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Question 72
Let x be the least number which when divided by 15, 18, 20 and 27, the remainder in each case is 10 and x is a multiple of 31. What least number should be added to x to makeit a perfect square?
Solution
LCM of 15, 18, 20 and 27 = 540
Number = $$\frac{540k +Â 10}{31}$$
= $$\frac{527k + 13k +Â 10}{31}$$
Value of k = 4
So, number =Â 540 $$\times $$ 4Â + 10 = 2160 +Â 10 = 2170
Square of 47 = 2209
Number added to make it perfect square = 2209 - 2170 =Â 39
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