Henry Ernest Dudeney/Modern Puzzles/198 - A Card Trick/Solution
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Modern Puzzles by Henry Ernest Dudeney: $198$
- A Card Trick
- Take an ordinary pack of playing-cards and regard all the court cards as tens.
- Now, look at the top card -- say it is a seven -- place it on the table face downwards and play more cards on top of it, counting up to twelve.
- Thus, the bottom card being seven, the next will be eight, the next nine, and so on, making six cards in that pile.
- Then look again at the top card of pack -- say it is a queen -- then count $10$, $11$, $12$ (three cards in all) and complete the second pile.
- Continue this, always counting up to twelve, and if at last you have not put sufficient cards to complete a pile, put these apart.
- Now, if I am told how many piles have been made and how many unused cards remain over,
- I can at once tell you the sum of all the bottom cards in the piles.
- I simply multiply by $13$ the number of piles less $4$, and add the number of cards left over.
- Thus, if there were $6$ piles and $4$ cards over, then $13$ times $2$ (i.e. $6$ less $4$) added to $5$ equals $31$, the sum of the bottom cards.
- Why is this?
- This is the question.
Solution
Let there be $n$ piles.
Let $p_k$ denote the number of cards in pile $k$.
Let $b_k$ denote the value of the bottom card.
Let $m$ be the number of cards left over.
We have that:
- $p_k = 13 - b_k$
for all $1 \le k \le n$.
\(\ds p_k\) | \(=\) | \(\ds 13 - b_k\) | by definition | |||||||||||
\(\ds 52\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n p_k + m\) | that is, the total number of cards in the pack | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {13 - b_k} + m\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 13 n - \sum_{k \mathop = 1}^n b_k + m\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 13 n + m - 52\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n b_k\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 13 \paren {n - 4} + m\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n b_k\) |
That is, the sum of all the bottom cards in the piles equals the number of piles minus $4$, all times $13$, plus the ones left over.
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $198$. -- A Card Trick
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $183$. A Card Trick