Find the value of X such that: 1∙5∙9∙13∙17∙...∙2021≡X(mod1000)


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Question Number 139737 by mathsuji last updated on 30/Apr/21

$F i n d t h e v a l u e o f X s u c h t h a t :$ $1 \cdot 5 \cdot 9 \cdot 13 \cdot 17 \cdot . . . \cdot 2021 \equiv X (m o d 1000)$
	Answered by mindispower last updated on 01/May/21
	$A=Π_(k=0) ^(505) (1+4k)≡X[1000] 1000=8.5^3 k=2n⇒1+4k≡1[8] k=2n+1⇒1+4k≡5[8] Π_(k=0) ^(505) (1+4k)≡Π_(k=0) ^(252) (8k+1).Π_(k=0) ^(252) (5+8k)≡Π_(k=0) ^(252) 5[8] =5^(252) [8] 5^2 =1[8]⇒5^(252) ≡1[8] ⇒A≡1[8] 1+4k=5s⇒5s−4k=1 (s,k)=(1,1) solution ⇒5(s−1)−4(k−1)=0⇒k=5l+1 ⇒1+4(5l+1)=5(4l+1)=1+4k l=0,l=1,l=2⇒5,45,25∈{1,5,9,13,17^� .....} ⇒(1.5.9......2021)≡0[125] A≡0[125] A≡1[8] A=8k+1=125d 125d−8k=1 (d,k)=(5,78) solution k=125m+78 A≡8(125m+78)+1=1000m+625[1000]≡X[100] X=625$
	$A = \underset{k = 0}{\prod^{505}} (1 + 4 k) \equiv X [1000]$ $1000 = 8 . 5^{3}$ $k = 2 n \Rightarrow 1 + 4 k \equiv 1 [8]$ $k = 2 n + 1 \Rightarrow 1 + 4 k \equiv 5 [8]$ $\underset{k = 0}{\prod^{505}} (1 + 4 k) \equiv \underset{k = 0}{\prod^{252}} (8 k + 1) . \underset{k = 0}{\prod^{252}} (5 + 8 k) \equiv \underset{k = 0}{\prod^{252}} 5 [8]$ $= 5^{252} [8]$ $5^{2} = 1 [8] \Rightarrow 5^{252} \equiv 1 [8]$ $\Rightarrow A \equiv 1 [8]$ $1 + 4 k = 5 s \Rightarrow 5 s - 4 k = 1$ $(s, k) = (1, 1) s o l u t i o n$ $\Rightarrow 5 (s - 1) - 4 (k - 1) = 0 \Rightarrow k = 5 l + 1$ $\Rightarrow 1 + 4 (5 l + 1) = 5 (4 l + 1) = 1 + 4 k$ $l = 0, l = 1, l = 2 \Rightarrow 5, 45, 25 \in {1, 5, 9, 13, 1 \bar{7} . . . . .}$ $\Rightarrow (1.5.9 . . . . . . 2021) \equiv 0 [125]$ $A \equiv 0 [125]$ $A \equiv 1 [8]$ $A = 8 k + 1 = 125 d$ $125 d - 8 k = 1$ $(d, k) = (5, 78) s o l u t i o n$ $k = 125 m + 78$ $A \equiv 8 (125 m + 78) + 1 = 1000 m + 625 [1000] \equiv X [100]$ $X = 625$
	Commented by mathsuji last updated on 02/May/21

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