Die Hard #01

${ 7 }^{ 9999 }\quad \equiv \quad x\quad (mod\quad 1000)\\ By\quad Euler's\quad Theorem,\quad { a }^{ \phi (m) }\quad \equiv \quad 1\quad (mod\quad m).\quad \\ \phi (m)\quad =\quad m(1-\cfrac { 1 }{ a } )(1-\cfrac { 1 }{ b } ).\quad .\quad .\quad .\quad .\quad \quad \quad \quad \quad \\ m\quad =\quad { a }^{ p }.{ b }^{ q }{ .c }^{ r }.......\quad ,\quad where\quad a,b,c,...\quad are\quad primes.\quad \quad \\ So,\quad \phi (1000)\quad =\quad 1000(1-\cfrac { 1 }{ 2 } )(1-\cfrac { 1 }{ 5 } )\quad =\quad 400.\quad \\ { 7 }^{ 400 }\quad \equiv \quad 1\quad (mod\quad 1000)\quad \Longrightarrow \quad { 7 }^{ 10000 }\quad \equiv \quad 1\quad (mod\quad 1000).\quad \\ { 7 }^{ 10000 }\quad \equiv \quad 7x\quad (mod\quad 1000)\quad \Longrightarrow \quad 7x\quad \equiv \quad 1\quad (mod\quad 1000).\quad \\ \Longrightarrow \quad x\quad =\quad 143.\quad \quad \quad$

\text{It can be done binomially too}\implies 7^{9999}=343.7^{9996}=343[2400+1]^{249}\ =343[\dbinom{249}{0} 2400^{249}+\dbinom{249}{1} 2400^{248}\cdots \cdots \dbinom{249}{248} 2400+\dbinom{249}{249} 1]\ =343[2400^{249}+249.2400^{248}+\cdots \cdots \cdots +249.2400+1]\ \text{only last 2 terms are sufficient for last 3 digits}\ \implies 343[\cdots \cdots \cdots +597600+1]\ \implies 343.597600+343\ \text{last 3 digits are }\implies [.....800+343]=143