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

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

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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?

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20x

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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.

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

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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.

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