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| Question 962491:  What is the smallest positive integer that gives a remainder of 1 when divided
 by 4, a remainder of 2 when divided by 5, and a remainder of 3 when divided by
 6?
 Answer by Edwin McCravy(20064)
      (Show Source): 
You can put this solution on YOUR website! What is the smallest positive integer that gives a remainder of 1 when divided by 4, a remainder of 2 when divided by 5, and a remainder of 3 when divided by
 6?
 
 
Suppose the solution is N.
In order to leave a remainder of 1 when divided by 4, N must be 1 more
than a multiple of 4. So there exists an integer a such that
N = 4a+1
In order to leave a remainder of 2 when divided by 5, N must be 2 more
than a multiple of 5. So there exists an integer b such that
N = 5b+2
In order to leave a remainder of 3 when divided by 6, N must be 3 more
than a multiple of 6. So there exists an integer c such that
N = 6c+3
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So we have:
 N  =  4a+1  =  5b+2  =  6c+3
Add 3 to each of those
N+3 =  4a+4  =  5b+5  =  6c+6
Factor the gcf out of the last three:
N+3 = 4(a+1) = 5(b+1) = 6(c+1) 
This shows that N+3 must be a multiple of 4,5, and 6.
The smallest multiple of those is their LCM = 60
Therefore N+3 = 60
            N = 57
Checking:
    14       11        9
  4)57     5)57     6)57
    4        5        54 
    17        7        3 
    16        5
     1        2
Answer: 57
Edwin
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