When 5, 6, 8, 9 and 12 divide the least number 'x', the remainder of each case is 1, but x is divisible by 13. What will be the remainder when x is divided by 31?

When 5, 6, 8, 9 and 12 divide the least number 'x', the remainder of each case is 1.

LCM of 5, 6, 8, 9 and 12 ::

LCM of 5, 6, 8, 9 and 12 = $$2\times2\times2\times3\times3\times5$$

= 360

So x = (360a+1)    Eq.(i)

(360a+1) is completely divisible by 13.

$$\frac{(360a+1)}{13}$$

$$\frac{(351a+9a+1)}{13}$$

$$27a+\frac{(9a+1)}{13}$$

So from the above, we can say that (9a+1) should be completely divisible by 13. For that we can try to put a = 1, 2, 3...

a = 1 then (9a+1) = 10

a = 2 then (9a+1) = 19

a = 3 then (9a+1) = 28

a = 4 then (9a+1) = 37

a = 5 then (9a+1) = 46

a = 6 then (9a+1) = 55

a = 7 then (9a+1) = 64

a = 8 then (9a+1) = 73

a = 9 then (9a+1) = 82

a = 10 then (9a+1) = 91

The least value of a = 10, when (9a+1) will be completely divisible by 13.

From Eq.(i) x = (360a+1)

= $$(360\times10+1)$$

= (3600+1)

3601

When x=3601 is divided by 31, the remainder is 5.