The sum of the numbers from $100$ to $999$ that do not have the digit $0$ as well as do not have repeated digit.

Question

Considering the numbers from $100$ to $999$. Excluding numbers that have the digit $0$, also excluding numbers that have repeated digit. What is the sum of remaining numbers?

That is, we need to find

$123+124+125+\cdots+129+132+134+135+136+\cdots+139+142+143+145+146+147+\cdots+987$

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Using Microsoft Excel, I found that the answer is $279720$. But I do not know if the calculation can be done in a purely mathematical way (without using any software). I hope you can provide some hints.

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Your help would be appreciated. THANKS!

Answer 1

Each digit runs from $1$ to $9$, which sums to $\binom{10}2=45$; for each value there are $8\cdot7=56$ admissible values of the other digits; and the total value of the digits is $111$. So the sum is

$$ 45\cdot56\cdot111=279720\;. $$

Answer 2

Approach that slightly differs from the one provided by Joriki.

If $abc$ denotes such number then so does $a'b'c'$ where $a+a'=b+b'=c+c'=10$.

This indicates that the average of these numbers is: $$\frac12(abc+a'b'c')=555$$

There are $9\times8\times7=504$ numbers that satisfy the conditions.

So the sum of these numbers should be:$$504\times555=279720$$