Earlier today I set up the following two puzzles. Here they are again with solutions.
The first is by Randall Munroe, cartoonist of the webcomic xkcd and author of the best-selling book what if
1. Now all together
Find all the possible ways, if any, to fill in the three blanks to make all three statements true.
Solution There are three ways, some more interesting than others!
Let the spaces be represented by x, y and z.
If we only consider the numbers in the spaces – there is only one solution x = y = z = oh The ‘joke’ is to realize that we can include the numbers of the panels. In this case there are two solutions.
The panels are three comparisons: x = 5+ y + z, y = √3xzand z = min (x, y, 1)
Let’s say few (x, y, 1) = y. Therefore z = j therefore x = 2y + 5, and y = √3xy. It simplifies to y2 + 3y = 0. We know y is greater than or equal to 0, since it is the value of a square root, therefore y= o. Give us one solution: x = 5, y = z = oh
Let’s say little now (x, y, 1) = x. Therefore z = xand y = –5, what a contradiction since y is greater than or equal to 0.
When few (x, y, 1) = xwe get the second solution: x = 12, y = 6, z = 1.
2. One to Venn
The four squares below form a Venn diagram with ten regions, labeled A to J. The letters A to J each stand for a number between 1 and 10, so that no two regions share the same number and all numbers from 1 up to 10 are used. The regions in each square add up to the same number. What value is D, the intersection of all four squares?
Solution: D = 7
Since the sum of the regions is the same for each square, when two squares intersect, the regions that do not intersect must be equal. So, for the red and green squares, illustrated below, A = G + H + I.
Similarly, F = B + E + I, and J = B + C + G.
So A+F+J = G+H+I+B+E+I+B+C+G
= 2(G+I+B) +C+E+H.
The highest possible value of A+F+G is 10 + 9 + 8 = 27.
The lowest possible value of 2(G+I+B) +C+E+H is 2(1 + 2 + 3) + 4 + 5 + 6, which is also 27.
It follows that both these values are 27. The only number missing is 7, and the only missing label is D, so D = 7.
I hope you enjoyed today’s puzzles. I’ll be back in two weeks.
Both of today’s puzzles are taken from The Mathematical Playground: People and Problems from 31 Years of Math Horizons. Math Horizons is the undergraduate magazine of the Mathematical Association of America, and the book, a compilation of its best puzzles, is full of great material. For more information see here.
what if (10th anniversary edition) by Randall Munroe will be released in the UK on November 14th.
My latest book is Think twice: solve the simple puzzles (almost) all get wrong. (In the US it is called Puzzle Me Twiceand out tomorrow.) To support the Guardian and Observer, order your copy from guardianbookshop.com. Delivery charges may apply.
I’ve been doing a puzzle here on alternate Mondays since 2015. I’m always on the lookout for great puzzles. If you want to suggest one, email me.
