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
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”?
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.
Let be three digits, where is the avergare of and . The digit is determined uniquely by and . Hence we consider the pair and .
are either both even or both odd, since is an integer. We can choose pairs from and pairs from .
Thus, there are possible triples ; the number of rearrangements is .
We need to count four pairs with zero: . They determine four triples, each with four possible (not having a leading zero) rearrangements, so there are 16 more numbers. Adding up gives possible three digit numbers.
I'm not sure, I'm afraid I might miss something...