It’s a Puzzlement
Dividing the Rashers

by John P. Robertson

The following is from Oswald Jacoby and William Benson, Intriguing Mathematical Problems, Dover 1996.

At Harmony House it is Mrs. Crabbe’s custom to set up a platter of breakfast bacon so that her lodgers may help themselves as they come down in the morning. On this particular morning, had all [five] lodgers shared alike, there would have been a whole number of rashers each, but the Oldest Boarder, who came down last, found only half a rasher left for him. Jones always takes one rasher; Robinson takes his fair share of what remains; Brown is greedy, takes his fair share [of what remains] and then half a rasher extra; Wag likes three rashers, but, being superstitious, always leaves at least one for those who follow him.

How many rashers did Wag take?

Light Bulbs
The problem was to determine which of three switches controls which of three light bulbs in another room. You know that the up position of each switch turns its light on, and the down position turns the light off. You want to make your determination without going back and forth between the rooms. Mary Corbett sent in a solution that was the most popular among solvers. Namely, put two of the switches up, and one down. Wait a few minutes, and put down one of the two that are up. Move quickly to the other room. The light that is on is controlled by the switch that is up. For the two lights that are off, the cooler one is controlled by the switch originally down, and the warmer one is controlled by the switch that was up for awhile.

For the extra credit problem, one did not know whether the up position for a switch turned the light on or off. Bob Hallstrom’s solution is to do the same as the above. If one light is on, then the up position turns the light on, and the above solution applies. If two lights are on, then the down position turns the light on. The light that is off is controlled by the switch that is up. Of the two that are on, the cooler one is controlled by the switch changed from up to down last.

Solutions to the main problem were submitted by Gary Colton, Dale Evans, Bill Leslie, Eric Ruud, Bob Spitzer, and Desmond Sutherland. Both Mary Corbett and Bob Hallstrom submitted solutions to both parts, as did Richard Amundson, Jennifer Capute, Greg Cuzzi, D. A. Elia, Tim Hansen, Joe Kilroy, Tom Kozik, Thomas Lee, Philip Lew, Oleg Movchan, Barry Savage, Jim Shoenfelt, Mike Singer, Russell Wenitsky, and Jim Wickwire.

Number Theory Problems
One of Gary Venter’s problems was to investigate the possible lengths of sequences of consecutive integers each of which can be written as a product of two-digit numbers. Tom Struppeck observes that the maximum length of such a sequence is no more than 100 for the very simple reason that every 101st integer is divisible by 101. Since a sequence of length 8 is known, the maximum length of such a sequence is at least 8 and no more than 100. Can anyone narrow this down further?

A related question is whether the number of pairs of consecutive integers such that each can be written as a product of two-digit numbers is finite or infinite. This leads quickly to connections to the ABC-conjecture, which is considered just as intractable as Fermat’s Last Theorem was a few years ago.