Ladies' Diary/Money Counting Problem
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Puzzle
- In how long a time would a million of millions of money be in counting,
- supposing one hundred pounds to be counted every minute without intermission,
- and the year to consist of $365$ days, $5$ hours, $45$ minutes?
Solution
The number of minutes $m$ in the year is calculated as:
\(\ds m\) | \(=\) | \(\ds \paren {365 \times 24 \times 60} + \paren {5 \times 60} + 45\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 525 \, 600 + 300 + 45\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 525 \, 945\) |
The number of $100$s $h$ in a million of millions is calculated as:
\(\ds h\) | \(=\) | \(\ds \dfrac {1 \, 000 \, 000 \times 1 \, 000 \, 000} {100}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 10 \, 000 \, 000 \, 000\) |
This gives the total number of minutes to count the money.
The total number of years $t$ needed to count the money is therefore:
\(\ds t\) | \(=\) | \(\ds \dfrac {10 \, 000 \, 000 \, 000} {525 \, 945}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 19013 \text { years}, 207 \, 715 \text { minutes}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 19013 \text { years}, 3461 \text { hours}, 55 \text { minutes}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 19013 \text { years}, 144 \text { days}, 5 \text { hours}, 55 \text { minutes}\) |
$\blacksquare$
Historical Note
This was the first mathematical question to be published in The Ladies' Diary.
It was posed in the year $1707$.
Sources
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): The Ladies' Diary or Woman's Almanac, $\text {1704}$ – $\text {1841}$: $136$