Thread: Averaging Digit...

I am having difficulty figuring this out....

1) how many four digit numbers are composed of three distinct digits (with no leading 0's), such that one digit is the average of other three...?

2) How many such four digit numbers with repeated digit are possible?

thanks for the help!

I can't interpret #1 unless it contains a typo. First it says how many four digit numbers and then composed of three distinct digits, suggesting that a maximum of one repeat is allowed. Then #2 says with repeated digit, which seems to me the exact same question as #1???

Can I assume that the poorly worded #1 means to ask "How many four-digit numbers with distinct digits are possible..." ???

Anyway, the question #1 is equivalent to solving in where for all , and then taking every possible reordering of these digits as long as the leading digit is nonzero. Question #2 is the same, except removing the criteria that each value must be distinct.

To be honest, I don't see a "clever" way to do this. The brute-force method would be listing out all the solutions to the equation, eliminating the ones with repeated digits, multiplying by the 4! different orderings, and individually eliminating solutions with a leading zero. Similar approach for #2...

Originally Posted by Vedicmaths

I am having difficulty figuring this out....

1) how many four digit numbers are composed of three distinct digits (with no leading 0's), such that one digit is the average of other three...?

2) How many such four digit numbers with repeated digit are possible?

thanks for the help!

... a "clever" way to do this...

(digit1 + digit2 + digit3) /3 = the fourth digit.
Only the three digit numbers divisible by 3 are useful.
The smallest would be 123
The largest would be 987

987-123 = 864
864/3 = 288

Does that help?

.

Aidan: Yes, there are 288 3-digit numbers divisible by three. But that number includes those with non-distinct digits, like 333, 255, etc. And the smallest would not be 123, it would be 102.