Find the least number which when divided by 6, 8, and 15 leaves a remainder 5, but when divided by 13, leaves no remainder.

Let the number be N

N is 5 more than a multiple of 6, so N=6p+5 for some integer p.
N is 5 more than a multiple of 8, so N=8q+5 for some integer q.
N is 5 more than a multiple of 15, so N=15r+5 for some integer r.
N is a multiple of 13, so N=13t

So

(1)   N = 6p+5 = 8q+5 = 15r+5 = 13t

Since N = 15r+5 = 5(3r+1), is a multiple of 13, 3r+1 is also multiple of 13.

So 3r+1 = 13u for some integer u
     3r = 13u-1
So  15r = 65u-5

So (1) becomes

N = 6p+5 = 8q+5 = 65u-5+5 = 13t
N = 6p+5 = 8q+5 = 65u = 13t

We can dispense with the 13t, since N is a multiple of 65, it's 
automatically a multiple of 13. So (1) becomes

(2)   N = 6p+5 = 8q+5 = 65u

Since 6p+5 = 8q+5
        6p = 8q
        3p = 4q

Write 4q as 3q+q

        3p = 3q+q

Divide through by 3

         p = q+

So     p-q = 

Since p-q is an integer, so is , say the integer v

Then p-q = v and  = v

So  = v
   q = 3v

p-q = v

p-3v = v
   p = 4v 

Substituting in (2)

   N = 6(4v)+5 = 8(3v)+5 = 65u
   N = 24v+5 = 65u

Since  24v+5 = 65u
         24v = 65u-5
         24v = 5(13u-1)

v is a multiple of 5, so v = 5w for some integer w

      24(5w) = 5(13u-1)
         14w = 13u-1
           1 = 13u-14w   

Write 14 as 13+1

           1 = 13u-(13+1)w
           1 = 13u-13w-w

Divide through by 13

           = u-w-
      = u-w

The right side is an integer so the left side must be also.
Suppose that integer is x.  Then

 = x and u-w = x

1 + w = 13x
    w = 13x-1

Substituting in 

  u-w = x
  u-(13x-1) = x
  u-13x+1 = x
       u = 14x-1 

Since N = 65u
      N = 65(14x-1)
      N = 910x-65

The smallest value x can take on is 1, so

      N = 910(1)-65
      N = 910-65
      N = 845 

Edwin