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Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.


Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)

Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

Mega Quadratic Equations

Age 16 to 18
Challenge Level

We received lots of solutions to this problem! Thank you to Amrit, Agathiyan, Guruvignesh and Adithya from Hymers College, Daniel from Eastbury Community School, Minhaj from St Ivo School, Anas from St John's Primary School, Thomas from St Catherine's College, Timothy from Gosforth Academy, Pablo, Yang, Jie and Marc from King's College Alicante in Spain, Siddhant from Indus Internation School of Bangalore in India and Kevin from Harrow International School in Hong Kong for sending us your solutions.

Here is Minhaj's solution to the first equation:

To solve $(n^2 - 5n + 5)^{(n^2 - 11n + 30)} = 1$, I considered the following three situations:
a^0&=1 \\
1^m&=1 \\
(-1)^z&=1 \quad \text{(where $z$ is even)}
\end{align}$$Using the ideas above we can solve $n^2 - 5n + 5 = 1$ or $n^2 - 11n + 30 = 0$ or $n^2 - 5n + 5 = -1$. Solving these three quadratics we find:
n^2 - 5n + 5 = 1 &\iff n^2 - 5 + 4 = 0 \iff (n-1)(n-4) = 0 \iff n=1, 4 \\
n^2 - 11n + 30 = 0 &\iff (n - 5)(n-6) = 0 \iff n=5, 6 \\
n^2 - 5n + 5 = -1 &\iff n^2 - 5n + 6 = 0 \iff (n-2)(n-3) = 0 \iff n = 2, 3
\end{align}$$For the last case, to ensure that $m$ is even, we can substitute $n=2$ and $n=3$ into $n^2 - 11n + 30$, which gives us $m=12$ and $m=6$ respectively.

So the six solutions to this equation are $n=1, 2, 3, 4, 5, 6$.

Agathiyan showed that the power in the first equation is always even using the following method:

We can prove that $n^2-11n+30$ will always be even when $n$ is an integer. If $t$ is any positive integer, then $2t$ is any even number. If we substitute that even number into $n^2-11n+30$ we get:
\end{align}$$So when $n$ is even, our quadratic will also be even.

If $n$ is an odd number, say $2t+1$, we get:
(2t+1)^2-11(2t+1)+30&=(4t^2+4t+1) -(22t+11)+30\\
\end{align}$$So when $n$ is odd, our quadratic will still be even. Therefore $n^2-11n+30$ is always even when $n$ is an integer.

Here is Timothy's solution to the second equation:

Applying these cases of $x^y=1$ to the second equation, $(n^2 - 7n + 11)^{(n^2 - 13n + 42)} = 1$, gives us three equations and six solutions:

For $1^y=1$:
n^2 - 7n + 11 &= 1 \\
n^2 - 7n + 10 &= 0 \\
(n-2)(n-5)&=0 \\
n&=2, 5
For $(-1)^y=1$ when $y$ is even:
n^2 - 7n + 11 &= -1 \\
n^2 - 7n + 12 &= 0 \\
(n-3)(n-4)&=0 \\
n&=3, 4
\end{align}$$We should also check that these give us an even $y$:
$$3^2-13\times 3 + 42=12 \\
4^2-13\times 4 + 42=6 \\$$Therefore these are valid solutions.

For $x^0=1$:
n^2 - 13n + 42 &= 0 \\
(n-6)(n-7)&=0 \\
\end{align}$$So our solutions are $n=2,3,4,5,6,7$.

Pablo used the following method to generate more Mega Quadratic Equations:

The pattern of solutions tells us that the next Mega Quadratic Equation should have solutions $n=3,4,5,6,7,8$.

The equation for the power equal to $0$ has given us the two biggest solutions, so working backwards:
n^2 - 15n + 56 &= 0
\end{align}$$So the power is $n^2 - 15n + 56$.

Next, for the base we need two pairs of equations with solutions $n=3,6$ and $n=4,5$.
n^2 - 9n + 18 &= 0 \\
n^2 - 9n + 19 &= 1\\
n^2 - 9n + 20 &= 0 \\
n^2 - 9n + 19 &= -1
\end{align}$$So our base is $n^2 - 9n + 19$.

Therefore the next Mega Quadratic Equation is $(n^2 - 9n + 19)^{(n^2 - 15n + 56)}=1$.

Kevin found a general method for generating Mega Quadratic Equations:

In order to find a pattern for these Mega Quadratic Equations I looked at how I found the values of $n$ for both equations. I noticed that when I was making equations for the base to be equal to $1$, the values I found for $n$ in the first question were $1$ and $4$. The values in the second question were $2$ and $5$. The values of $n$ were increased by one!

I did the same thing with the power equation and found in the first question and second question, I got $n=5,6$ and $n=6,7$ respectively. Again they increased by one!

I tried finding relationships with the values of $n$ with the mega quadratic equation and I found out that the values of $n$ that I found by making the base equal $1$ help make the coefficients in the equation. For example, consider $n^2−5n+5$. The values of $n$ I found for this question ($1$ and $4$) make the quadratic like this:
n^2 - 5n + 4 &= 0 \\
n^2 - 5n + 5 &= 1
\end{align}$$So the $-5n$ comes from adding our solutions together and the $5$ comes from multiplying our solutions together and adding $1$ (because that's what we wanted the base to equal).

For the power equation, you get:
n^2 - 11n + 30 &= 0
\end{align}$$So the $-11n$ comes from adding our solutions together and the $30$ comes from multiplying our solutions together.

From what I found, I made an equation to make a Mega Quadratic Equation. If we let the smallest solution for $n$ be $x$, then our six solutions will be $n=x$, $x+1$, $x+2$, $x+3$, $x+4$ and $x+5$. This gives us the Ultimate Quadratic Equation:
$$(n^2 - ((x)+(x+3))n+((x)(x+3)+1)) ^{(n^2 - ((x+4)+(x+5))n+(x+4)(x+5))}$$Which simplifies to:
$$(n^2 - (2x+3)n+(x^2+3x+1) )^{(n^2 - (2x+9)n+(x^2+9x+20))}$$For example, the first question's values of $n$ are $1$, $2$, $3$, $4$, $5$ and $6$. $1$ is the smallest solution, so that is our value of $x$.

Using this equation, you can find an infinite amount of Mega Quadratic Equations like:
$$(n^2−9n+19)^{(n^2−15n+56)}=1 \qquad\qquad \text{when }x=3 \\
(n^2-11n+29)^{(n^2-17n+72)}=1 \qquad\qquad \text{when }x=4$$You can find many Mega Quadratic Equations and their answers by using the equation!