Diophantus of Alexandria/Arithmetica/Book 1/Problem 8
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Problem
- What number must be added to $100$ and to $20$ (the same number added to each) so that the sums are in the ratio $3 : 1$?
Solution
- $20$
Proof
Let $x$ be the number added.
Then we have:
\(\ds 100 + x\) | \(=\) | \(\ds 3 \paren {20 + x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 100 - 60\) | \(=\) | \(\ds 2 x\) | rearranging | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 20\) | simplifying |
$\blacksquare$
Sources
- c. 250: Diophantus of Alexandria: Arithmetica: Book $\text I$: Problem $8$
- 1910: Sir Thomas L. Heath: Diophantus of Alexandria (2nd ed.): The Arithmetica: Book $\text {I}$: $8$
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): The First Pure Number Puzzles: $23$