The least number which when divided by 3, 4 and 7 leaves a remainder 2, but when divided by 5 leaves no remainder, is:

145

170

250

335

Solution: L.C.M. of 3, 4 ,7 = 84. .'. Required number is of the form 84x+2. Least value of x for which (84x+2) is divisible by 5 is x = 2. .'. Required number = (84*2+2) = 170.

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