Negative powers

Well done to everyone who submitted their work on this problem. We received many good answers with explanations of how you approached your work with these expressions, including several solutions from students at Hymers College in England and Doha College in Qatar, Daniyal from Greenfield International School in India, Yashneal from Higham Lane School in UK, Vihaan from King George V School in Hong Kong, Ci Hui from Queensland Academies for Science Mathematics and Technology in Australia, Jenny from Tapton Secondary School in UK and Samantha from The Steele School, and Helen  from Lady Margaret School.

We thought this problem was a good way to get you thinking about combining negative powers, which you might not be so familiar with. As Jenny from Tapton Secondary School noted in her solutions, this was her first time investigating negative indices.

Several solutions observed that for finding the smallest value if could be useful to think about the size of the denominator when rewriting these expressions as a fraction. As Lamees from Doha College explained:

We know the larger the denominator, the smaller the fraction (if the numerator is the same). e.g. $\dfrac{1}{2}>\dfrac{1}{3}$.

This is good thinking. If we knew that all the possible values had the same sign (all positive, or all negative) it would allow us to find the value closest to zero. However, as several of you noticed, it is possible to get positive and negative values for these expressions. Therefore, as Ahmed from Doha College and Yashneal from Higham Lane School explained clearly, we need to think about the sign as well as the size of the expression. Here is Yashneal's explanation:

$$((−a^{−b} )^{−c})^{−d} = (−1)^{cd}a^{−bcd}$$

The sign (positive or negative) is given by $(−1)^{cd}$. 

• If $cd$ is even, $(−1)^{cd} = +1$ so the result is positive.
• If $cd$ is odd, $(−1)^{cd} = −1$ so the result is negative. 

The size (magnitude) is given by $a^{−bcd} = \dfrac{1}{a^{bcd}}$

• The bigger $a^{bcd}$ makes the final number closer to zero.
• The smaller $a^{bcd}$ makes the final number much larger in magnitude. 

Building on this, Yashneal outlined a strategy for working out the different values:

Now we need to consider $a = 1, 2, 3, 4$. For each $a$, the product $bcd$ is the product of the three remaining numbers

If only positive values were possible, then putting $-3$ first would give $\dfrac{1}{6561}$, which is the value closest to zero. However, Ahmed and Yashneal, as well as Ci Hui from Queensland Academies for Science Mathematics and Technology, Vihaan from King George V School, all found they could obtain a smaller value by making the final two powers odd. 

For example, Vihaan noticed that swapping the order of $3$ and $2$ in the original expression (being mindful to use interpretation B) gives $\left(\left(-4^{-2}\right)^{-3}\right)^{-1}=-\dfrac{1}{4096}$. This is the smallest possible value.

Ayat and Gazal from Doha College and Ci Hui all identified $1$ as the highest possible value of the expression, but they obtained $1$ using different expressions. 

Gazal made the value of $1$ using the expression $((-1^{-2})^{-3})^{-4}$

while Ayat found that $((-1^{-4})^{-3})^{-2} =1$.

Ci Hui used a general algebraic form, giving 

$((-1^{-\triangle})^{-\square})^{-\triangledown}=(-1)^{\square \times \triangledown}$ 

Ci Hui then noted that 

If $\square \times \triangledown$ is even, raising $-1$ to this power will give a value of $1$, and if $\square \times \triangledown$ is odd, raising $-1$ to this power will give a value of $-1$. [This is a sound general statement, but in this problem the only numbers available are $2, 3$ and $4$, so can you get an odd product $\square \times \triangledown$?] 

As these examples show, and as Jenny from Tapton Secondary School commented, the order of some powers can be changed without changing the value of the expression. This gives a clue to the last part of problem as several different rearrangements of $1,2,3$ and $4$ give expressions that have the same value. Some of you seemed worried that you might have to work out 24 different values (using 'brute force'), but actually there are far fewer values that you have to find. Yashneal and Ahmed both used their algebraic approaches from above to find the full sent of possible values. Here is Yashneal's summary of the results:

The values that can occur are:

$$1, \dfrac{1}{4096}, \dfrac{-1}{4096}, \dfrac{1}{6561}$$

So there are $4$ distinct numbers possible.

Therefore the largest value is $1$ and the smallest value is $\dfrac{-1}{4096}$.

I decided to check all possible values and using that I discovered that:

$1$ occurs $6$ times

$\dfrac{1}{4096}$ occurs $8$ times

$\dfrac{-1}{4096}$ occurs $4$ times

$\dfrac{1}{6561}$ occurs $6$ times

You can click here to read Yashneal's full solution.

 
Can you explain how and why these different numbers occur multiple times? How many calculations do you actually need to do? Hint: What are $2^{-4}$ and $4^{-2}$?

As these solutions show, there is a lot to think about when working with complicated nested expressions such as these. Thank you to everyone who sent in their results and ideas.