Q : ( 26 ! )^( 58) + k ≡^( 29) 0 and , k ∈ N , find: Min ( k )=?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 173411 by mnjuly1970 last updated on 11/Jul/22

$Q :$ ${(26!)}^{58} + k \overset{29}{\equiv} 0$ $and, k \in N, find :$ $Min (k) = ?$
	Answered by Rasheed.Sindhi last updated on 11/Jul/22
	$(29−1)!≡−1(mod 29) [Wilson′s theorem] 28!≡−1+29(mod 29) 28!≡28(mod 29) 27!≡1(mod 29) 27!≡1+29×13(mod 29) 27!≡378(mod 29) 26!≡14(mod 29) (26!)^2 ≡14^2 ≡22(mod 29).....(i) (26!)^(28) ≡14^(28) ≡1(mod 29) {(26!)^(28) }^2 ≡1^2 (mod 29)........(ii) (i)×(ii) (26!)^(56) ∙(26!)^2 ≡1∙22(mod 29) (26!)^(58) ≡22−29=−7(mod 29) (26!)^(58) +7≡0(mod 29) (26!)^(58) +7+29m≡0(mod 29) k=7+29m [k∈N⇒m≥0] m=0: min(k)=7$
	$(29 - 1)! \equiv - 1 (m o d 29) [{W i l s o n}^{'} s t h e o r e m]$ $28! \equiv - 1 + 29 (m o d 29)$ $28! \equiv 28 (m o d 29)$ $27! \equiv 1 (m o d 29)$ $27! \equiv 1 + 29 \times 13 (m o d 29)$ $27! \equiv 378 (m o d 29)$ $26! \equiv 14 (m o d 29)$ ${(26!)}^{2} \equiv 14^{2} \equiv 22 (m o d 29) . . . . . (i)$ ${(26!)}^{28} \equiv 14^{28} \equiv 1 (m o d 29)$ ${{(26!)}^{28}}^{2} \equiv 1^{2} (m o d 29) . . . . . . . . (i i)$ $(i) \times (i i)$ ${(26!)}^{56} \cdot {(26!)}^{2} \equiv 1 \cdot 22 (m o d 29)$ ${(26!)}^{58} \equiv 22 - 29 = - 7 (m o d 29)$ ${(26!)}^{58} + 7 \equiv 0 (m o d 29)$ ${(26!)}^{58} + 7 + 29 m \equiv 0 (m o d 29)$ $k = 7 + 29 m [k \in N \Rightarrow m ⩾ 0]$ $m = 0 : m i n (k) = 7$
	Commented by mnjuly1970 last updated on 11/Jul/22

	$thanks alot sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum