Congratulations to the Key Stage 3 Maths
Club at Strabane Grammar School. We give below their methods of
finding A where [3(230 + A)] ^{2}
= 492 80A.

The number on the right must be divisible by 9 because of the 3 which is squared on the left. This means its digits must add to give a multiple of 9 - (9, 18, 27, 36,...... etc).

4+9+2+8 = 23

23 + A must give 27 or 36 or 45 etc.

27 is the only total we can reach since A is a digit so A =
4.

We also wrote a program in Basic to check this.

> 10 FOR A = 0 TO 9

> 20 IF ((230 + A)*3)^2 = 492800 + A PRINT A

> 30 NEXT A

Escape

> RUN

4

Elizabeth of Ipswich High School checked the square roots of 49280A to find which one was an integer. James of Hethersett High School, and S.Kuo of Eaton School Norwich, and groups from Ousedale School, Newport Pagnell and Madras College, St Andrew's and Bithian and Guobin from The Chinese High School, Singapore, used mixtures of logical deduction, algebra and trial and error, homing in between answers that turned out too low and too high.