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Knowledge You Can Build On — Est. 1998

Friday Puzzler -- Clamps

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Friday Puzzler -- Clamps

#1

Friday Puzzler -- Clamps

Alex Y

An old woodworker had many clamps and many sons.

In his will, he specified how the clamps were to be distributed:

The eldest son was to choose 1 clamp and get 1/7 of the remaining clamps, chosen by his brothers.

Son #2 was to choose 2 clamps and get 1/7 of the remaining clamps, also chosen by his brothers.

Son # 3 was to choose 3 ...

The youngest son was upset with this scheme until he figured out that each son would receive the same number of clamps.

How many sons and how many clamps did this old woodworker have?

Re: Friday Puzzler -- Clamps

#2

Re: Friday Puzzler -- Clamps

Henry Higginbotham

I made a couple of unsuccessful runs at this, done in by remainders that weren't divisible by seven. Then it occurred to me: It Can't Be Done! Because, as everyone knows, the number of clamps you need is always one more than the number you have.

Re: Friday Puzzler -- Clamps

#3

Hypothetical!

Alex Y

LOL! Good point, Henry. This whole puzzler is strictly hypothetical and not very realistic!

But it is solvable, without trial and error.

Re: Friday Puzzler -- Clamps

#4

Re: Hypothetical!

Larry Barrett

I like Henry's answer. But here is a question - when the nth son takes his turn, he first takes n clamps from the pile. At that point, are there zero clamps left? so that when he next takes 1/7 of the remainder, he takes 1/7 of zero, and the game is over?

Re: Friday Puzzler -- Clamps

#5

Yes

Alex Y

The will provides for distributing all of the clamps to his sons, so there are none left over after the last son gets his.

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