\[\large { \left( \sqrt { 10x+1 } +1 \right) }^{ 10x+1 }-{ \left( \sqrt { 10x+1 } -1 \right) }^{ 10x+1 }\]

Find and generalise the last digit of the expression above to \(x\) where \(x\) is a whole number.

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:

TopNewestWhat do you mean? Is \(x\) supposed to be a fixed integer?

Or is the last digit of that expression independent of \(x\)? What if \(x\) is a real number?

Log in to reply

20x

Log in to reply

When \sqrt{10x-1} is rational,the last digit is always 2. Also ,x must be even. This would be because the of 10x+1 always ends with a 1, Some basic number theory says that sqrt(10x+1) ends with a 9 or 1. If 9,the expression is equivalent to (9+1)^(10x+1)-(9-1)^(10x+1) =0-8^(2x+1) =2 (mod 10) If 1, (1+1)^(10x+1)-(1-1)^(10x+1) =2^(2x+1)-0 =2-0 =2 mod(10). Hope this answers your question.

Log in to reply

Not true, The last digit of \({ \left( \sqrt { 91 } +1 \right) }^{ 91 }-{ \left( \sqrt { 91 } -1 \right) }^{ 91 }\) is 8

Log in to reply

I did state that the above is true if \[\sqrt{10x+1}\]is rational. But now you mentioned it,I will try to generalize the cases. Thanks anyway.

Log in to reply

..

Log in to reply