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 make it a perfect square?
39
Step 1 — LCM of divisors:
\(15=3\cdot5,\;18=2\cdot3^2,\;20=2^2\cdot5,\;27=3^3\), so \(\text{LCM}=2^2\cdot3^3\cdot5=540\).
Step 2 — Form of x:
A number leaving remainder 10 with each divisor has the form \(x = 540k + 10\).
Step 3 — Impose divisibility by 31:
Since \(540 \equiv 13 \pmod{31}\), we need \(13k + 10 \equiv 0 \pmod{31}\), i.e. \(13k \equiv 21 \pmod{31}\). Testing gives \(k=4\) (since \(13\times 4 = 52 \equiv 21\)).
Step 4 — Compute x and nearest square:
\(x = 540(4)+10 = 2170\). Since \(46^2=2116\) and \(47^2=2209\), the least number to add is \(2209 - 2170 = \mathbf{39}\).
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