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Why do we deal with Euler's Theorem and perform calculation on mod 100,1000,etc. while solving questions like: What are the last three digits of \(7^{999}\) ?

Note by Swapnil Das 3 years, 1 month ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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Euler theorem helps us to deal with large congruences like mod 100 and mod 1000 with ease.

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Hey! I am launching a contest! ON BRILLIANT! Would u participate?

Ok bro !

@Karthik Venkata – It is just mini RMO contest on BRILLIANT!

But why mod 100, 1000?

For finding out the last and last but one digits , we work in mod 10 and mod 100 systems respectively.

@Karthik Venkata – Oh thanks!

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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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TopNewestEuler theorem helps us to deal with large congruences like mod 100 and mod 1000 with ease.

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Hey! I am launching a contest! ON BRILLIANT! Would u participate?

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Ok bro !

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But why mod 100, 1000?

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For finding out the last and last but one digits , we work in mod 10 and mod 100 systems respectively.

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