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Number Theory Level 4

What is the remainder when the last 3 digits of 3 5 7 9 { 3 }^{ 5^{ { 7 }^{ 9 } } } is divided by 43?


The answer is 13.

Lance Joseph Esquejo by Lance Joseph Esquejo , 25, Philippines

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1 solution

F i r s t , 7 9 m o d 1000 = 607 5 607 m o d 1000 = 125 s i n c e 5 n m o d 10 3 = 5 3 3 125 m o d 1000 = 443 T h e r e f o r e , 443 m o d 43 = ( 430 + 13 m o d 43 ) = 13 First,\\ { 7 }^{ 9 }mod1000=607\\ { 5 }^{ 607 }mod1000=125\quad since\quad { 5 }^{ n }mod{ 10 }^{ 3 }={ 5 }^{ 3 }\\ { 3 }^{ 125 }mod1000=443\\ Therefore,\\ 443mod43=(430+13mod43)=13\quad

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