April 22, 2015 11:04:19 am
You must have come across it by now, this little puzzle that went viral a couple of weeks ago. We couldn’t let that pass without matching it in “Problematics”, could we? So here’s my version.
Take three children and call them Albert, Bernard and Cheryl, in recognition of the inspiration they provided. Show them a set of five labels, “GIRL”, “GIRL”, “GIRL”, “BOY”, “BOY”. Then hide away two of the labels, so that the children don’t know which two are hidden and which three remain with you. Finally, paste the three remaining labels on their foreheads @ 1 label/forehead/child.
Mind you, there is no connection between any child’s label and his/her gender. Cheryl, though a girl, may have been labelled either “GIRL” or “BOY”; the same holds for the two boys. That way, not one of the children knows what the label on his/her head says. But each one can see the labels on the foreheads of the other two.
Let the guessing game begin.
Albert: I can see the labels on the foreheads of Bernard and Cheryl… I wish I had more information, though. As of now, I have no idea what is on my forehead.
Bernard: Albert’s statement isn’t enough. Although I can see Albert’s and Cheryl’s labels, I still don’t know what is written on my forehead.
Cheryl: At first, I didn’t know either. Thank you, boys, for your helpful statements. Now I know that the label on my forehead reads ____.
Puzzle#7A: How does Cheryl know, and what’s her label?
Open & shut cases
We could do with an easier one now. The year is 2035 and you throw a party for six guests — Albert, Bernard, Cheryl (all three now grown up) and their respective spouses. Keep their presents ready, a tie for each man, a scarf for each woman, all based on their favourite colours. Albert and Ms A like red. Bernard and Ms B like blue. Cheryl likes blue but Mr C likes red.
Take nine gift boxes, three large and six small. The six presents go into the smaller boxes — red tie, red scarf; blue tie, blue scarf; blue scarf, red tie. Don’t label these boxes. Put each pair of boxes into a larger box. These you label simply as “A”, “B” and “C”.
Too late, you realise everything is wrong. The boxes, however, look identical and you can no longer tell which box contains what. All you know is that no box carries the right label. And you need to fix that.
Puzzle#7B: You don’t want to open all nine boxes. To be able to replace every label correctly, which boxes do you need to open?
What you wrote
Four people solved both of last week’s puzzles. From their letters, here are the excerpts that show their various approaches to Puzzle#6A.
In Puzzle#6A, the triplet other than (5, 12, 13) is (6, 8, 10). I picked the general entities (a^2 – b^2), (2ab) and (a^2 + b^2) as the sides of right-angled triangles. Now, perimeter = area.
a^2 – b^2 + 2ab + a^2 + b^2 = ½(a^2 – b^2)(2ab)
=> 2a(a + b) = ab(a^2 – b^2)
=> 2 = b(a – b)
For the product b(a – b) to become 2,
Either b should be 2 and (a – b)=1 which yields a = 3 and b = 2 and thus triplet (5, 12, 13);
Or b= 1 and (a – b)= 2, which yields b = 1 and a = 3, thus triplet (6, 8, 10).
Tushar Gupta, DoMS, IIT Madras
I solved Puzzle#6A using Euclid’s formula [a = 2mn, b = m^2 – n^2, c = m^2 + n^2].
Area = A = ½ ab
Perimeter = P = a + b + c
=> P (a + b – c) = (a + b + c) (a + b – c)
=> P (a + b – c) = (a + b)^2 – c^2
=> P (a + b – c) = a^2 + b^ – c^2 + 2ab
=> P (a + b – c) = 2ab = 4A
Since P = A,
a + b = 4 + c
=> 2mn + m^2 – n^2 = 4 + m^2 + n^2 [from Euclid’s formula]
=> 2m(m – n) = 4
=> m(m – n) = 2
Solution 1: (m, n) = (3, 2); then a = 12, b = 5, c = 13
Solution 2: (m, n) = (3, 1); then a = 6, b = 8, c = 10
Harsha T R (NIT Karnataka, batch of 2012)
X,Y being the sides meeting at 90º, Z being the side opposite (X^2 + Y^2 = Z^2),
X + Y + Z = XY/2
=> 2Z = XY – 2(X + Y)
=> 4Z^2 = (XY)^2 + 4(X + Y)^2 – 4(XY)(X + Y)
Substituting Z^2 = X^2 + Y^2 and solving, we get,
X = 4 + 8/(Y – 4)
It can be seen that, if X has to be a positive integer, then (Y – 4) can have only four possible values: 1, 2, 4, 8. We get four solutions for (X, Y) — (5,12), (6,8), (8,6), (12,5). Only two triangles are possible, which are (6, 8, 10) and (5, 12, 13).
Sampath Kumar V (IIM Kozhikode alumnus)
The triangle in Puzzle#6A has sides 6, 8, 10. I used hit and trial.
For Puzzle#6B(i), if we have coins/notes of denominations 1, 2, 5, 10, 20, 50, then we can’t have 4, 9, 14, 19, 24, 29, 34, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 59, 64, 69, 74, 79, 84, 89-100. Any number with the digit 4 or 9 can’t be put together.
For Puzzle#6B(ii), we need 1, 2, 2, 10, 20, 20, 50 (total 100) to make all numbers between 1 and 100.
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