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How many positive integers less than or equal to 6*7*8*9 solve the system of congruences:

 

m = 5 (mod 6)

m = 4 (mod 7)

m = 3 (mod 8)

m = 2 (mod 9)

m = 1 (mod 10)

 Feb 18, 2022
 #1
avatar
+1

m = 5 (mod 6)

m = 4 (mod 7)

m = 3 (mod 8)

m = 2 (mod 9)

m = 1 (mod 10)

 

LCM[6, 7, 8, 9, 10] ==2,520

 

The smallest integers that will satisfy all five congruences < 3,024  are:

 

11  and  2,521 (2,520 + 1)

 Feb 18, 2022
 #2
avatar+117224 
+1

 

Guest, you made a small mistake.  I like seeing your answers though. :)

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I'm going to try and solve in pairs

m=-1 mod 6 and 1 mod 10

11  works

 

m=4 mod 7 and 1 mod 10

11 works

 

now I look, 11 works for all of them

 

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Mod number Prime factors
6 2 3      
7       7  
8 2*2*2        
9   3*3      
10 2   5    
prime factors needed 2*2*2=8 3*3=9 5 7 5*7*8*9

 

So the solutions will be of the form    m=(5*7*8*9)k + 11

the first 2 are 11 and  (5*7*8*9) + 11

the 3rd one is         2(5*7*8*9) + 11   and this is bigger than    6*7*8*9

so

There are just 2          they are     11 and  2531

 Feb 19, 2022

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