The Pill Scale (Part 2)

A variation on yesterday’s puzzle:

Suppose there are six bottles of pills, and more than one of them may contain defective pills that weigh 6 grams instead of 5. How can we identify the bad bottles with a single weighing?

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Take 1 pill from the first bottle, 2 from the second, 4 from the third, 8 from the fourth, 16 from the fifth, and 32 from the sixth. This gives us 63 pills that ought to weigh 315 grams. As before, the bad pills will add some excess weight, but now that weight can tell us which combination of bottles is bad. Suppose the 63 pills weigh 336 grams, or 21 grams too much. That extra 21 grams can only have come from the fifth bottle (16 grams), the third (4 grams), and the first (1 gram). The same is true for any overage: Because any integer is uniquely expressible as the sum of powers of 2, the excess weight will always identify uniquely any combination of bad bottles.

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February 16, 2014February 23, 2014 | Puzzles

The Pill Scale (Part 1)

An efficiency-minded pharmacist has just received a shipment of 10 bottles of pills when the manufacturer calls to say that there’s been an error — nine of the bottles contain pills that weigh 5 grams apiece, which is correct, but the pills in the remaining bottle weigh 6 grams apiece. The pharmacist could find the bad batch by simply weighing one pill from each bottle, but he hits on a way to accomplish this with a single weighing. What does he do?

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He numbers the bottles from 1 to 10, then takes one pill from bottle one, two pills from bottle two, and so on. This gives him a pile of 55 pills that ought to weigh 275 grams. Its difference from this weight can tell him which bottle is bad: If the pile is 1 gram overweight then the bad pills are in bottle one, if 2 grams then they’re in bottle two, and so forth.

February 15, 2014February 23, 2014 | Puzzles

Black and White

kuznecov and plaskin chess problem

This week’s puzzle has a twist: Imagine that the board has been rolled into a cylinder so that the a- and h-files are joined and pieces can move across the boundary. How can White mate in two moves?

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1. Ra5! The rook tours the fifth rank and returns to its original square, forcing 1. … Kc1 2. Ra1#. The normal-seeming 1. Rxa6, which would ordinarily lead to the same mate, here fails because Black can take the rook. By Kuznecov and Plaskin, from Miodrag Petković, Mathematics and Chess, 1997.

February 14, 2014February 23, 2014 | Puzzles

Letters and Numbers

The first 10 letters of the alphabet, ABCDEFGHIJ, form a cipher that conceals the name of a number less than 100. What is the number?

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Eighty-four.

February 10, 2014February 16, 2014 | Puzzles

Groaners

How does a deaf man indicate to a hardware clerk that he wants to buy a saw?
How can you aim your car north on a straight road, drive for a hundred yards, and find yourself a hundred yards south of where you started?
What runs fore to aft on one side of a ship and aft to fore on the other?
A very fast train travels from City A to City B in an hour and a quarter. But the return trip, made under identical conditions, requires 75 minutes. Why?
Does Canada have a 4th of July?
Exhausted, you go to bed at 8 p.m., but you don’t want to miss an appointment at 10 a.m. the next day, so you set your alarm clock for 9. How many hours do you sleep?

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He says, “I want to buy a saw.” Drive backward. Its name. An hour and a quarter is 75 minutes. Yes. One.

February 9, 2014February 16, 2014 | Puzzles

Black and White

shinkman chess problem

By William Anthony Shinkman. White to mate in two moves.

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After the quiet 1. Qf2!, the queen will mate next move. From the Dubuque Chess Journal, July 1871.

February 7, 2014February 16, 2014 | Puzzles

The Rolling Die

rolling die puzzle

Imagine a die that exactly covers one square of a checkerboard. Place the die in the top left corner with the 6 uppermost. Now, by tipping the die over successively onto each new square, can you roll it through each of the board’s 64 squares once and arrive in the upper right, so that the 6 is exposed at the beginning and end but never elsewhere?

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The solution above is unique. I think this puzzle was devised by John Harris of Santa Barbara, Calif., in 1964; I found it in Miodrag Petkovic’s Mathematics and Chess, 1997.

February 5, 2014February 16, 2014 | Puzzles

Perspective

In a photograph, is there a way to distinguish between a landscape and its reflection?

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Yes. Because the camera is above the reflective surface, the two images are not identical. If the camera lens is, say, 5 feet above the water, then the reflected image appears as if viewed from 5 feet below it.

February 4, 2014February 9, 2014 | Puzzles

Black and White

holst chess problem

By Viktor Holst. White to mate in two moves.

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1. Qe4! and, one way or another, the queen will mate. From the Illustrated Family Journal, 1917.

January 31, 2014February 9, 2014 | Puzzles

In a Word

manzil
n. the distance between two stopping places

Another puzzle by Sam Loyd: Two ferry boats ply the same route between ports on opposite sides of a river. They set out simultaneously from opposite ports, but one is faster than the other, so they meet at a point 720 yards from the nearest shore. When each boat reaches its destination, it waits 10 minutes to change passengers, then begins its return trip. Now the boats meet at a point 400 yards from the other shore. How wide is the river?

“The problem shows how the average person, who follows the cut-and-dried rules of mathematics, will be puzzled by a simple problem that requires only a slight knowledge of elementary arithmetic. It can be explained to a child, yet I hazard the opinion that ninety-nine out of every hundred of our shrewdest businessmen would fail to solve it in a week. So much for learning mathematics by rule instead of common sense which teaches the reason why!”

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At their first meeting, 720 yards from one shore, the boats together have traveled a distance equal to the width of the river. When each reaches its destination, the combined distance is twice the width of the river. At their second meeting, 400 yards from the other shore, the combined distance is three times the river’s width. So at the second meeting each boat has gone three times as far as it had at the first meeting. At the first meeting one boat had traveled 720 yards, so at the second meeting it has gone three times this distance, or 2,160 yards. At this point it has covered 400 yards since doubling back, so the river is 2,160 – 400 = 1,760 yards, or one mile, wide. “The amount of time each ship consumed at the landing does not affect the problem.”

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January 30, 2014February 9, 2014 | Language · Puzzles