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FWIW, I got the same answers. 1. 12 edges * 6 non-corner pieces per edge. 2. There are 111 digits there. You can't keep six nines and have enough digits left.
by monktastic1 7y ago
FWIW, I got the same answers.
1. 12 edges * 6 non-corner pieces per edge.
2. There are 111 digits there. You can't keep six nines and have enough digits left. So you keep five (from 9, 19, 29, 39, 49), and now have six digits to keep from 5051525354555657585960. It can't start with 8 (or you still won't have enough digits left), so you choose 785960.
3. First notice that 0+...+10 = 55. To make it end in 2, you have to discard the 3. In general if you have n+...+(n+10) you discard n+3. Now n+...+(n+10) = 5.5*(n+(n+10)) by Gauss' trick, and subtracting (n+3) you have 10n + 52 = 2012.
Very much doubt I could have solved them in elementary school(!) though.
- asdffdsa 7y agoCould you explain why you can't just do 99999988888? 100 digits in any order, what am I missing?
- dan-robertson 7y agoYou’re allowed to delete whichever digits you like. But you aren’t allowed to reorder the digits which are there. I.e. if you had to delete one digit you could go from 189 to 89 but not to 98.