# When I’m 65

“Clearly a product of five and thirteen
But if you must contrive
Arithmeticians, multiply Gaussians
And still get sixty-five”

(with apologies to Lennon and McCartney)

Part of the peterjamesthomas.com Maths and Science archive

Origins

Arithmetical puzzles pervade the Internet [1], one I came across recently was:

 The product of two numbers is 65, what is their sum?

A certain lack of clarity in the question left some loopholes through which some much more interesting Mathematics could be explored. I propose to cover this in the following article. I’ll start with the answers that the problem poser probably had in mind when setting their puzzle.

Obvious

The obvious place to start is to break $65$ down into factors (things that when multiplied together get us back to $65$). Given my Mathematical background (and the nature of the number in question) breaking it down into prime factors [2] is what I proceeded to do.

Well as $65$ ends in a $5$, it is divisible by $5$. Also $5$ itself is a prime number of course. When we divide $65$ by $5$ we get $13$. Again $13$ is a prime number and so we are done, i.e.

$65=5\times 13$

Given this, the first answer to our puzzle is:

$5 + 13 = 18$

But we are just starting, even on the obvious answers. Also staring us in the face is the following:

$65=1\times 65$

$1 + 65 = 66$

At this point we may begin to feel somewhat pleased with ourselves, but there are two further simple solutions which have been overlooked. As a pointer to these, the question referred to “numbers”, not what sort of numbers. We leapt – perhaps not unreasonably – into thinking about positive whole numbers (Natural Numbers in the Mathematical terminology), what about thinking about negative whole numbers (expanding our solution set to the Integers)? Well immediately we can see that:

$65=-5\times -13$

So a third solution is:

$-5 + -13 = -18$

and of course also:

$65=-1\times -65$

$-1 + -65 = -66$

We now have as solutions the set $\{\pm18,\pm66\}$ [3].

Surely that’s a bit more than the person posing the question had envisaged. However, there is more to come. Again we can get new solutions by broadening our concept of number to now include fractions (bearing in mind once more than there will be positive and negative solutions).

Generalised

Most of brainteasers similar to the one we are considering tend to have solutions in the Natural Numbers. However nothing in the question itself rules out looking for answers in any other number system. We have already fruitfully considered the Integers, next we think about solutions that are Rational Numbers (aka fractions of whole numbers like $\dfrac{1}{2}$ or $\dfrac{22}{7}$). If we label our two numbers $x$ and $y$, we have:

$65=xy$

Which we can rearrange as:

$y=\dfrac{65}{x}$

We can see that (with the exception of $x=0$ for obvious reasons) for any value $\dfrac{a}{b}$ of $x$ we can always generate a value of $y$ that satisfies our product equation.

If we now want to consider the sum, $x+y$, we can substitute for $y$ to get:

$x+\dfrac{65}{x}$

If we plot this graph, then for each chosen value of $x$, this plots the answer to our question of what is the sum of $x$ and $y$.

So instead of four solutions we now have a countably infinite number of solutions, something of a shift in perspective.

[Incidentally if we plug each of $\pm1,\pm5$ into the above equation, we recover our initial four Integer solutions]

Obscure

Well that seems pretty comprehensive, doesn’t it? However, again, a broader knowledge of number systems opens the gate to yet more potential solutions. To understand these, we need the concept of Complex Numbers. I provide a whistle-stop tour of these in A Brief Taxonomy of Numbers (see the link in the previous sentence) and a much more leisurely review in Glimpses of Symmetry (Chapter 7 – Imaginary Battleships). Given this, I will only sketch what Complex Numbers are here and refer readers to my other works for additional background.

To leap right in, Complex Numbers are an extension of the Real Numbers [4]. Hang on a second! We haven’t as yet met the Real Numbers! Well these the next class of numbers up from the Rationals Numbers, which we met in the last section. The difference between the Rationals and the Reals is that the latter includes numbers, such as $\sqrt{2}$ which cannot be expressed as fractions of whole numbers [5]. If you consider a continuous number line stretching to infinity in both directions from zero, then any inhabitant of this line will be a Real Number. If you metaphorically stick a pin in the line, the resulting number may be an Integer, it may be a Rational Number, but it will always be a Real Number. There are no gaps in between Real Numbers, which is why they are often described as forming a continuum.

So with that definition under our belt, what about Complex Numbers. These can be viewed as two copies of the Real Numbers spliced together. The way that this happens is by defining a new number, $i$ and saying that any Complex Number is of the form $a + ib$, where both $a$ and $b$ are Real Numbers. The special number $i$ arises from elementary equations as one of the two solutions [6] to the equation:

$x^2+1=0$

or equivalently

$x^2=-1$

That is $i=\sqrt{-1}$ or $i^2=-1$. While this formulation can seem counter-intuitive to some, it is a perfectly robust and valid definition. The concept of number is somewhat slippery, something I explore in A Brief Taxonomy of Numbers, but it can be convincingly argued that $i$ is no less concrete a number than $65$ [7].

So back to the question at hand, let’s consider two related Complex Numbers, $8+i$ and $8-i$ [8]. What happens when we multiply these? Well, if we multiply out the terms long-hand, we have:

$(8+i)(8-i)=$ $(8\times 8)+(8\times -i)+(i\times 8)+(i\times -i)$

Which – given that $(8\times -i)$ and $(i\times 8)$ cancel out – can be simplified to:

$(8\times 8)+(i\times -i)$

which clearly gives us:

$64-i^2=64- -1=64+1$ (by the definition of $i$)

Which of course yields the final answer of, you have guessed it, $65$!

We can use these two new numbers whose product is 65 to generate another solution to our initial problem:

$(8+i)+(8-i)=8+8=16$

By the same logic we applied at the beginning, the negatives of the above numbers will also work, so:

$(-8-i)(-8+i)=65$

and

$(-8-i)+(-8+i)=-8+ -8=-16$

 Aside: To be clear here, we could get these answers from the formula we established in the last section, by setting the result of our sum to be 16, i.e.: $x+\dfrac{65}{x}=16$ Which we can write as: $x^2+65=16x$ Rearranging: $x^2-16x+65=0$ If we then use the standard formula for finding the roots of a quadratic in the form $ax^2+bx+c=0$, namely: $\displaystyle x=\dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$ We get: $\displaystyle x=\dfrac{16 \pm\sqrt{16^2-4\times 1\times 65}}{2\times 1}$ Which simplifies to: $\displaystyle x=\dfrac{16 \pm\sqrt{256-260}}{2}$ and then: $\displaystyle x=\dfrac{16 \pm\sqrt{-4}}{2}$ and then: $\displaystyle x=\dfrac{16 \pm 2i}{2}$ Which finally yields: $\displaystyle x=8 \pm i$ Which is the same as the first Complex solution we found above [9].

So that is a rather interesting result, an obvious question is, can we generalise solutions in the Complex Numbers in the same way that we did for Rational Numbers. Well let’s explore this idea.

Recondite

So if we have two Complex Numbers, $z=a+bi$ and $w=c+di$, where each of $a,b,c,d \in\mathbb{R}$ [10], then we know that:

$\displaystyle z\times w=(a+bi)(c+di)=65$

What does this tell us? Well let’s first of all multiply out the brackets and also write $65$ as $65+0i$:

$\displaystyle (a+bi)(c+di)=(ac-bd)+(ad+bc)i=$ $65+0i$

The only way that this can work is if the Real and Imaginary parts (i.e. the term without an $i$ and the term with one) of the multiplied out brackets are equal, so this implies two things:

$\displaystyle ac-bd=65 \hspace{20mm} (1)$

and

$\displaystyle ad+bc=0 \hspace{20mm} (2)$

$(2)$ we can rewrite as:

$\displaystyle \dfrac{c}{a}=-\dfrac{d}{b}$

Which means that the ratio of the Real and Imaginary parts of our two multiplicands is the same (allowing for the negative sign). We can call this ratio $k$ and then write:

$\displaystyle c=ka \hspace{20mm} d=-kb \hspace{20mm} (3)$

Which in turn means that our two multiplicands are:

$\displaystyle (a + bi)$ and $(ka -kbi) = k(a-bi)$

This of course fits the pattern of our $\displaystyle (8 + i)$ and $(8 - i)$ solution above.

We can also use $\displaystyle (3)$ to substitute in equation $(1)$ above to get:

$\displaystyle aka- -bkb=k(a^2+b^2)=65$

or:

$\displaystyle a^2+b^2=\dfrac{65}{k}\hspace{20mm} (4)$

Which is the equation of a circle of radius $\sqrt{\dfrac{65}{k}}$ in the Complex Plane [11].

So, to summarise, our generic Complex solutions are pairs of the format,

$(a+bi),k(a-bi)$ where $a,b,k\in\mathbb{R}$ and $\displaystyle a^2+b^2=\dfrac{65}{k}$

In words, pairs of conjugate Complex Numbers, where one is potentially modified by a constant and the unmodified numbers lie on a circle of radius $\sqrt{\dfrac{65}{k}}$ in the Complex Plane.

What percentage of the Complex Plane is covered by such numbers is left as an exercise for the reader.

The sum of any of these pairs of numbers is obviously:

$(a+bi)+k(a-bi)=(1+k)a+(1-k)bi$ where $a,b,k\in\mathbb{R}$ and $\displaystyle a^2+b^2=\dfrac{65}{k}$

We can use $\displaystyle (4)$ to state this just in terms of $\displaystyle a$ and $\displaystyle k$ as follows:

$(1+k)a+(1-k)i\sqrt{\dfrac{65}{k}-a^2}$ where $a,k\in\mathbb{R}$

With this finding, we have reached the end of today’s foray into the world of Mathematics. It is perhaps instructive to note that an apparently quotidian question can lead to unexpected places and deeper results if looked at in just the right way. This phenomenon is emblematic of how Maths works in general; deep connections and fundamental insights can follow from looking at something apparently more simple from a different angle.

Acknowledgements

The following people’s input is acknowledged:

• James Kua, who wrote an excellent review of the 65 question on Q&A site Quora and with whom I discussed the specific Complex solution.
• Doug Michael, who spotted a schoolboy error in my original reworking of the Beatles’ classic (last time I checked $15\times 3\ne 65$).
• Bryan Wolf, who suggested that I tidy up my text where referring to Rational and Real solutions.

Of course any [other] errors and omissions remain the responsibility of the author.

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Notes

 [1] I have partaken in this myself, as in: Solve if u r a genius, Facebook squares “puzzle” and The triangle paradox. [2] If you need a refresher on Prime Numbers and prime factorisation, you can start with A Brief Taxonomy of Numbers and consider also skimming a section called “Primed for Action” in Glimpses of Symmetry (Chapter 8 – Simplicity). [3] The curly brackets indicate that what falls between them are the members of a set, e.g. $A=\{1,2,3,4,5\}$, $B=\{1,3,5\}$, $C=\{Apples,Pears\}$ etc. The empty set can be written $\{\}$, but is more typically denoted by $\O$. [4] As an aside, exactly the same approach we applied to the Rationals will give us solutions in the Real Numbers, so the argument is not repeated here. Of note however is that while the number of Rational solutions is countably infinite, the number of Real solutions is uncountably infinite. [5] For a proof (dating to antiquity) that $\sqrt{2}$ cannot be expressed as a whole number fraction, see a footnote in Glimpses of Symmetry (Chapter 4 – Rationality and Reality). However tread with care, Hippasus of Metapontum was allegedly murdered by fellow members of the Pythagorean Society for divulging this information, which was also contrary to a central tenet of their cult. [6] The other solution is of course $-i$. [7] Again see A Brief Taxonomy of Numbers for more on this. [8] Technically each is the Complex Conjugate of each other, see a section of Glimpses of Symmetry (Chapter 7 – Imaginary Battleships). [9] Is we instead pick $-16$ we get the other two results from above. [10] $\in$ stands for “in”, or “is a member of”, $\mathbb{R}$ is shorthand for the Real Numbers. [11] For an explanation of the Complex Plane, see Glimpses of Symmetry (Chapter 7 – Imaginary Battleships).
 Text & Images: © Peter James Thomas 2018. Published under a Creative Commons Attribution 4.0 International License.