[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 596

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Two college friends (programmers, of course) met each other. One asked the other how many children he had. The latter replied that there were three. When asked about their ages, he said that the product of the ages of the children was his age, and the sum of the ages of the children was the number of the group in which they were once studying at the institute. The retort about insufficient information was followed by the announcement that the youngest was a redhead. After that the problem was successfully solved.
Shall we go?
 
TheXpert:
10101
The answer counts.
Actually A+B=15
 
IDLER: 3 takeaways if not too much cognac drunk.

Describe the sequence briefly. Of course, what matters is what the other point guard carries, catching the mizzer.

I have 3 possible as well.

 
Aleksander:

Listen... how about one more time... :-) I'm having second thoughts... - your king's path to black pawns is blocked by rook on line 5

If the rook stays on the 5th rank, it faces b8F, the only square that prevents that is b5, but that doesn't save the pass to the a-side: even though

- pawns in a row, but they're going to hit the bishop...

it's enough for White to get to a7, then b8 is taken under control, and a pawn from b7 goes there. And the black rook can't hit the passing pawn on the vertical a, as in this case the control of the b7 pawn is lost.

 
Question for chess fans - can any of you play go?
 
TheXpert:
Question for chess fans - can any of you play go?
I just want to be able to))
 
Mathemat:

Describe the sequence briefly. Of course, what matters is what the other point guard carries, catching the mizzer.

I also have 3 possible ones.


We go into the diamonds 4 times. On 2 and 3 the left-hand man brings out a spade, on 4 he looks at the suit of clubs or hearts the unfortunate will bring out and carries the opposite. Then a spade pass, a rebuke and you get it.
 
MikeM:
Two college friends (programmers, of course) met each other. One asked the other how many children he had. The latter replied that there were three. When asked about their ages, he said that the product of the ages of the children equals his age, and the sum of the ages of the children equals the number of the institute group the friends were once in. The retort about insufficient information was followed by the announcement that the youngest was a redhead. After that, the problem was successfully solved.
Shall we go?

was in this thread
 
Mischek2:

was in this thread.
Then I apologise profusely. I didn't read the whole thread.
 
MikeM:
The answer counts.
Actually A+B=15

Explain to me.
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