Set of 5 Triplets whose Sums and Products are Equal

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Theorem

The following set of $5$ triplets of integers have the property that:

the sum of the integers in each triplet are equal

and:

the product of the integers in each triplet are equal:
$\tuple {6, 480, 495}$, $\tuple {11, 160, 810}$, $\tuple {12, 144, 825}$, $\tuple {20, 81, 880}$, $\tuple {33, 48, 900}$

The sum is $981$, and the product is $1 \, 425 \, 600$.


This is the only known such set of $5$ triplets of integers with this property.


Proof

\(\ds 6 + 480 + 495\) \(=\) \(\ds 981\)
\(\ds 11 + 160 + 810\) \(=\) \(\ds 981\)
\(\ds 12 + 144 + 825\) \(=\) \(\ds 981\)
\(\ds 20 + 81 + 880\) \(=\) \(\ds 981\)
\(\ds 33 + 48 + 900\) \(=\) \(\ds 981\)


\(\ds 6 \times 480 \times 495\) \(=\) \(\ds \paren {2 \times 3} \times \paren {2^5 \times 3 \times 5} \times \paren {3^2 \times 5 \times 11}\)
\(\ds \) \(=\) \(\ds 2^6 \times 3^4 \times 5^2 \times 11\)
\(\ds \) \(=\) \(\ds 1 \, 425 \, 600\)
\(\ds 11 \times 160 \times 810\) \(=\) \(\ds 11 \times \paren {2^5 \times 5} \times \paren {2 \times 3^4 \times 5}\)
\(\ds \) \(=\) \(\ds 2^6 \times 3^4 \times 5^2 \times 11\)
\(\ds \) \(=\) \(\ds 1 \, 425 \, 600\)
\(\ds 12 \times 144 \times 825\) \(=\) \(\ds \paren {2^2 \times 3} \times \paren {2^4 \times 3^2} \times \paren {3 \times 5^2 \times 11}\)
\(\ds \) \(=\) \(\ds 2^6 \times 3^4 \times 5^2 \times 11\)
\(\ds \) \(=\) \(\ds 1 \, 425 \, 600\)
\(\ds 20 \times 81 \times 880\) \(=\) \(\ds \paren {2^2 \times 5} \times 3^4 \times \paren {2^4 \times 5 \times 11}\)
\(\ds \) \(=\) \(\ds 2^6 \times 3^4 \times 5^2 \times 11\)
\(\ds \) \(=\) \(\ds 1 \, 425, 600\)
\(\ds 33 \times 48 \times 900\) \(=\) \(\ds \paren {3 \times 11} \times \paren {2^4 \times 3} \times \paren {2^2 \times 3^2 \times 5^2}\)
\(\ds \) \(=\) \(\ds 2^6 \times 3^4 \times 5^2 \times 11\)
\(\ds \) \(=\) \(\ds 1 \, 425 \, 600\)

$\blacksquare$


Sources