Can we simplify calculating large sums of numbers?

[Ilikebugs]

View attachment 6242 uhh, how would we get a better way?

 

[MarkFL]

I think I would express one particular number that can be made as:

\(\displaystyle 100A+10B+C\)

There are $7\cdot6\cdot5=210$ different numbers that can be made, with each digit appearing $30$ times. Hence the sum $S$ will be given by:

\(\displaystyle S=30\cdot100\cdot\sum_{k=1}^{7}(k)+30\cdot10\cdot\sum_{k=1}^{7}(k)+30\cdot\sum_{k=1}^{7}(k)=30(100+10+1)\cdot\sum_{k=1}^{7}(k)=30\cdot111\cdot\sum_{k=1}^{7}(k)\)

Use the formula:

\(\displaystyle \sum_{k=1}^{n}(k)=\frac{n(n+1)}{2}\)

To complete the computation of the sum. :)

 

[Ilikebugs]

so is the answer 30*111*28?

 

[MarkFL]

so is the answer 30*111*28?

You are one step closer...now carry out the multiplication to get the number. :D

 

[Ilikebugs]

Is the sum 93240

 

[MarkFL]

Is the sum 93240

That's what I get. :D

As a follow-up, let's generalize a little here and say that the list of possible digits is $\{1,2,\cdots,n\}$ and we are going to sum up all the $m$ digit numbers having distinct digits, where $m\le n\le9$. Can you verify that:

\(\displaystyle S=\frac{1}{n}\prod_{k=n-m+1}^{n}(k)\cdot\sum_{k=0}^{m-1}\left(10^k\right)\cdot\sum_{k=1}^{n}(k)\)

 

[Ilikebugs]

No, I'm bad at proving things