Question Number 103209 by nimnim last updated on 13/Jul/20

Answered by bramlex last updated on 13/Jul/20

n = 7k+1 ...(1)×6  n= 6p+3 ...(2)×7  n = 5q +2...(3)  ⇒ (2)−(1)  ⇒n = 42(p−k)+15 ...(4)  set p−k = l  consider   5q+2 = 42l+15   ⇒2 (mod 5) = 42l +15  ⇒2l = 2 (mod 5), l = 1 (mod 5)  ⇔ l = 5t+1  n = 42l +15 = 42(5t+1)+15  n = 210t + 57

Commented bynimnim last updated on 13/Jul/20

thanks

Answered by floor(10²Eta[1]) last updated on 13/Jul/20

(1)x≡1(mod 7)⇒x=7a+1  (2)x≡3(mod 6)  (3)x≡2(mod 5)  (2):7a+1≡3(mod 6)⇒a≡2(mod 6)  ⇒a=6b+2  ⇒x=7(6b+2)+1=42b+15  (3):42b+15≡2(mod 5)⇒2b≡2(mod 5)  ⇒b≡1(mod 5)⇒b=5c+1  ⇒x=42(5c+1)+15  ⇒x=210c+57, c∈Z

Commented bynimnim last updated on 13/Jul/20

Thanks

Answered by 1549442205 last updated on 13/Jul/20

From the condition that number when  divided by 6 gives remainder 3 and  when divided by 5 gives remainder 2  it follows that if adding that number   to 3 we get a multiple of 30.Hence  Denoting by n the  number we need find  then n=30k−3.On the other hands,n=7m+1.We  infer 30k+3=7m+1⇒m=((30k−4)/7)=4k+((2k−4)/7)  Put ((2k−4)/7)=p⇒k=((7p+4)/2)=3p+2+(p/2)  Put (p/2)=q⇒p=2q⇒k=7q+2  m=30q+8⇒n=210q+57.Thus,the   number we need find is  n=210q+57(q∈N)

Commented bynimnim last updated on 13/Jul/20

Thanks