What fractions can you find between the square roots of 65 and 67?
Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.
The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?
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
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).
So the answer is 64.
(PS: I presume that when Mr Jones asked Mr Smith whether Mr Smith's
house number had 1 as the second digit, Mr Smith would have answered
Here is Agathe's solution:
To start this problem off, I wrote all the perfect squares and cubes
between 13 and 1300. I decided that Mr Smith had replied 'yes' to all
the questions to start my working. This meant Mr Smith' s house number
was between 500 and 1300, and that it had to be either 512, 529, 576,
625, 676, 729, 784, 841, 900, 961, 1000, 1024, 1089, 1156, 1225, or
1296. But Mr Jones had 2 numbers left so Mr Smith' s answers could not
all be 'yes'.
I then tried another answer to the perfect square question by making Mr
Smith say 'no'. I was left with 2 numbers, 512 and 1000, and 512's
second digit is one so this worked.
But because Mr Smith had lied to the first and second question, the real
number was between 13 and 500, a perfect square and a perfect cube. The
only number that perfect squares and cubes have in common between 13 and
500 is 64. Therefore Mr Smith' s house number must be 64.