The least multiple of 10, which leaves a remainder of 2, when divided by 3, 9, 16 and 18 is:

260

290

320

380

Solution: L.C.M. of 3, 9, 16 and 18 is 144. Let required number be 144x+2, which is a multiple of 10. Least value of x for which (144x+2) is divisible by 10 is x=2. .'. Required number = 144*2+2 = 290.

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