Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.

364

LCM of 12, 18, 24, 30 = 360

Required number = 360K + 4 [which is exactly divisible by 7 for certain value of K]

Lets find out that number:

⇒ \(\frac{360K\;+\;4}{7}\) = \(\frac{357K\;+\;(3K\;+\;4}{7}\) = \(\frac{3K\;+\;4}{7}\)

Put K = 1

= \(\frac{3\;+\;4}{7}\) = \(\frac{7}{7}\) = 0 remainder

Therefore,

⇒ Number = (360 × 1) + 4 = 364 [which is exactly divisible by 7]