let-p-a-prime-number-s-t-p-7-and-a-333-3-p-1-times-Show-that-11-a-

Question Number 111208 by Aziztisffola last updated on 02/Sep/20

let p a prime number s . t p ⩾ 7 and

a = \underset{p - 1 t i m e s}{333 \dots \dots 3}

Show that 11 ∣ a .

Commented by mr W last updated on 02/Sep/20

p {d o e s n}^{'} t n e e d t o b e p r i m e, p c a n b e

a n y o d d n u m b e r p ⩾ 3 .

Commented by mr W last updated on 02/Sep/20

p = o d d n u m b e r, s a y p = 2 n + 1 w i t h

n ⩾ 1 .

a = \underset{2 n t i m e s}{3333 . . 33} = 3 \times \underset{2 n t i m e s}{1111 . . 11}

= 3 \times (11 + 1100 + 110000 + \dots + 1100 . . 00)

= 3 \times (1 + 100 + 10000 + \dots + 100 . . 00) \times 11

\Rightarrow 11 ∣ a

Commented by Aziztisffola last updated on 02/Sep/20

yes sir thank you .

Answered by Aziztisffola last updated on 02/Sep/20

My answer

a = 3 + 3 \times 10 + \dots + 3 \times 10^{p}

= 3 (1 + 10 + 10^{2} + \dots + 10^{p})

= 3 (\frac{10^{p + 1} - 1}{9}) = \frac{10^{p + 1} - 1}{3}

\Rightarrow 3 a = 10^{p + 1} - 1

10 \equiv - 1 [11] \Rightarrow 10^{p + 1} \equiv 1 [11] (p + 1 i s e v e n)

\Rightarrow 10^{p + 1} - 1 \equiv 0 [11]

\Rightarrow 3 a \equiv 0 [11] \Rightarrow 11 ∣ 3 a \overset{3 \land 11 = 1}{\Rightarrow} 11 ∣ a