Consider the following results.

\(99^{1} = 99\)

\(99^{2} = 9801\)

\(99^{3} = 970299\)

\(99^{4} = 96059601\)

\(99^{5} = 9509900499\)

Prove that \(99^{n}\) ends in 99 for odd n.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

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

Sort by:

TopNewest\(99\equiv(-1)\pmod{100} \Rightarrow 99^{n}\equiv(-1)^{n}\pmod{100}.\)But,whenever \(n\) is an odd integer\(, (-1)^{n}=(-1) \Rightarrow 99^{n}\equiv(-1)\equiv(99)\pmod{100}\) for all odd \(n.\)

Log in to reply

Seriously. And Obviously -_- This was easy. Welll Done Adarsh Kumar

Log in to reply

For this , let us consider two cases when \(n\) is odd and when it is even ,

When \(n\) is even :

QEDLog in to reply

This is similar to prove that

\((100-1)^n\equiv 99 \mod 100\)

For \(n\) is an odd number.

Noticed that every term of the expansion is divisible by \(100\) except for the last term, \(-1\). Hence we get

\((100-1)^n\equiv -1 \mod 100\)

\((100-1)^n\equiv 99 \mod 100\)

Log in to reply

While for \(n\) is an even number, a similar approach can prove that \(99^n\) ends in \(01\).

Log in to reply