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Note by Llewellyn Sterling 3 years, 3 months ago

Easy Math Editor

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We can easily prove that \(3^{444}+4^{333}\) is divisible by \(5\) using Modular Congruences.

Observe that

\[3^{444} \equiv 9^{222} \equiv (-1)^{222} \equiv 1 {\pmod 5}\]

Similarly,

\[4^{333} \equiv (-1)^{333} \equiv -1 {\pmod 5}\]

Adding up congruences, we get

\[3^{444}+4^{333} \equiv 1 + (-1) \equiv 0 {\pmod 5} \quad _\square\]

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestWe can easily prove that \(3^{444}+4^{333}\) is divisible by \(5\) using Modular Congruences.

Observe that

\[3^{444} \equiv 9^{222} \equiv (-1)^{222} \equiv 1 {\pmod 5}\]

Similarly,

\[4^{333} \equiv (-1)^{333} \equiv -1 {\pmod 5}\]

Adding up congruences, we get

\[3^{444}+4^{333} \equiv 1 + (-1) \equiv 0 {\pmod 5} \quad _\square\]

Log in to reply