F2 U' f F2 (cross 4/4)
R' U R (F2L1 3/7)
U2 u' R U' R' (F2L1.5 5/12)
D' L' U L (F2L2.5 4/16)
d' D2 R U R' (F2L3.5 5/21)
d L' U L d R U' R' (F2L4 8/29)
y F U R U' R' F' (OLL cross 6/35)
U' r U R' U' r' F R F' (OLL corners 9/44)
U' R U R' F' R U2 R' U2 R' F R U R U2 R' U' (Rperm 17/61)
next U' F'...
I think you mean R' M' r; R' M r does nothing to a solved cube. That, with the little fingerpull trick, is basically how many people execute M2, and it's the same exact alg.
I feel really stupid now for not noticing the perfect cubes part. I did feel the 12, 18, therefore 24 thing was a stretch, but it fit the solution.
Although I sort of share your sentiment about sequence problems, I have a rather reasonable one:
Find the next term in
2, 3, 10, 21, 55, 104, 221...
The differences between terms are 7, 19, 37; the differences between these are 12 and 18; the difference between 37 and the next increased amount is thus 24, so the next increased amount is 61, which added to 63 gets 124. This is not a 5-10 second problem for 9-11 year olds, but it is certainly...
Some tips from my own experience administering forums:
1. As someone said, pick your moderators carefully. You should know the person you appoint as a moderator well and should be on the same page with them regarding what needs moderated, and you should get along with them quite well. That...
Heh. I think you may've been confused because I tried to make some sense of joey's 720 number, being the number of uniquely identifiable color schemes (30) times the number of uniquely identifiable orientations (24)
Given white on top, yellow on botton, red on front, we can have, in order of R-B-L faces:
blue orange green (normal scheme)
blue green orange
green orange blue
green blue orange
orange blue green
orange green blue
all of which are uniquely identifiable. 30 is correct, I have already accounted...
Nobody said 6 standard colors, but if we say we can choose from as many colors as we want, there could be infinitely many combinations and trying to calculate a number becomes futile. However, if we want the number of colorings possible using n>6 colors, we just multiply 30 by (n choose 6) -...
The first color can go anywhere, then we put one of five colors opposite it.
The third color also can go in any of the remaining four spots, and one of three goes opposite that one.
The fifth color can go in either of the remaining two spots, and the sixth goes opposite it.
We have 5, 3, and...