7 ... 7's remainder

N = 2 200 + 3 300 + 4 400 \large N = 2^{200} + 3^{300} + 4^{400}

Find the remainder when N N is divided by 7.


The answer is 2.

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3 solutions

Method 1: Note that

  • (i) 2 3 1 ( m o d 7 ) 2 200 = 2 3 66 + 2 2 2 ( m o d 7 ) 4 ( m o d 7 ) 2^{3} \equiv 1 \pmod{7} \Longrightarrow 2^{200} = 2^{3*66 + 2} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7} ;

  • (ii) 3 3 1 ( m o d 7 ) 3 300 = 3 3 100 ( 1 ) 100 ( m o d 7 ) 1 ( m o d 7 ) 3^{3} \equiv -1 \pmod{7} \Longrightarrow 3^{300} = 3^{3*100} \equiv (-1)^{100} \pmod{7} \equiv 1 \pmod{7} ;

  • (iii) 4 400 = 2 800 = 2 3 266 + 2 2 2 ( m o d 7 ) 4 ( m o d 7 ) 4^{400} = 2^{800} = 2^{3*266 + 2} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7} .

Thus N ( 4 + 1 + 4 ) ( m o d 7 ) 2 ( m o d 7 ) N \equiv (4 + 1 + 4) \pmod{7} \equiv \boxed{2} \pmod{7} .

Method 2: Using Fermat's Little Theorem we have that

  • (i) 2 6 1 ( m o d 7 ) 2 200 = 2 6 33 + 2 2 2 ( m o d 7 ) 4 ( m o d 7 ) 2^{6} \equiv 1 \pmod{7} \Longrightarrow 2^{200} = 2^{6*33 + 2} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7} ;

  • (ii) 3 6 1 ( m o d 7 ) 3 300 = 3 6 50 1 ( m o d 7 ) 3^{6} \equiv 1 \pmod{7} \Longrightarrow 3^{300} = 3^{6*50} \equiv 1 \pmod{7} ;

  • (iii) 4 6 1 ( m o d 7 ) 4 400 = 4 6 66 + 4 4 4 ( m o d 7 ) 2 2 ( m o d 7 ) 4 ( m o d 7 ) 4^{6} \equiv 1 \pmod{7} \Longrightarrow 4^{400} = 4^{6*66 + 4} \equiv 4^{4} \pmod{7} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7} .

Once again we have N ( 4 + 1 + 4 ) ( m o d 7 ) 2 ( m o d 7 ) N \equiv (4 + 1 + 4) \pmod{7} \equiv \boxed{2} \pmod{7} .

Chew-Seong Cheong
Dec 11, 2016

N 2 200 + 3 300 + 4 400 (mod 7) 2 3 × 66 + 2 + 3 3 × 100 + 4 3 × 133 + 1 (mod 7) 8 66 × 4 + 2 7 100 + 6 4 133 × 4 (mod 7) 4 ( 7 + 1 ) 66 + ( 28 1 ) 100 + 4 ( 63 + 1 ) 133 (mod 7) 4 ( 1 66 ) + ( 1 ) 100 + 4 ( 1 133 ) (mod 7) 4 + 1 + 4 9 2 (mod 7) \begin{aligned} N & \equiv 2^{200} + 3^{300} + 4^{400} \text{ (mod 7)} \\ & \equiv 2^{3\times 66+2} + 3^{3 \times 100} + 4^{3 \times 133+1} \text{ (mod 7)} \\ & \equiv 8^{66}\times 4 + 27^{100} + 64^{133}\times 4 \text{ (mod 7)} \\ & \equiv 4(7+1)^{66} + (28-1)^{100} + 4 (63+1) ^{133} \text{ (mod 7)} \\ & \equiv 4(1^{66}) + (-1)^{100} + 4 (1^{133}) \text{ (mod 7)} \\ & \equiv 4 + 1 + 4 \equiv 9 \equiv \boxed{2} \text{ (mod 7)} \end{aligned}

Viki Zeta
Dec 10, 2016

2 2 4 ( m o d 7 ) ( 2 2 ) 2 2 ( m o d 7 ) 2 4 2 ( m o d 7 ) ( 2 4 ) 5 2 5 ( m o d 7 ) 4 ( m o d 7 ) 2 100 4 ( m o d 7 ) 2 ( m o d 7 ) 2 200 2 2 ( m o d 7 ) 4 ( m o d 7 ) Similarly, 3 300 1 ( m o d 7 ) 4 400 4 ( m o d 7 ) { 2 200 + 3 300 + 4 400 } ( 1 + 4 + 4 ) ( m o d 7 ) 9 ( m o d 7 ) 2 ( m o d 7 ) 2^2 \equiv 4 \mathrm{(mod 7)} (2^2)^2 \equiv 2 \mathrm{(mod 7)} \\ 2^4 \equiv 2 \mathrm{(mod 7)} \\ (2^4)^5 \equiv 2^5 \mathrm{(mod 7)} \equiv 4 \mathrm{(mod 7)} \\ 2^{100} \equiv 4 \mathrm{(mod 7)} \equiv 2 \mathrm{(mod 7)} \\ 2^{200} \equiv 2^2 \mathrm{(mod 7)} \equiv 4 \mathrm{(mod 7)} \\ \text{Similarly,} \\ 3^{300} \equiv 1 \mathrm{(mod 7)} \\ 4^{400} \equiv 4 \mathrm{(mod 7)} \\ \color{#3D99F6}{\boxed{\therefore \{ 2^{200} + 3^{300} + 4^{400} \} \equiv (1 + 4 + 4) \mathrm{(mod 7)} \equiv 9 \mathrm{(mod 7)} \equiv 2 \mathrm{(mod 7)}}}

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