# three digit numbers

• Nov 19th 2010, 08:20 AM
Plato
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”?
• Nov 19th 2010, 08:36 AM
Unknown008
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
• Nov 19th 2010, 08:52 AM
Plato
That is a huge undercount.
• Nov 19th 2010, 08:55 AM
Unknown008
Ok, back to work then.
• Nov 19th 2010, 09:02 AM
Unknown008
Quote:

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?
• Nov 19th 2010, 09:11 AM
Plato
A huge overcount.
• Nov 19th 2010, 09:14 AM
Unknown008
Oh my (Rofl)

Ok, getting to work again.
• Nov 19th 2010, 09:18 AM
Unknown008
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. >.<
• Nov 19th 2010, 09:20 AM
Plato
You are getting close. But on ring yet.
• Nov 19th 2010, 09:38 AM
Unknown008
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.
• Nov 19th 2010, 09:42 AM
Plato
All three digits must be distinct.
• Nov 19th 2010, 09:50 AM
Unknown008
Must be my english (didn't know what distinct really implied >.<) and removing those I'd get 112!
• Nov 19th 2010, 11:59 AM
tonio
Quote:

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
• Nov 19th 2010, 12:02 PM
melese
An attepmt.
Quote:

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 $\displaystyle a, b, m$ be three digits, where $\displaystyle m$ is the avergare of $\displaystyle a$ and $\displaystyle b$. The digit $\displaystyle m$ is determined uniquely by $\displaystyle a$ and $\displaystyle b$. Hence we consider the pair $\displaystyle a$ and $\displaystyle b$.

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

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

I'm not sure, I'm afraid I might miss something...
• Nov 19th 2010, 12:06 PM
Plato
@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.