Diophantus of Alexandria/Arithmetica/Book 1/Problem 17

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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 1

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.

$\blacksquare$


Proof 2

Let $a$, $b$, $c$ and $d$ be the numbers requested.

Then we have:

\(\text {(1)}: \quad\) \(\ds b + c + d\) \(=\) \(\ds 22\)
\(\text {(2)}: \quad\) \(\ds a + c + d\) \(=\) \(\ds 24\)
\(\text {(3)}: \quad\) \(\ds a + b + d\) \(=\) \(\ds 27\)
\(\text {(4)}: \quad\) \(\ds a + b + c\) \(=\) \(\ds 20\)
\(\ds \leadsto \ \ \) \(\ds a - b\) \(=\) \(\ds 2\) $(2) - (1)$
\(\ds b - c\) \(=\) \(\ds 3\) $(3) - (2)$
\(\ds d - c\) \(=\) \(\ds 7\) $(4) - (3)$


We use the method of false position.

Set $c = 1$.

Then:

\(\ds b\) \(=\) \(\ds 1 + 3 = 4\)
\(\ds d\) \(=\) \(\ds 1 + 7 = 8\)
\(\ds a\) \(=\) \(\ds 4 + 2 = 6\)
\(\ds \leadsto \ \ \) \(\ds b + c + d\) \(=\) \(\ds 13\)
\(\ds a + c + d\) \(=\) \(\ds 15\)
\(\ds a + b + d\) \(=\) \(\ds 18\)
\(\ds a + b + c\) \(=\) \(\ds 11\)


These sums are all $9$ less than what they should be.

So we have to add $3$ to each number, to make:

\(\ds a\) \(=\) \(\ds 9\)
\(\ds b\) \(=\) \(\ds 7\)
\(\ds c\) \(=\) \(\ds 4\)
\(\ds d\) \(=\) \(\ds 11\)

$\blacksquare$


Sources

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