Rationals Between

What fractions can you find between the square roots of 56 and 58?

Root to Poly

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

Consecutive Squares

The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?

Smith and Jones

Stage: 4 Challenge Level:

What an excellent problem - and quite demanding, so it was good to see so many of you have a go. Some of you simply sent in answers but no explanation and we do like to know how you got there. NRICH is about mathematical thinking and so we publish solutions that give an insight into your thought processes. The aim is that other students can look at the solutions and learn from the way the problem has been approached. There was an excellent solution from Trevor of Riccarton High School. Well done Trevor. Other solutions included those from Agathe of Stamford High School and Ian of the William Lovell School. Agathe's solution is also given after Trevor's.

The first clue we have is that the number is between 13 -1300. The second clue we have is that it may be above 500. The third clue is that the answer may be a square number and the fourth that the number may be a cube.

So, to follow the third and fourth clues, Trevor listed all the square numbers and all the cube numbers between 13 -1300. You could generate these very quickly on a spreadsheet.

The fifth clue was that whatever Mr Smith said, Jones believed. So Jones followed what he thought was true and came up with two answers. One of which had "1" as its second digit.

If Mr Jones was told that the number was more than 500, and that the number wasn't a square number or a perfect cube, then we would end up with over 500 answers, which doesn't work. The same goes if Mr Jones was told that the number was less than 500, and that the number wasn't a square number or a perfect cube. It doesn't work either.

So let's presume that Mr Jones was told that the number was more than 500 and that the number was a square but not a cube. It's impossible because no square number bigger than 500 has 1 as a second digit.

If Mr Jones was told that the number was smaller than 500, and that the number was a square number but not a perfect cube, then we have met the criteria that one of the solutions has 1 as the second digit, but there are a lot more than just 2 solutions, which is what we were looking for.

Let's presume that Mr Smith said that the number was more than 500, and that the number wasn't a square but a perfect cube. 27,64,125, 216,343 are out because they are smaller than 500. Looking at the list of square numbers, 729 is out because it appears there. Leaving 512, 1000 left. It meets the criteria that there are two solutions left. And 512 has a 1 as a second digit.

This means that Mr Smith told Jones that the number was more than 500, not a square number but a perfect cube.

We know that Mr Smith lied to Jones about whether the number was 500 or not, and whether the number was a square or not. Hence, we can come up with the conclusion that:

Mr Smith's house is less than 500

His house is a square number and a perfect cube

Looking through the solutions we can get, there is only one solution, and that Mr Smith's house number is 64 (less than 500, square number, and a perfect cube).