let p a prime number s.t p≥7 and a=333......3_(p−1 times) Show that 11∣a.


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

$let p a prime number s . t p ⩾ 7 and$ $a = \underset{p - 1 t i m e s}{333 . . . . . . 3}$ $Show that 11 ∣ a .$
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

$p = o d d n u m b e r, s a y p = 2 n + 1 w i t h$ $n ⩾ 1 .$ $a = \underset{2 n t i m e s}{3333 . . 33} = 3 \times \underset{2 n t i m e s}{1111 . . 11}$ $= 3 \times (11 + 1100 + 110000 + . . . + 1100 . . 00)$ $= 3 \times (1 + 100 + 10000 + . . . + 100 . . 00) \times 11$ $\Rightarrow 11 ∣ a$
Commented by Aziztisffola last updated on 02/Sep/20

$yes sir thank you .$
	Answered by Aziztisffola last updated on 02/Sep/20

	$My answer$ $a = 3 + 3 \times 10 + . . . + 3 \times 10^{p}$ $= 3 (1 + 10 + 10^{2} + . . . + 10^{p})$ $= 3 (\frac{10^{p + 1} - 1}{9}) = \frac{10^{p + 1} - 1}{3}$ $\Rightarrow 3 a = 10^{p + 1} - 1$ $10 \equiv - 1 [11] \Rightarrow 10^{p + 1} \equiv 1 [11] (p + 1 i s e v e n)$ $\Rightarrow 10^{p + 1} - 1 \equiv 0 [11]$ $\Rightarrow 3 a \equiv 0 [11] \Rightarrow 11 ∣ 3 a \overset{3 \land 11 = 1}{\Rightarrow} 11 ∣ a$
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