Bachet's Recreational Problems

[aeh49]

Please help me with any or all of the following problems from Claude Gaspar Bachet. My text book does not explain how to go about solving these questions.

1) Ask a person to choose secretly a number, and then to treble it. Inquire if the product is even or odd. If it is even, ask him to take half of it; if it is odd, ask him to add 1 and then take half of it. Tell him to multiply the result by 3 and to tell you how many times, say n, 9 will divide integrally into the product. Then the number originally chosen was 2n or 2n+1, according as the result in the first step was even or odd. Prove this.

2) Tell A to choose secretly any number, greater than 5, of counters, and B to take 3 times as many. Ask A to give 5 counters to B, and then ask B to transfer to A 3 times as many counters as A has left. You may now tell B that he has 20 counters. Explain why this is so and generalize to the case where the 3 and 5 are replaced by p and q.

3) Ask someone to think of an hour, say m, and then to touch on a watch the number that marks some other hour, say n. If, beginning with the number touched, he taps successively in the counterclockwise direction the numbers on the watch, meanwhile mentally counting the taps as m, m+1, and so on, until he reaches the number n + 12, then the last number tapped will be that of the hour he originally thought of. Prove this.

 

[soroban]

Hello, aeh49!

I think I've got the first two . . .

1) Ask a person to choose secretly a number, and then to treble it.
Inquire if the product is even or odd.
. . If it is even, ask him to take half of it.
. . If it is odd, ask him to add 1 and then take half of it.
Tell him to multiply the result by 3 and to tell you how many times,
say \(\displaystyle n\), 9 will divide integrally into the product.

Then the number originally chosen was \(\displaystyle 2n\) or \(\displaystyle 2n+1\),
according as the result in the first step was even or odd.
Prove this.

\(\displaystyle \text{The initial number is either [1] even or [2] odd.}\)


\(\displaystyle \text{[1] The number is even; it is of the form: }2n\)

. . .\(\displaystyle \text{Multiply by 3: }\:3(2n) \:=\:6n\)

. . .\(\displaystyle \text{This number is even, so we take half: }\:6n \div 2 \:=\:3n\)

. . .\(\displaystyle \text{Multiply by 3: }\:3(3n) \:=\:9n\)

. . .\(\displaystyle \text{Divide by 9: }\:9n \div 9 \;\to\;n\)

. . .\(\displaystyle \text{The first step was "even" . . . the original number was }2n.\)


\(\displaystyle \text{[2] The number is odd; it is of the form }2n+1.\)

. . .\(\displaystyle \text{Multiply by 3: }\:3(2n+1) \:=\:6n+3\)

. . .\(\displaystyle \text{This number is odd: add 1 and take half: }\:\frac{(6n+3)+1}{2} \:=\:3n+2\)

. . .\(\displaystyle \text{Multiply by 3: }\:3(3n+2) \:=\:9n + 6\)

. . .\(\displaystyle \text{Divide by 9: }\9n+6) \div 9 \to n\)

. . .\(\displaystyle \text{The first step was "odd" . . . The original number was }2n+1.\)

2) Tell A to choose secretly any number of counters, greater than 5, and B to take 3 times as many.
Ask A to give 5 counters to B, and then ask B to give to A 3 times as many counters as A has left.
You may now tell B that he has 20 counters.

Explain why this is so and generalize to the case where the 3 and 5 are replaced by p and q.

\(\displaystyle \text{Suppose A takes }x\text{ counters. }\text{ Then B takes }3x\text{ counters. }\;\begin{pmatrix}A\\B\end{pmatrix} \:=\:\begin{pmatrix}x\\3x\end{pmatrix}\)

\(\displaystyle \text{A gives 5 counters to B: }\;\begin{pmatrix}A\\B\end{pmatrix} \:=\:\begin{pmatrix}x-5 \\ 3x+5\end{pmatrix}\)

\(\displaystyle \text{B gives }3(x-5)\text{ counters to A: }\:\begin{pmatrix}A\\B\end{pmatrix} \:=\:\begin{pmatrix}x-5 + 3(x-5) \\ (3x+5) - 3(x-5) \end{pmatrix}\)


. . \(\displaystyle \text{Therefore, B has: }\3x+5) - 3(x-5) \:=\:3x + 5 - 3x + 15 \:=\:20\text{ counters.}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


\(\displaystyle \text{At the beginning, A takes }x\text{ counters, and B takes }px\text{ counters: }\;\begin{pmatrix}A\\B\end{pmatrix}\:=\:\begin{pmatrix}x \\ px\end{pmatrix}\)

\(\displaystyle \text{A gives }q\text{ counters to B: }\:\begin{pmatrix}A\\B\end{pmatrix} \:=\:\begin{pmatrix}x-q\\px+q\end{pmatrix}\)

\(\displaystyle \text{B gives }p(x-q)\text{ counters to A: }\;\begin{pmatrix}A\\B\end{pmatrix} \:=\:\begin{pmatrix}(x-q) + p(x-q) \\ (px+q)-p(x-q) \end{pmatrix}\)


\(\displaystyle \text{Therefore, B has: }\;(px+q) - p(x-q) \;=\;px + q - px + pq\)

. . . . . . . . . . . . \(\displaystyle =\;q + pq \;=\;q(p+1)\text{ counters.}\)