Category Archives: Puzzles and Problems

Locating the Loot

This is a straight-forward problem from Geoffrey Mott-Smith in 1954.

“A brown Terraplane car whizzed past the State Police booth, going 80 miles per hour. The trooper on duty phoned an alert to other stations on the road, then set out on his motorcycle in pursuit. He had gone only a short distance when the brown Terraplane hurtled past him, go­ing in the opposite direction. The car was later caught by a road block, and its occupants proved to be a gang of thieves who had just robbed a jewelry store.

Witnesses testified that the thieves had put their plunder in the car when they fled the scene of the crime. But it was no longer in the car when it was caught. Reports on the wild ride showed that the only time the car could have stopped was in doubling back past the State Police booth.

The trooper reported that the point at which the car passed him on its return was just 2 miles from his booth, and that it reached him just 7 minutes after it had first passed his booth. On both occasions it was apparently making its top speed of 80 miles per hour.

The investigators assumed that the car had made a stop and turned around while some members of the gang cached the loot by the roadside, or perhaps at the office of a “fence.” In an effort to locate the cache, they assumed that the car had maintained a uniform speed, and allowed 2 minutes as the probable loss of time in bringing the car to a halt, turning it, and regaining full speed.

On this assumption, what was the farthest point from the booth that would have to be covered by the search for the loot?”

Answer.

See Locating the Loot for solutions.

Mystery Dice Question

This is a relatively simple probability question from Presh Talwalkar that becomes an excuse to describe a powerful tool.

“Amazon’s Mystery Dice Interview Question

You are given a normal die and a blank die. (Each die is six-sided and equally likely to show each face). Label the blank die using the numbers 0 to 6 so that when you roll the two die the sum shows each whole number from 1 to 12 with equal chance. You can use a number more than once, or not at all, so you could label the faces 1, 2, 3, 4, 4, 5. But you do have to label all six faces of the blank die.”

See the Mystery Dice Question

Spy Gift Problem

This is a recent Alex Bellos problem that supposedly can be solved by 12-year-olds!

“Today’s problems come from Axiom Maths, a charity that that takes high-attaining primary school children and provides them with maths enrichment during secondary school.

One of Axiom’s main activities is to organise ‘maths circles’, in which small groups of pupils get together to tackle fun problems. Such as the ones below, which are aimed at children aged 11/12, and form the basis for further explorations.

Really Secret Santa

A group of nine secret agents: 001, 002, 003, 004, 005, 006, 007, 008 and 009 have organised a Secret Santa. The instructions are coded, to keep the donors secret.

  • Agent 001 gives a present to the agent who gives a present to agent 002
  • Agent 002 gives a present to the agent who gives a present to agent 003
  • Agent 003 gives a present to the agent who gives a present to agent 004
  • and so on, until
  • Agent 009 gives a present to the agent who gives a present to agent 001

Which agent will agent 007 get her present from?”

Answer.

See the Spy Gift Problem for solutions.

(Update 9/7/2024) Rigorous Solution Continue reading →

Timing the Car

This is yet another simple problem from Henry Dudeney.

“57. TIMING THE CAR

“I was walking along the road at three and a half miles an hour,” said Mr. Pipkins, “when the car dashed past me and only missed me by a few inches.”

“Do you know at what speed it was going?” asked his friend.

“Well, from the moment it passed me to its disappearance round a corner I took twenty-seven steps and walking on reached that corner with one hundred and thirty-five steps more.”

“Then, assuming that you walked, and the car ran, each at a uniform rate, we can easily work out the speed.” ”

Answer.

See Timing the Car for a solution.

Distance to Flag Problem

The following puzzle is from the Irishman Owen O’Shea.

“The figure shows the location of three flags [at A, B, and C] in one of the fields on a neighbor’s farm.  The angle ABC is a right angle.  Flag A is 40 yards from Flag B.  Flag B is 120 yards from flag C.  Thus, if one was to walk from A to B and then on to C, one would walk a total of 160 yards.

Now there is a point, marked by flag D, [directly] to the left of flag A.  Curiously, if one were to walk from flag A to flag D and then diagonally across to flag C, one would walk a total distance of 160 yards.

The question for our puzzlers is this: how far is it from flag D to flag A?”

This problem has a simple solution.  But it also suggests a more advanced alternative approach.

Answer.

See the Distance to Flag Problem for a solution.

More Right Triangle Magic

James Tanton asked to prove the following surprising property of a right triangle and its circumscribed and inscribed circles.

“Every triangle is circumscribed by some circle of diameter D, say, and circumscribes another circle of smaller diameter d. For a right triangle, d + D equals the sum of two side lengths of the triangle. Why?”

See More Right Triangle Magic

Two Containers Mixing Puzzle

This is a slightly different type of a mixture problem from Dan Griller.

“Two containers A and B sit on a table, partially filled with water.  First, 40% of the water in A is poured into B, which completely fills it.  Then 75% of the water in B is poured into A, which completely fills it.  80% of the water in A is poured into B, which completely fills it.  Calculate the ratio of the capacity of container A to the capacity of container B, and the fraction of container A that was occupied by water at the start.”

Answer.

See the Two Containers Mixing Puzzle for solution.

Circles in Circles

Here is another problem from the “Challenges” section of the Quantum magazine.

“Inside a circle there are two intersecting circles. One of them touches the big circle in point A, the other in point B. Prove that if segment AB meets the smaller circles at one of their common points, then the sum of their radii equals the radius of the big circle. Is the converse true?  (A. Vesyolov)”

See Circles in Circles

  1. Pages:
  3. 1
  4. ...
  5. 6
  6. 7
  7. 8
  8. 9
  9. 10
  10. 11
  11. 12
  12. ...
  13. 34