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Rationals Between

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

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Root to Poly

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

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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?

Power Countdown

Stage: 4 Challenge Level: Challenge Level:1

For making 8, everyone noticed that the only ways of doing this are $2^3$, and $64^\frac{1}{2}$, which is $4^\frac{3}{2}=2^3$, and so these are the same.

Zack, Haren and Niall from JAPS made $125$ by doing $5^3$ where $3=81^\frac{1}{4}$

Charlotte and Jai from Tytherington High School sent us the following solutions:

$16^5=1048576$
$1048576^\frac{1}{2}=1024$

$16^\frac{1}{2}=4$
$243^\frac{1}{5}=3$
$4^3=64$

or alternatively:
$243^\frac{1}{5}=3$
$16^3=4096$
$4096^\frac{1}{2}=64$

Can you see why these two methods are equivalent?

Daniel, Alex, Nick and George from Tytherington gave these solutions:

$243^\frac{1}{5}=3$
$512^\frac{1}{3}=8$
$8^2=64$

$243^\frac{1}{5}=3$
$343^\frac{1}{3}=7$
$7^2=49$

Tom from Devenport Boys gave an explanation of why the last two values are unobtainable;

$89$ is a prime number, and so its only factors are $1$ and $89$, which also means that all its powers, such as $89^2$, will only have factors $1, 89, 7921$ and so on, so it is not a power of any other numbers, and so cannot be achieved.

$216=6^3$, and there is no way of achieving 6 with the values given, with a similar factor method.