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7 77 7 77 ( m o d 777 ) = ? \Large 7^{777^{77}} \pmod{777} =\, ?


The answer is 112.

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1 solution

Otto Bretscher
Mar 14, 2016

Let's first deal with the exponent, 77 7 77 777^{77} . We have 77 7 77 1 ( m o d 4 ) 777^{77}\equiv 1 \pmod{4} and 77 7 77 0 ( m o d 9 ) 777^{77}\equiv 0\pmod{9} since 777 is divisible by 3, so 77 7 77 9 ( m o d 36 ) 777^{77}\equiv 9 \pmod{36} . Since λ ( 111 ) = 36 \lambda(111)=36 , we have 7 77 7 77 1 7 8 7^{777^{77}-1}\equiv 7^8 16 ( m o d 111 ) \equiv 16 \pmod{111} and, multiplying through with 7, we find 7 77 7 77 112 ( m o d 777 ) 7^{777^{77}}\equiv \boxed{112} \pmod{777}

Could you please explain lamda function? I am not aware of it and thus not able to appreciate the solution provided by you. What is this lamda function called?

Kaustubh Miglani - 5 years, 2 months ago

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Hi :-)

The function that @Otto Bretscher is referring to is the Carmichael function .

Martin Sergio H. Faester - 5 years, 2 months ago

77 7 77 0 ( m o d 777 ) 7 0 1 ( m o d 777 ) 777^{77} \equiv 0 \pmod {777}\\ 7^0 \equiv 1 \pmod {777}

Where is the problem with this reasoning?

Alex Spagnoletti - 5 years, 2 months ago

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You can't just work modulo 777 in the exponent ... consider some simple examples. That's where the Lambda function comes in!

Otto Bretscher - 5 years, 2 months ago

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got it, thanks

Alex Spagnoletti - 5 years, 2 months ago

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