tougher problems on LCM

A gardener had a number of shrubs to plant in rows. At first he tried to plant 5 in each row, then 6, then 8, then 9 and then 12, but had always 1 left. On trying 13 he had none left. What is the smallest number of shrubs that he could have had?    

Answer:

           

The LCM of 5,6,8,9 and 12 is 360.

So, the required number is 360y + 1

Given that 360y + 1 does not leave a remainder with 13.

360 = 13 x 27 + 9

So, 360y + 1 = (13 x 27 + 9)y + 1

= 13x27y + 9y+1

Since there is no remainder, 9y + 1 should be equal to 13 or its multiple.

For, y = 1, 9y +1 = 10 ..... (x)

Y = 2, 9y + 1 = 19 ..............(x)

Y = 3, 9y +1 = 28 ...............(x)

Y = 4, 9y +1 = 37 ...............(x)

Y = 5, 9y +1 = 46 ...............(x)

Y = 6, 9y +1 = 55 ...............(x)

Y = 7, 9y +1 = 64 ...............(x)

Y = 8, 9y +1 = 73 ...............(x)

Y = 9, 9y +1 = 82 ...............(x)

Y = 10, 9y +1 = 91 ...............(√) divisible by 13

So, the number is 360y + 1 = 360x10 + 1 = 3601