...

This is what I did:
9 8 7 6 5 4 3 2 1 0

First of all, 999, 888, 777, etc making 9 numbers.
Then, I make some sort of 3 pinned fork, grabbing 987, 876, 765, etc giving [(3! x 7) + 4] numbers
Then, I repeat using a larger fork, getting 975, 864, 753, etc giving [(3! x 5) + 4] numbers
Repeat using even larger forks, getting [(3! x 3) + 4] numbers
And [(3! x 1) + 4] numbers

For a total of 9 + 46 + 34 + 22 + 10 = 121

Unless you are not counting the 999, 888, 777, I don't find what I'm missing.

If the triples (999, 888, etc) are indeed included, I'll wait for someone else.

All three digits must be distinct.

12. Must be my english (didn't know what distinct really implied >.<) and removing those I'd get 112!

13. Originally Posted by Plato
Here is a challenge problem I read in a recent publication of the MAA for undergraduates. I found it interesting that the author’s way of solving is completely different from mine.

“How many three digit numbers (no leading zeros) are there if the digits are distinct and one of the three digits is the average of the other two”?

I did it the good'ol way: counting:

Spoiler:
There are 112: begin with 1, then two, three, etc. and count the numbers that can give an integer average (for odds only odds, for evens only evens), and count up: with 1 we get: 132, 153, 174, 195 and all their permutations (6 each), with 2 we get 201 (and only 4 permutations since zero cannot be leading), 243, 264, 285 times 6 each, with 3 there are 354, 375, 396 times 6 each (3 and 1 is already counted for in the list of 1), etc.

Or, as many other times, I'm wrong, of course

Tonio

14. ## An attepmt.

Originally Posted by Plato
Here is a challenge problem I read in a recent publication of the MAA for undergraduates. I found it interesting that the author’s way of solving is completely different from mine.

“How many three digit numbers (no leading zeros) are there if the digits are distinct and one of the three digits is the average of the other two”?
Let $a, b, m$ be three digits, where $m$ is the avergare of $a$ and $b$. The digit $m$ is determined uniquely by $a$ and $b$. Hence we consider the pair $a$ and $b$.

$a, b$ are either both even or both odd, since $(a+b)/2$ is an integer. We can choose $\binom{5}{2}$ pairs from $\{1, 3, 5, 7, 9\}$ and $\binom{4}{2}$ pairs from $\{2, 4, 6, 8 \}$.
Thus, there are $10+6=16$ possible triples $\{a, b, m\}$; the number of rearrangements is $16\cdot3!=96$.

We need to count four pairs ${a, b}$ with zero: $\{0, 2\}, \{0, 4\}, \{0, 6\}, \{0, 8\}$. They determine four triples, each with four possible (not having a leading zero) rearrangements, so there are 16 more numbers. Adding up gives $96+16=112$ possible three digit numbers.

I'm not sure, I'm afraid I might miss something...

15. @Tonio
That is exactly the way I would do it.
I posted this to see how others might do it.
The author noted that each such triple forms an arithmetic progression.
And proceeded to count.