Brain-Teaser 848: The magic class
From The Sunday Times, 16th October 1977 [link]
The class only had nine pupils.
Three girls sat in the front row, three boys in the back one, while in the middle rows the sexes sat alternately.
Altogether, including the teacher, the sexes were equally divided.
When their home-work was returned marks were compared and the children were surprised to discover that the total marks gained by those in each row were the same, as were also those for each column from front to back and for each diagonal of the square in which they sat. Excitedly they pointed out that fact to the teacher who replied that, when checking their work, which had been marked out of ten, he noticed that every digit had been used once and once only.
Three was the lowest mark awarded to a girl.
What was the highest mark given to a boy?
This puzzle is included in the book The Sunday Times Book of Brain-Teasers: Book 1 (1980). The puzzle text above is taken from the book.
[teaser848]
Frits, 
GeoffR, and 
S2T2 
Jim Randell 10:05 am on 2 September 2021 Permalink |
The possible marks are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. For a set of nine of them to use all ten digits requires 10 to be used, which means 0 and 1 cannot be used, so the marks are 2-10.
These marks sum to 54, so the magic constant is 18, and the central square must be 6.
The teacher is revealed to be male so the layout must be (front to back): (f f f) (f m f) (m m m).
The following run file executes in 70ms.
Run: [ @replit ]
Solution: 9 was the highest mark awarded to a boy.
The layout looks like this (or reflected about a vertical axis):
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GeoffR 4:08 pm on 2 September 2021 Permalink |
# Using the same integer (A - I) and m/f references from itertools import permutations for p1 in permutations(range(2, 11)): A, B, C, D, E, F, G, H, I = p1 if (A + B + C == D + E + F == G + H + I == A + D + G == B + E + H == C + F + I == A + E + I == C + E + G): if min(D, F, G, H, I) == 3: #lowest girl mark HBM = max(A, B, C, E) # highest boy mark print('All Marks ( A to I): ', A, B, C, D, E, F, G, H, I) print('Highest Boy Mark: ', HBM) # All Marks ( A to I): 7 2 9 8 6 4 3 10 5 # Highest Boy Mark: 9 # All Marks ( A to I): 9 2 7 4 6 8 5 10 3 # Highest Boy Mark: 9LikeLike
Frits 4:06 pm on 3 September 2021 Permalink |
The magic constant is 18, and the central square must be 6.
All corners can’t be even as using odd marks for all the sides would make odd total marks.
Only two opposite corners can’t be even as then the other 2 corners would have to be odd.
To make total even marks we would need odd marks for all the sides as well.
So all corners must be odd and all sides even.
As 3 is in a bottom corner 9 must be in a top corner (male).
The 10 mark must be adjacent to the 3 mark and thus female.
So we can conclude that the highest mark given to a boy is a 9.
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