Question 1146503
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There must be positive integers A,B,C,D,E,F, such that

2A+1 = 3B+2 = 4C+3 = 5D+4 = 6E+5 = 7F = the answer

We will solve linear Diophantine equations by the standard rules, which are:

1. Find the least coefficient of a variable in absolute value, say it's P.
2. Write all other numerical integers in terms of their nearest multiple of P.
3. Divide all terms by P.
4. Move all fractions to one side, and all non-fractions to the other side.
5. Set both sides equal to a new integer, say, Q, making 2 equations.
6. This will give two new Diophantine equations.  
6. For the side that has the fractions, clear it of its fractions and solve.
7. Substitute to find F in terms of the new integer.

    6E + 5 = 7F
6E + 6 - 1 = 6F + F
E + 1 -1/6 = F + F/6
 E + 1 - F = F/6 + 1/6 = G 
 E + 1 - F = G;  F + 1 = 6G
                     F = 6G - 1
E + 1 - (6G - 1) = G
  E + 1 - 6G + 1 = G
     E + 2  - 6G = G
               E = 7G - 2
          6E + 5 = 7F
   6(7G - 2) + 5 = 7F
    42G - 12 + 5 = 7F
         42G - 7 = 7F
          6G - 1 = F

     5D + 4 = 7F
     5D + 4 = 7(6G - 1)
     5D + 4 = 42G - 7
    5D + 11 = 42G 
5D + 10 + 1 = 40G + 2G
D + 2 + 1/5 = 8G + 2G/5
 D + 2 - 8G = 2G/5 - 1/5 = H
 D + 2 - 8G = H;   2G/5 - 1/5 = H
                       2G - 1 = 5H
                       2G - 1 = 4H + H
                      G - 1/2 = 2H + H/2
                       G - 2H = H/2 + 1/2 = J
                  G - 2H = J;  H + 1 = 2J
                                   H = 2J - 1
           G - 2(2J - 1) = J  
              G - 4J + 2 = J
                   G + 2 = 5J
                       G = 5J - 2

F = 6G - 1
F = 6(5J - 2) - 1
F = 30J - 12 - 1
F = 30J - 13

             4C + 3 = 7F
             4C + 3 = 7(30J - 13)
             4C + 3 = 210J - 91
            4C + 94 = 210J
        4C + 92 + 2 = 208J + 2J
       C + 23 + 2/4 = 52J + 2J/4
       C + 23 - 52J = 2J/4 - 2/4 = J/2 - 1/2 = K
       C + 23 - 52J = K;               J - 1 = 2K
                                           J = 2K + 1
C + 23 - 52(2K + 1) = K
 C + 23 - 105K - 52 = K
      C - 29 - 104K = K
             C - 29 = 105K
                  C = 105K + 29

          4C + 3 = 7F
4(105K + 29) + 3 = 7F
  420K + 116 + 3 = 7F
      420K + 119 = 7F

We can stop here, since 7F is the answer, and it's odd, so if we divide by 2,
it will leave remainder 1. The minimum positive value for 7F is when K = 0.

Answer = 420(0) + 119 = 119

Checking: 
 
7)119 gives 17 with remainder 0

6)119 gives 19 with remainder 5

5)119 gives 23 with remainder 4

4)119 gives 29 with remainder 3

3)119 gives 39 with remainder 2

2)119 gives 59 with remainder 1

Edwin</pre>