Tag Archives: Presh Talwalkar

Three Dutchmen Puzzle

Presh Talwalkar presented an interesting puzzle that originated in the Ladies’ Diary of 1739-40, was recast by Henry Dudeney in 1917, and further modified using American money.

“Each of three Dutchmen, named Hendrick, Elas, and Cornelius has a wife. The three wives have names Gurtrün, Katrün, and Anna (but not necessarily matching the husband’s names in that order). All six go to the market to buy hogs.

Each person buys as many hogs as he or she pays dollars for one. (1 hog costs $1, 2 hogs are $2 each, 3 hogs cost $3 each, etc.) In the end, each husband has spent $63 more than his wife. Hendrick buys 23 more hogs than Katrün, and Elas 11 more than Gurtrün. Now, what is the name of each man’s wife?”

See the Three Dutchmen Puzzle 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

String of Beads Puzzle

This is a nifty problem from Presh Talwalkar.

“This is from a Manga called Q.E.D. I thank Sparky from the Philippines for the suggestion!

A string of beads is formed from 25 circles of the same size. The string passes through the center of each circle. The area enclosed by the string inside each circle is shaded in blue, and the remaining areas of the circles are shaded in orange. What is the value of the orange area minus the blue area? Calculate the area in terms of r, the radius of each circle.”

Answer.

See the String of Beads Puzzle for solutions.

Square In A Quarter Circle

Another puzzle by Presh Talwalkar.

“Thanks to John H. for the suggestion!

A square is inscribed in a quarter circle such that the outer vertices are on the arc of the quarter circle. If the quarter circle has a radius equal to 1, what is the area of the square?

I am told this was given to 7th grade students (ages 12-13), and I think it is a very challenging problem for that age group. In fact I think it is a good problem for any geometry student.”

Answer.

See the Square in Quarter Circle for solutions.

Putnam Ellipse Areas Problem

This is a nifty problem from Presh Talwakar.

“This is adapted from the 1994 Putnam, A2. Thanks to Nirman for the suggestion!

Let R be the region in the first quadrant bounded by the x-axis, the line y = x/2, and the ellipse x2/9 + y2 = 1. Let R‘ be the region in the first quadrant bounded by the y-axis, the line y = mx and the ellipse. Find the value of m such that R and R‘ have the same area.”

Answer.

See the Putnam Ellipse Areas Problem for solution.

Incredible Trick Puzzle

Here is another typical sum puzzle from Presh Talwalkar.

“Solve the following sums:

_____1/(1×3) + 1/(3×5) + 1/(5×7) + 1/(7×9) + 1/(9×11) =

_____1/(4×7) + 1/(7×10) + 1/(10×13) + 1/(13×16) =

_____1/(2×7) + 1/(7×12) + 1/(12×17) + … =”

The only reason I am including this puzzle is that Talwalkar gets very excited about deriving a formula that can solve sums of this type.  This gives me an opportunity to discuss the “formula vs. procedure” way of doing math.

Answer.

See the Incredible Trick Puzzle for solutions.

Triangle Stripes Problem

This is a fairly straight-forward problem from Presh Talwalkar.

“A triangle is divided by 8 parallel lines that are equally spaced, as shown below. Starting from the top small triangle, color each alternate stripe in blue and color the remaining stripes in red. If the blue stripes have a total area of 145, what is the total area of the red stripes?”

Answer.

See the Triangle Stripes Problem for solutions.

Pinocchio’s Hats

This problem in logic from Presh Talwalkar recalled an article I wrote a while ago but did not publish.  So I thought I would post it as part of the solution.

“Assume that both of the following sentences are true:

  1. Pinocchio always lies;
  2. Pinocchio says, “All my hats are green.”

We can conclude from these two sentences that:

  • (A) Pinocchio has at least one hat.
  • (B) Pinocchio has only one green hat.
  • (C) Pinocchio has no hats.
  • (D) Pinocchio has at least one green hat.
  • (E) Pinocchio has no green hats.”

Actually, the question is which, none or more, of statements (A) – (E) follow from the two sentences?

Answer.

See Pinocchio’s Hats for solutions.

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