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Six Is the Sum

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In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?


Age 7 to 11 Challenge Level:

I looked at this one as soon as I got on the site and started off by writing down all of the ones up to $199$:
$111$, $113$, $115$, $117$ ... $197$, $199$.
I got the answer to all of them which was $3875$.
I then changed the first digit to $3$ to make it $311$, $313$ ... $399$.
Then I did the same for all the rest.
My final answer when I had added it all up was $69 375$.

Nicely done Adam from Poltair Community School and Sports College, Cornwall!

A big thank you too to Michael for your elegant method of finding the sum of all 3 digit numbers each of whose digits is odd:

I think the answer is $69 375$.
1. I estimated the total c $50 000$.
2. I wrote out all the three-digit numbers $100$-$199$ which had odd digits only, and observed a pattern: value of $100$ occurred $25$ times.
3. Decided that if all those numbers were written out, the values of $100$, $300$, $500$, $700$, and $900$ would each occur $25$ times; the values of $10$, $30$, $50$, $70$, and $90$ would do the same; and the values of $1$, $3$, $5$, $7$, and $9$ similarly.
4. Therefore, the sum can be simplified to: $25(100+300+500+700+900) + 25(10+30+50+70+90) + 25(1+3+5+7+9) = 69 375$
5. This is still very large!
We can simplify it further to: $25(111+333+555+777+999) = 69 375$