Question 50

Find the least multiple of 7 that leaves the remainder 4 when divided by 6, 9, 15 and 18.

Solution

$$6=3\times2,\ 9=3\times3,\ 15=3\times5,\ 18=3\times3\times2.$$

So, $$LCM\ of\ \left(6,9,15,18\right)=9\times5\times2=90.$$

So, the required number should be in the form of $$\left(90\times7\times k+4\right)\ \ or\ \ \left(360k+4\right).$$

So, the least number is 364 when k=1.

D is correct choice.

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