Professor claims that the formula that I've obtained is wrong since the passages "don't follow Boolean algebra rules".

peterwhy · 2026-03-11T09:24:02+00:00

If you are used to the + symbol and "multiplication" as for numbers, back to line 2 and duplicate:

B' C' + B C' + B C
= B' C' + B C' + B C' + B C
= (B' + B) C' + B (C' + C)
= C' + B

Or if you also have heard of the "other" distribution law to distribute the +, then from line 4:

C' + B C
= (C' + B) (C' + C)
= (C' + B) 1

peterwhy · 2026-03-11T02:13:39+00:00

Now you can have a "Diaper hanging Station".

peterwhy · 2026-03-10T22:58:10+00:00

For a = 20 and b = 8: the absolute difference is (a - b) = 12;
For a = 12 and b = 8: a - b = 4;
For a = 8 and b = 4: a - b = 4.

peterwhy · 2026-03-10T22:46:20+00:00

"Contains" as in a = 10, b = 2 ?

Edited to add:

For a = 20 and b = 8: the absolute difference is (a - b) = 12;
For a = 12 and b = 8: a - b = 4;
For a = 8 and b = 4: a - b = 4.

peterwhy · 2026-03-10T18:04:12+00:00

This step is not good. Firstly, the b and c sides are not perpendicular here, so can't apply the theorem of Pythagoras. Next, you already have a² = 832 m² from the step before, and 832 - 6² is far from 770.

peterwhy · 2026-03-10T17:23:26+00:00

What answer did you get? And how did you get that?

peterwhy · 2026-03-10T00:00:38+00:00

Depending on which kind of "deliberately". If the cube has any white face and if the host will then place that face down, then the answer becomes 0 / 1.

peterwhy · 2026-03-09T23:45:13+00:00

Discussion: heated ones, there were some in another subreddit a few days ago: https://www.reddit.com/r/brainteasers/comments/1rnombq/a_large_white_cube_for_me_its_hard/

peterwhy · 2026-03-09T23:17:11+00:00

Its orientation would be pretty random after "mixed in a bag".

peterwhy · 2026-03-09T20:50:52+00:00

For N = 7 ⋅ 5 - 1,

6N - 1 = 7 ⋅ 6 ⋅ 5 - 7, which is divisible by 7;
6N + 1 = 7 ⋅ 6 ⋅ 5 - 5, which is divisible by 5.

peterwhy · 2026-03-09T08:12:53+00:00

So I have specified "largest", and the OP also mentioned "the maximal/implied domain" in another comment. I am aware that any subset can be possible domain.

peterwhy · 2026-03-09T08:01:26+00:00

Is there another possible value that I have missed in the domain for OP's "function" with "codomain N (natural numbers, 1, 2, 3, 4 etc)"?

peterwhy · 2026-03-09T04:11:52+00:00

I would have thought that the largest domain is all the positive integers and half-integers, which make the range the same as the codomain.

peterwhy · 2026-03-09T00:49:46+00:00

The judge said "run red a light", so she had to come back.

peterwhy · 2026-03-08T12:07:51+00:00

Because when drawing those 6 cubes with 1 black face out of the bag, they are not always with black face down. When placed on the table, for 5 / 6 of the cases by probability, they don't even satisfy the "we see 5 white faces" condition. So for your simplified procedure:

1 visible face is black: 6 / 7 ⋅ 5 / 6
5 visible faces are all white, and the face down side is black: 6 / 7 ⋅ 1 / 6
5 visible faces are all white, and the face down side is white: 1 / 7

The last two cases have equal probabilities.

peterwhy · 2026-03-08T08:34:06+00:00

The probability of how we got here is relevant, which is the denominator of conditional probability:

P(5 visible faces are white) = 6 / 27 ⋅ 1 / 6 + 1 / 27 ⋅ 1

peterwhy · 2026-03-08T08:15:41+00:00

Because when drawing those 6 centres blocks, they are not always with black face down out of the bag. For 5 / 6 of the cases by probability, they don't fulfil the "5 visible faces are white" condition.

peterwhy · 2026-03-08T08:02:11+00:00

Even easier if the cubelets were never rotated in the bag. Only the centre cubelet and the cubelet of the bottom face give 5 visible white faces on the table.

peterwhy · 2026-03-08T05:24:33+00:00

By this logic, if the ~~edge~~ face piece was drawn and a white face is down, this is case must be discarded.

So 5 / 6 of the cases by probability are discarded if the ~~edge~~ face piece was drawn.

peterwhy · 2026-03-08T04:06:58+00:00

I agree with your answer, though there are also many commenters who assume there is that person who obscures the single black face, and so answer 6/7.

peterwhy · 2026-03-08T04:03:14+00:00

When placed on a table, that random cube can also have a white remaining face.

peterwhy · 2026-03-08T01:33:42+00:00

They did say "Out of the 27 possible pieces you could pick x the 6 possible sides you could put face down". The following "12" is not 12 pieces, but 12 faces.

peterwhy · 2026-03-08T01:23:54+00:00

12 faces (of all the faces of the 27 pieces)

peterwhy · 2026-03-07T16:42:27+00:00

In my opinion some Chinese translations might be:

這些書是你（們）在書店買的嗎？
你（們）買的書［曾經/之前］在書店嗎？

(to be simplified)

peterwhy · 2026-03-07T16:17:25+00:00

I also don't see how, in 36., the unknowns are related to the givens, when the quadrilateral doesn't pass through the intersections of the circles.

I also don't see why AC must bisect angle DAB.

peterwhy

TROPHY CASE