Dr Math and his family question. How to solve without trial and error?

Question

Dr Math told his family to write 4 different integers from 1 to 9 on to the 4 ]

Answer 1

a), b) look good.

c) in order to produce 4 consecutive integers, one coin must have consecutive integers and one coin must have numbers separated by 1, (e.g. 3,5 or 6,8). the first coin has one even number, the second coin either has 0 or 2.

d) Make the powers of 2 work for you.

$1,2- (2-1 = 1 = 2^0)\\ 3,5 -(5-3 = 2 = 2^1)\\ 6,10 - (10-6 = 4 = 2^2)\\ 11,19- (19-11 = 8 = 2^3)\\ $

Can you reason why this works?

Answer 2

For problem d, the $16$ is a dead give-away: You will be forming these 16 numbers out of combinations of 4 binary digits. That is, the differences between the lesser and greater face of coin $k$ should be $2^{k-1}$. These differences are 1, 2, 4, and 8. If we were not allowed to go outside of the range 1 to 9, one of the coins would have to have a $1$ and a $9$ on its two faces.

There are plenty of solutions; a clean one is for the coins to be $$\pmatrix{(1,2)\\(1,3)\\(1,5)\\(1,9)} $$ which allows you to form any integer between 4 and 19.

Let's work backward to problem c. The same reasoning says that those two coins will have to have face-differences of 1 and 2 respectively. So the coin faces are $\{a, a+1, b, b+2\}$. Coin A thus has one odd and one even face, and coin B has either both odd or both even. Add the number of even faces and you get either 1 or 3.

Answer 3

There are two errors in part a.

You have two pairs of coins that share numbers, namely $4|5\ 4|6$ and $5|6\ 3|5$. With these removed we now have only three pairs of coins. Fortunately you've also missed a few: $6|7\ 2|4$ and $7|8\ 1|3$ both work, and do not have duplicates.

Everybody's throwing their hat in the ring for d, so here's mine: $1|9\ 2|4\ 3|7\ 5|6$. This provides every number from 11 to 26, and uses only one-digit numbers on the coins.