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)
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)
+118687
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
