A-n-2-n-3-n-4-n-5-n-B-n-100-n-101-n-102-n-103-n-1-find-values-of-n-while-7-A-n-2-show-that-B-n-A-n-7-

Question Number 151256 by pticantor last updated on 19/Aug/21

A_{n} = 2^{n} + 3^{n} + 4^{n} + 5^{n}

B_{n} = 100^{n} + 101^{n} + 102^{n} + 103^{n}

1) f i n d v a l u e s o f n w h i l e 7 ∣ A_{n}

2) s h o w t h a t B_{n} \equiv A_{n} [7]

Answered by Rasheed.Sindhi last updated on 20/Aug/21

1)

2^{3} \equiv 1 (m o d 7) \Rightarrow 2^{6} \equiv 1 (m o d 7) \Rightarrow 2^{6 k} \equiv 1 (m o d 7)

3^{6} \equiv 1 (m o d 7) \Rightarrow 3^{6 k} \equiv 1 (m o d 7)

4^{3} \equiv 1 (m o d 7) \Rightarrow 4^{6} \equiv 1 (m o d 7) \Rightarrow 4^{6 k} \equiv 1 (m o d 7)

5^{6} \equiv 1 (m o d 7) \Rightarrow 5^{6 k} \equiv 1 (m o d 7)

{\begin{cases} (2^{6 k} \equiv 1 (m o d 7)) \times 2^{u} \\ (3^{6 k} \equiv 1 (m o d 7)) \times 3^{u} \\ (4^{6 k} \equiv 1 (m o d 7)) \times 4^{u} \\ (5^{6 k} \equiv 1 (m o d 7)) \times 5^{u} \end{cases}; u \in {0, 1, 2, \dots, 5}

\Rightarrow {\begin{cases} 2^{6 k + u} \equiv 2^{u} (m o d 7) \\ 3^{6 k + u} \equiv 3^{u} (m o d 7) \\ 4^{6 k + u} \equiv 4^{u} (m o d 7) \\ 5^{6 k + u} \equiv 5^{u} (m o d 7) \end{cases}; u \in {0, 1, 2, \dots, 5}

n = 6 k + u \Rightarrow

A_{n} = 2^{n} + 3^{n} + 4^{n} + 5^{n} =

2^{6 k + u} + 3^{6 k + u} + 4^{6 k + u} + 5^{6 k + u}

\equiv 2^{u} + 3^{u} + 4^{u} + 5^{u} (m o d 7)

u = 1 : 2^{1} + 3^{1} + 4^{1} + 5^{1} = 14 = 7 \times 2 ✓

u = 2 : 2^{2} + 3^{2} + 4^{2} + 5^{2} = 54 \Rightarrow 7 ∤ 54 \times

\dots

u = 1, 3, 5 s a t i s f y t h e a b o v e c o n g r u e n c e

H e n c e

∴ F o r n = 6 k + 1, 6 k + 3, 6 k + 5, 7 ∣ A_{n}

\forall k \in {0, 1, 2, \dots}

Answered by Rasheed.Sindhi last updated on 20/Aug/21

2)

100 \equiv 2 (m o d 7) \Rightarrow 100^{n} \equiv 2^{n} (m o d 7) \dots (i)

101 \equiv 3 (m o d 7) \Rightarrow 101^{n} \equiv 3^{n} (m o d 7) \dots (i i)

102 \equiv 4 (m o d 7) \Rightarrow 102^{n} \equiv 4^{n} (m o d 7) \dots (i i i)

103 \equiv 5 (m o d 7) \Rightarrow 103^{n} \equiv 5^{n} (m o d 7) \dots (i v)

(i) + (i i) + (i i i) + (i v) :

100^{n} + 101^{n} + 102^{n} + 103^{n}

\equiv 2^{n} + 3^{n} + 4^{n} + 5^{n} (m o d 7)

B_{n} \equiv A_{n} (m o d 7)