## PRIZE PUZZLE FOR JUNE 2007## PERFECT SQUARESThree of the following four numbered statements are TRUE. One of them is FALSE. What is the value of x? 1. x is a perfect square.
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## Answer: 863.1. x is not a perfect square so this statement is FALSE.
To show that there is no other answer we must try to (a) show that statement 1 is false with another answer that keeps statements 2, 3 & 4 true or (b) find solutions when statements 2, 3 or 4 are false with the others true. We start off by making the assumption that statement 1 is true; i.e. that x is a perfect square, and try to prove that it is or is not. To find solutions to equations like b*b + n = t*t we note
that For statements 1 & 2, the mean value of (t+b) and (t-b) is t
which is 19. If x = 324, statement 3 is false because 324+98 = 422 which is not a perfect square. So x = 324 is not a solution to the main problem. If we investigate the possibility that statements 1 and 3 are true we come up against the fact that the number 98 has factors that sum to an odd number and therefore the value of t cannot be an integer. So there is no value of x that satisfies statements 2 and 3. That leaves the only possible answer that statements 2, 3 and 4
are true and 1 is false. In this case we must use an offset of
37 to make our algebraic symbols match the problem so that n =
98-37 = 61. The only factors of 61 are 61 and 1 that leads to t
= 31 (the mean value of 61 and 1) and b = 30 with our offset "x"
= 900. I received five correct answers. They were from John Bownas, John Stafford, Michael Hodge, Richard Burkill and Clem Robertson (who won the draw). Well done all. |