Diophantus of Alexandria/Arithmetica/Book 1/Problem 8

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: $23$