[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 597
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Explain to me.
A (hexadecimal) = 10 (decimal)
B (hexadecimal) = 11 (decimal)
15 (Hexadecimal) = 21 (decimal) = 10101 (binary)
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?
The man has 10 equations with 3 unknowns. Theoretically it's possible to solve.
Age1 * Age2 * Age3 = Age4
Age1 + Age2 +Age3 = Group Number
Age1 < Age2
Age1 <Age3
Age1 < Age4
Group number > 0
Age1 > 0
Age2 > 0
Age3 > 0
Age4 > 0
Arithmetic in hexadecimal:
A (hexadecimal) = 10 (decimal)
B (hexadecimal) = 11 (decimal)
15 (hexadecimal) = 21 (decimal) = 10101 (binary)
there was a thought in this direction when Andrew answered explicitly in binary )
The man has 10 equations with 3 unknowns. It is theoretically possible to solve.
Age1 * Age2 * Age3 = Age4
Age1 + Age2 +Age3 = Group Number
Age1 <> Age2
Age1 <>Age3
Age1 <> Age4
Group number > 0
Age1 > 0
Age2 > 0
Age3 > 0
Age4 > 0
But I somehow came to the conclusion that this is not necessary to find the solution (I may have overreacted).
Actually, in the condition the work is known. 36.
It's easier then. The other sons are probably 6 years old. And the youngest is 1 year old. What is missing from this problem is that the ages of the two sons are the same.
We go to the diamonds 4 times. On 2 and 3 the left hand brings out a spade, on 4 looks which suit a club or a heart will bring the unfortunate and carries the opposite one. Then a spade pass, a rebound and you get it.
О! If South realizes on turn 4 that he won't be released alive, he can take out a spade and then he'll get only one bribe. There's another exotic theme. If Yug made an illogical demolition and left 2 clubs and one heart, he'd never get caught.
The unfortunate one may cheat and carry a spade on the tambourine of the East. Oh, come on. 2 is the minimum there, and we still have to think about 3.
P.S. Ahead of you. But even if the unfortunate bids a spade, it's still a 2.
There was a puzzle they wanted badly. Here, solve it.
[The problem is rated 4 points, i.e. difficult.]
Black's move. Which piece stands on g4?
Another one, a three-point. The weighing is only one. The scales allow you to see the exact difference in the weight of the coins.
There are exactly 50 fake coins among the 101 coins. The weights of all the real coins are the same; the weight of each counterfeit coin is 1 gram more or less than the real coin (counterfeit coins may have different weights). How does one weighing on a two-cup scale with an arrow and a scale (without weights) determine whether a given coin is fake?