[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 395
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the windows of the house are not informative, almost not at all
The word itself indicates that all three are of different ages
the corridor of the friend's age is hardly informative
2y 3y 4y does not contradict
2g 3g 5 years does not contradict
2y 3y 4y does not contradict
2g 3g 5 years does not contradict
is it?
a) 2y 3y 4y, windows 9, friend's age 24, - does not contradict
b) 2y 3y 5y , windows 10y , friend's age 30 - not contradictory
a) 2y 3y 4y, windows 9, friend's age 24 - not contradictory
b) 2y 3y 5y, windows 10, friend's age 30 - not contradictory
why then, having counted the windows and knowing his own age, the friend still could not determine the age of the children?
yes i get it, i get it, )) it is necessary to search, as i have already written, the only possible age of this friend, to make it empty, but how to do it, not by overkill, i do not know
Yes, I got it, I got it, )) it is necessary to search, as I already wrote, the only possible age of this friend, whatevs it was empty, but how to do it, not by brute force, I do not know
1, 3, 9. Windows 13 (not much info), age 27 (still not much info).
Although theoretically it could also be 1, 5, 25 or 1, 2, 4. But unlikely: in the first case they don't live that long, in the second the friends are too disparate in age.
In short, the friend's age is the cube of a small prime number, p^3. This cube is decomposed into multipliers in several ways - either 1*1*p^3, or 1*p*p^2, or p*p*p. Only one case has the youngest one.
P.S. No, bad decision. Here, after the second father's answer, the natural solution is there and the only one.
An approximate layout, say, could be like this:
a1*b1*c1 = a2^2*c2 = Age
a1+b1+c1 = 2*a2+c2 = Windows
Here (a1, b1, c1) and (a2, a2, c2) are two possible solutions (the second one is discarded).
1, 3, 9. Windows 13 (not much info), age 27 (still not much info).
Although theoretically it could also be 1, 5, 25 or 1, 2, 4. But unlikely: in the first case they don't live that long, in the second the friends are too disparate in age.
In short, the friend's age is the cube of a small prime number, p^3. This cube is decomposed into multipliers in several ways - either 1*1*p^3, or 1*p*p^2, or p*p*p. Only one case has the youngest one.
P.S. No, bad decision. Here, after the second father's answer, the natural solution is there and the only one.
An approximate layout, say, could be like this:
a1*b1*c1 = a2^2*c2 = Age
a1+b1+c1 = 2*a2+c2 = Windows
Here (a1, b1, c1) and (a2, a2, c2) are two possible solutions (the second is discarded).
The windows are mixed up, there had better be trees in the garden, 13 windows is it a single storey barrack , or several storeys, but the windows are partly bricked up . If there is an indication of windows, then we are
Theoretically they are directed by multiplicity of floors and we must exclude prime numbers like 13 or 17