Henry Ernest Dudeney/Modern Puzzles/100 - Odds and Evens/Solution

Modern Puzzles by Henry Ernest Dudeney: $100$

Odds and Evens

Ask a friend to take an even number of coins in one hand and an odd number in the other.

You then undertake to tell him which hand holds the odd and which the even.

Tell him to multiply the number in the right hand by $7$ and the number in the left by $6$,

add the two products together, and tell you the result.

You can then immediately give him the required answer.

How are you to do it?

Solution

If the friend gives an odd total, the right hand holds the odd number.

Otherwise the left hand holds the odd number.

Proof

Let $r$ be the number of coins in the right hand, and $l$ be the number of coins in the left hand.

If $r$ is odd, then $7 r$ is also odd, while $6 l$ is always even.

Hence the total $7 r + 6 l$ is in this case odd.

If $r$ is even, then $7 r$ is also even, while again $6 l$ is always even.

Hence the total $7 r + 6 l$ is in this case even.

So if the friend gives an odd total, the right hand holds the odd number.

Otherwise the left hand holds the odd number.

$\blacksquare$

It is to be noted that with the low emphasis currently placed on ability to perform mental arithmetic, it is likely to be the case that a friend who is capable of multiplying the number of coins by $7$ and $6$ is also probably sufficiently mathematically sophisticated as to understand immediately how the trick works.

Also see

• Henry Ernest Dudeney: Puzzles and Curious Problems: $352$ - Locating the Coins, which revisits the same idea.

Sources

• 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $100$. -- Odds and Evens