We are looking for 1 more than some multiple of 2,3,4,5,6,and 8.

We prime factor them all:

 2 is prime
 3 is prime
 4 is 2x2
 5 is prime
 6 is 2x3
 8 is 2x2x2

So their LCM must have 3 factors of 2, sice 8 does.
It must have 1 factor of 3
It must have 1 factor of 5

So the LCM of 2,3,4,5,6,8, 2x2x2x3x5 = 120

120 or any multiple of 120 will leave a remainder of 0 
when divided by  2,3,4,5,6,8.

Therefore, 

1 more than any multiple of 120 will leave a remainder of 1 
when divided by 2,3,4,5,6, or 8.

We are seeking a number n which is 1 more than a multiple of 120 and 
which is also a multiple of 17.  Let the multiple of 120 than n is 1
more than be 120p and the multiple of 17 that n is be 17q.  So for 
our desired number n,

n = 120p+1 = 17q

Write 120 in terms of its nearest multiple of 17 which is 119, 
because 119 = 7*17.

(119+1)p+1 = 17q

119p+p+1 = 17q

Divide every term by 17



Isolate the fraction terms:



The right side is an integer, so the left side is too.

Let that integer by A, then





So we multiply through by 17 to clear of fractions:

p+1 = 17A

So p = 17A-1

We want p to be as small as possible, so we take A = 1

and p = 17(1)-1 = 16

And since q-7p = A

q-7(16) = 1
q-112 = 1
q = 113

So:

n = 120p+1 = 17q 
n = 120(16)+1 = 17(113)
n = 1920+1 = 1921
n = 1921 = 1921.

Answer: 1921

Edwin