When the smallest number x is divided by 5, 6, 8, 9 and 12, it gives remainder 1 in each case. But x is divisible by 13. What will be the remainder when x will be divided by 31 ?

5

LCM of 5, 6, 8, 9, 12 = 360

Required number = 360K + 1 [which is exactly divisible by 13 for certain value of K]

Lets find out that number:

⇒ \(\frac{360K\;+\;1}{13}\) = \(\frac{351K\;+\;(9K\;+\;1}{13}\) = \(\frac{9K\;+\;1}{13}\)

Put K = 10

= \(\frac{90\;+\;1}{13}\) = \(\frac{33}{11}\) = 0 remainder

Therefore,

⇒ x = (360 × 10) + 1 = 3601 [which is exactly divisible by 13]

Now,

When 3601 is divided by 31, the remainder = 5