Diophantus of Alexandria/Arithmetica/Book 1/Problem 17/Proof 1

Problem

The sums of $4$ numbers, omitting each of the numbers in turn, are $22$, $24$, $27$ and $20$ respectively

What are the numbers?

Solution

\(\ds a\)

\(=\)

\(\ds 9\)

\(\ds b\)

\(=\)

\(\ds 7\)

\(\ds c\)

\(=\)

\(\ds 4\)

\(\ds d\)

\(=\)

\(\ds 11\)

Proof

Let $x$ be the sum of all $4$ numbers.

Then the numbers individually are:

\(\ds a\)

\(=\)

\(\ds x - 22\)

\(\ds b\)

\(=\)

\(\ds x - 24\)

\(\ds c\)

\(=\)

\(\ds x - 27\)

\(\ds d\)

\(=\)

\(\ds x - 20\)

\(\ds \leadsto \ \ \)

\(\ds a + b + c + d = x\)

\(=\)

\(\ds 4 x - \paren {22 + 24 + 27 + 20}\)

\(\ds \leadsto \ \ \)

\(\ds 3 x\)

\(=\)

\(\ds 93\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds 31\)

The result follows.

$\blacksquare$

Sources

• c. 250: Diophantus of Alexandria: Arithmetica: Book $\text I$: Problem $17$
• 1910: Sir Thomas L. Heath: Diophantus of Alexandria (2nd ed.): The Arithmetica: Book $\text {I}$: $17$