| Adam Bonney
ODDLY:- 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 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 Brooker (age 9) 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 3-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! (And I don't mean the factorial!!)
Well done Michael! Good solutions were also sent in by Rebecca Pickering, Y7, Bourne Grammar School; Fraser Renwick from Madras College, St Andrew's; Ben Morrison, Monica Griffith-Jones, Natalie Clark, Emma Larner, and Bethany Doggett, Hethersett High School, Norfolk.
