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Find-the-value-of-X-such-that-1-5-9-13-17-2021-X-mod1000-




Question Number 139737 by mathsuji last updated on 30/Apr/21
Find the value of X such that:  1∙5∙9∙13∙17∙...∙2021≡X(mod1000)
FindthevalueofXsuchthat:15913172021X(mod1000)
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=505k=0(1+4k)X[1000]1000=8.53k=2n1+4k1[8]k=2n+11+4k5[8]505k=0(1+4k)252k=0(8k+1).252k=0(5+8k)252k=05[8]=5252[8]52=1[8]52521[8]A1[8]1+4k=5s5s4k=1(s,k)=(1,1)solution5(s1)4(k1)=0k=5l+11+4(5l+1)=5(4l+1)=1+4kl=0,l=1,l=25,45,25{1,5,9,13,17¯..}(1.5.92021)0[125]A0[125]A1[8]A=8k+1=125d125d8k=1(d,k)=(5,78)solutionk=125m+78A8(125m+78)+1=1000m+625[1000]X[100]X=625
Commented by mathsuji last updated on 02/May/21
thank you very much sir
thankyouverymuchsir

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