Teaser 2851: Knowing the lingo
From The Sunday Times, 14th May 2017 [link] [link]
A group of 64 students can speak, between them, French, German and Russian. There are three times as many French speakers and two times as many German speakers as there are Russian speakers. The number who speak all three languages is one fifth of the number who speak French and at least one other of these languages. The number who speak all three languages is also two ninths of the number who speak German and at least one other of these languages. Furthermore, the number who speak all three languages is also two fifths of the number who speak Russian and at least one other of these languages.
How many of the 64 students speak only French?
[teaser2851]
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Jim Randell 8:57 am on 17 February 2022 Permalink |
We assume that each of the students speaks at least one language (otherwise we can find many solutions), then we can represent the 7 internal areas of the Venn diagram: F, G, R, FG, FR, GR, FGR.
So we have:
And the additional conditions give 5 more equations.
But together these only give 6 equations in the 7 variables, so we need to choose a value for one of the regions to make a complete set of equations. The value for FGR region needs to be divisible by 2, so we can use that.
The following Python program runs in 54ms.
Run: [ @replit ]
from enigma import (Matrix, as_int, irange, printf) # the given equations (incomplete) eqs = [ # F G R FG FR GR FGR = k (( 1, 1, 1, 1, 1, 1, 1), 64), # 64 students in total ((-1, 0, 3, -1, 2, 3, 2), 0), # total F = 3(total R) (( 0, -1, 2, -1, 2, 1, 1), 0), # total G = 2(total R) (( 0, 0, 0, -1, -1, 0, 4), 0), # FGR = (1/5)(FG + FR + FGR) (( 0, 0, 0, -2, 0, -2, 7), 0), # FGR = (2/9)(FG + GR + FGR) (( 0, 0, 0, 0, -2, -2, 3), 0), # FGR = (2/5)(FR + GR + FGR) ] # choose a value for FGR for FGR in irange(0, 64, step=2): # make an additional equation for FGR # F G R FG FR GR FGR = k eq = ((0, 0, 0, 0, 0, 0, 1), FGR) # solve the equations for non-negative integers try: vs = tuple(as_int(x, "0+") for x in Matrix.linear(eqs + [eq])) except ValueError: continue (F, G, R, FG, FR, GR, FGR) = vs # there must be at least 1 speaker of each language if R + FR + GR + FGR == 0: continue # output solution printf("F={F} G={G} R={R} FG={FG} FR={FR} GR={GR} FGR={FGR}")Solution: 25 of the students speak only French.
There is only one possible arrangement (subject to the assumption above):
We can manually simplify the equations:
So given an even value for FGR = 2k, we can calculate the values for the other regions:
Run: [ @replit ]
from enigma import (irange, div, printf) # total number of students N = 64 # choose a value for FGR = 2k for k in irange(0, N // 2): n = N - 17 * k if n < 0: break R = div(n, 6) if R is None: continue # determine the other values FR = FGR = 2 * k GR = k FG = 3 * FGR F = 3 * R + 5 * k G = 2 * R + k # output solution printf("F={F} G={G} R={R} FG={FG} FR={FR} GR={GR} FGR={FGR}")If we do the steps manually we find we only need to check k = 0, 1, 2, 3, and only k = 2 gives viable values.
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GeoffR 11:45 am on 17 February 2022 Permalink |
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Jim Randell 4:24 pm on 17 February 2022 Permalink |
Although we are not told that each of the 7 subsets is non-empty (but we do know that there must be at least one student that speaks each of the languages).
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GeoffR 3:56 pm on 18 February 2022 Permalink |
Although the Solve Function is often used in Mathematica for solving sets of simultaneous equations, it did not work in this instance. It turns out that the FindInstance Function (ref Brian) was suitable. and can sometimes be used as an alternative solver in Mathematica.
FindInstance[{ F + G + R + FG + FR + GR + FGR == 64, F + FG + FR + FGR == 3 (R + FR + GR + FGR), G + FG + GR + FGR == 2 (R + FR + GR + FGR), 5 FGR == FG + FR + FGR, 9 FGR == 2 (FG + GR + FGR), 5 FGR == 2 (FR + GR + FGR), F + FG + FR + FGR > 0, G + FG + GR + FGR > 0, R + FR + GR + FGR > 0}, {F, R, G, FG, FR, GR, FGR}, Integers ] {{F -> 25, R -> 5, G -> 12, FG -> 12, FR -> 4, GR -> 2, FGR -> 4}}LikeLike