New user? Sign up

Existing user? Log in

What is the fastest way to show that \(102^2 \equiv 30\mod {247}\)?

Note by Axas Bit 2 years, 5 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

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

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

I suppose one way would be to first note that \(102 = 124 - 22\), and so

\(A = 102^{2} = (124 - 22)^{2} = 124^{2} - 2*22*124 + 22^{2} = 62*248 - 22*248 + 484.\)

Now \(248 \equiv 1 \pmod{247}\) and \(484 = 2*248 - 10 \equiv -10 \pmod{247}\), so

\(A \equiv (62*1 - 22*1 - 10) \pmod{247} \equiv 30 \pmod{247}\).

Log in to reply

Define "fastest".

In terms of no need to think about what to do: \( 102^2 = 42 \times 247 + 30 \).

'Fastest way' should definitely be checking if 247 divides \(102^2-30\). We find that it is true very fast.

Brian Charlesworth gave a standard answer. Here's an alternative approach:

Notice that \(50^2 = 2500 = 30 + 2470 = 30 + 247(10) \), then \[ 102^2 - 50^2 = (102 - 50)(102 + 50) = 52 \times 152 = 247 \times (\ldots ). \]

This tells us that \(102^2 - 50^2 \equiv 0 \pmod {247} \Rightarrow 102^2 \equiv 50^2 \pmod{247} \Rightarrow 10^2 \equiv 30 \pmod{247} \).

@Calvin Lin @Pi Han Goh @Otto Bretscher

@Brian Charlesworth Any thoughts?

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:

TopNewestI suppose one way would be to first note that \(102 = 124 - 22\), and so

\(A = 102^{2} = (124 - 22)^{2} = 124^{2} - 2*22*124 + 22^{2} = 62*248 - 22*248 + 484.\)

Now \(248 \equiv 1 \pmod{247}\) and \(484 = 2*248 - 10 \equiv -10 \pmod{247}\), so

\(A \equiv (62*1 - 22*1 - 10) \pmod{247} \equiv 30 \pmod{247}\).

Log in to reply

Define "fastest".

In terms of no need to think about what to do: \( 102^2 = 42 \times 247 + 30 \).

Log in to reply

'Fastest way' should definitely be checking if 247 divides \(102^2-30\). We find that it is true very fast.

Log in to reply

Brian Charlesworth gave a standard answer. Here's an alternative approach:

Notice that \(50^2 = 2500 = 30 + 2470 = 30 + 247(10) \), then \[ 102^2 - 50^2 = (102 - 50)(102 + 50) = 52 \times 152 = 247 \times (\ldots ). \]

This tells us that \(102^2 - 50^2 \equiv 0 \pmod {247} \Rightarrow 102^2 \equiv 50^2 \pmod{247} \Rightarrow 10^2 \equiv 30 \pmod{247} \).

Log in to reply

@Calvin Lin @Pi Han Goh @Otto Bretscher

Log in to reply

@Brian Charlesworth Any thoughts?

Log in to reply