17, 24, 31, 38, 45, 52

That is an arithmetic sequence with a1 = 17 and d = 7

The nth term of an arithmetic sequence is given by

an = a1 + (n - 1)d

an = 17 + (n - 1)(7)

an = 17 + 7n - 7

an = 10 + 7n

To be a multiple of 2, 3, 4, 5, 6, is to be a multiple of 2*2*3*5 = 60

So we are looking for the smallest integer k so that

10 + 7n = 60k

Write the 10 and the 60 in terms of their nearest multiples of 7.
10 is nearest 7 and 60 is nearest 63

(7 + 3) + 7n = (63 - 3)k

  7 + 3 + 7n = 63k - 3k

Divide through by 7

  1 +  + n = 9k - k

Isolate the fractional terms:

 + k = 9k - n - 1

Since the right side is an integer, the left side is also an integer.
Let that integer be A.

 + k = A    and   9k - n - 1 = A

Clear of fractions:

3 + 3k = 7A

Write the 7 in terms of its nearest multiple of 3, which is 6

3 + 3k = (6 + 1)A

3 + 3k = 6A + A

Divide through by 3

1 + k = 2A + 

Isolate the fraction:

1 + k - 2A = 

Since the left side is an integer, the right side is also an integer.
Let that integer be B.

 = B     and   1 + k - 2A = B

Clear of fractions:

A = 3B

Substituting in 1 + k - 2A = B

1 + k - 2(3B) = B

1 + k = 6B + B

1 + k = 7B

    k = 7B - 1

Substituting for A and k in 9k - n - 1 = A

9(7B - 1) - n - 1 = 3B

  63B - 9 - n - 1 = 3B

     63B - 10 - n = 3B

         60B - 10 = n

The smallest possible value of B that would produce

a positive value for n is 1, thus B = 1

       60(1) - 10 = n

               50 = n

and k = 7B - 1 = 7(1) - 1 = 6

Therefore the desired term is the 50th.  That term can be

gotten from either 

10 + 7n = 10 + 7(50) = 10 + 350 = 360

or from

60k = 60(6) = 360.

So the answer is 360, which is the 50th term.

Edwin