Cube Route

Created by Franz Armbruster in 1967, “Instant Insanity” was the Rubik’s Cube of its day, a simple configuration task with a dismaying number of combinations. You’re given four cubes whose faces are colored red, blue, green, and yellow:
Image: Wikimedia Commons

The task is to arrange them into a stack so that each of the four colors appears on each side of the stack. This is difficult to achieve by trial and error, as the cubes can be arranged in 41,472 ways, and only 8 of these give a valid solution.

One approach is to use graph theory — draw points of the four face colors and connect them to show which pairs of colors fall on opposite faces of each cube:
Image: Wikimedia Commons

Then, using certain criteria (explained here), we can derive two directed subgraphs that describe the solution:
Image: Wikimedia Commons

The first graph shows which colors appear on the front and back of each cube, the second which colors appear on the left and right. Each arrow represents one of the four cubes and the position of each of the two colors it indicates. So, for example, the black arrow at the top of the first graph indicates that the first cube will have yellow on the front face and blue on the rear.

This solution isn’t unique, of course — once you’ve compiled a winning stack you can rotate it or rearrange the order of the cubes without affecting its validity. B.L. Schwartz gives an alternative method, through inspection of a table, as well as tips for solving by trial and error using physical cubes, in “An Improved Solution to ‘Instant Insanity,'” Mathematics Magazine 43:1 (January 1970), 20-23.


For a puzzlers’ party in 1993, University of Wisconsin mathematician Jim Propp devised a “self-referential aptitude test,” a multiple-choice test in which each question except the last refers to the test itself:

1. The first question whose answer is B is question

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

2. The only two consecutive questions with identical answers are questions

(A) 6 and 7
(B) 7 and 8
(C) 8 and 9
(D) 9 and 10
(E) 10 and 11

3. The number of questions with the answer E is

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

The full 20-question test is here, the solution is here, and an interesting collection of solving routes is here.

(Jim Propp, “Self-Referential Aptitude Test,” Math Horizons 12:3 [February 2005], 35.)


A problem from the 1996 mathematical olympiad of the Republic of Moldova:

Twenty children attend a rural elementary school. Every two children have a grandfather in common. Prove that some grandfather has not less than 14 grandchildren in this school.

Click for Answer

Red or Black

Six sheets are set out in a room. Each identifies a different date in the same month. Weekdays are printed in black and Sundays in red. Six people will enter the room, one by one. Before the first one enters, one sheet is turned face down. Candidate 1 is then asked if she can deduce the color of the inverted sheet by examining the other sheets. Her answer, yes or no, is written on the back of the inverted sheet, followed by her number, 1. When 1 departs, a second sheet is turned face down. Candidate 2 enters and is asked whether she can deduce the color of the second sheet by considering her predecessor’s answer and the four face-up sheets. Her answer is noted in its turn, and this process continues — when the sixth candidate enters, she sees six face-down sheets, the first five of which bear the answers of the first five candidates. If all five of these answers are no, can Candidate 6 answer yes?

Click for Answer

Black and White

vielväter problem

In 1949 R.J. Darvall presented this position with a simple question: Who wins?

Click for Answer

The Ranchers’ Split

A problem from the Graham Dial, a publication of Graham Transmissions Inc., via L.A. Graham’s The Surprise Attack in Mathematical Problems (1968):

Little Euclid went with his father to visit a friend who ran a small ranch. The rancher told them that he and his partner used to herd cattle but had then sold the herd for as many dollars per head as there were head in the herd. With the proceeds they had bought as many sheep as they could at $10.00 per sheep, and with the money that remained they bought a dog. Later the rancher and his partner decided to split up evenly, so they added the dog to the herd and each of them took the same number of animals.

Euclid surprised him by saying, “I hope you paid him $2.00 to even things up.”

“Why, yes, I did,” the rancher said, “but how did you know? I haven’t told you any of the numbers or prices.” Can you prove that Little Euclid had to be correct?

Click for Answer

Next Up

What’s the next number in this sequence?

1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, …

Click for Answer

Custom Baking

From a Russian puzzle collection:

Is it possible to bake a cake that can be divided into four parts by a single straight cut?

Click for Answer