Question

Let x be the least mumber 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 make it a perfect square?​

Answer 1

$\rm\underline\bold{Question \purple{\huge{\checkmark}}}$

Let x be the least mumber 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 make it a perfect square?

$\huge{\tt{\colorbox{yellow}{AnswEr:}}}$

First take the LCM of 12, 15, 18 and 27

$\huge{\tt{\colorbox{blue}{Given:}}}$

You will get LCM 540

$\huge{\tt{\colorbox{orange}{Proof statement:}}}$

Now by observation you can clearly see that in each number there is a difference of 4

12 - 8 = 4

15 - 11 = 4

18 - 14 = 4

27 - 23 = 4

So just subtract the LCM by 4

You will get the required answer as 540 - 4 = 536