find x the least integers for simultaneous conguence x congruence 2(mod3); x?

Question

hi.. in the topic of our class on chinese remainder theorem, please help me:
find x the least integers for simultaneous conguence x congruence 2(mod3); x congruence 3(mod4); x congruence 4(mod5); x congruence 5 (mod 6)

thank you

Answer 1

So you're looking for the smallest (presumably positive) integer which is 3n+2, 4m+3, 5p+4, 6q+5?
I notice that all of these remainders are equivalent to -1:
So x=3n-1=4m-1=5p-1=6q-1.
Certainly -1 itself would work, but we're looking for positive integers.
Then you need to add a number to -1 which is simultaneously divisible by 3,4,5,6. Really you just need 3,4,5 because anything which is a multiple of 3 and 4 is also a multiple of 6.
The smallest such number is 3*4*5=60.
-1+60=59. So your answer is 59.
59=3*19+2
59=4*14+3
59=5*11+4
59=6*9+5