# The curious case of the number 3435

Posted by: Gary Ernest Davis on: April 17, 2011

The (base 10) number $3435$ has the curious property that $3^3+4^4+3^3+5^5 =3435$

It shares this property with $1$ since $1^1=1$.

What makes $3435$ even more curious is that it is the only number other than $1$ that has this property.

To prove this we need to talk a little more exactly about what is this property.

#### Baron von MÃ¼nchhausenThe MÃ¼nchausen property

Suppose $n$ is a positive integer and \$latex d_k, see ldots d_1, patient d_0\$ are the base 10 digits of n.

This means – a fact every elementary teacher of mathematics tries to get across to students when teaching place value – that

$n= d_k imes 10^k+ldots d_1 imes 10+d_0$

So given that $n=d_kd_{k-1}ldots d_1d_0$ (base 10) we form a new number:

$n^*=d_k^{d_k}+d_{k-1}^{d_{k-1}}+ldots d_1^{d_1}+d_0^{d_0}$

The property we are interested in – sometimes called the MÃ¼nchausen property – is that $n=n^*$.

So we see that $1^*=1 extrm{ and } 3435^*=3435$.

We will now see that there is no other positive integer $n$ for which $n=n^*$.

But first, store a little excursion into properties of the function $x^x$.

#### Some properties of $x^x, x>0$ and a convention for $0^0$

We should have discussed what happens if a digit of the positive integer $n$ is $0$: because, then, we have to assign a value to $0^0$.

There is no magic way to decide how to assign a value to $0^0$ – we will use the continuity of the function $F(x):=x^x$ for $x>0$:

Graph of F(x)=x^x for x>0

Because the limit of $F(x)=x^x$ as $x$ approaches $0$ with $x>0$ is $1$ we assign the value $1 extrm{ to } 0^0$.

There is nothing magic about this – it just turns out convenient for our purposes here to have $0^0=1$.

We can see the $F(x)=x^x$ initially decreases as $x$ increases, reaches a minimum value, and then increases.

We can see this from the derivative $frac{dx^x}{dx}$ of $x^x$, which we can calculate by first taking (natural) logarithms:

$log(x^x)=xlog(x)$ so $frac{frac{dx^x}{dx}}{x^x}=1+log(x)$ and therefore $frac{dx^x}{dx}=x^x(1+log(x))$

This is negative for $0 and positive for $x>frac{1}{e}$.

So, for non-negative integer values of $x$ the function $F$ is non-decreasing: in fact, apart from $F(0)=F(1)=1$, we have $F(n+1)>F(n)$ for all positive integers $ngeq 1$.

#### The number of digits of a positive integer

For a positive integer $n$ the number of (base 10) digits of $n$ is the floor of $log_{10}(n)+1$, the largest integer less than, or equal to, $log_{10}(n)+1$.

This means that the number of digits of $n$ is less than $log_{10}(n)+1$.

#### An inequality for $n^*$

If the base 10 digits of the positive integer $n$ are $d_k,ldots , d_1, d_0$ then:

1. $0leq d_i leq 9$ for $0leq ileq k$
2. The number of digits of $n$ is $k+1 leq log_{10}(n)+1$

So, $n^*=d_k^{d_k}+ldots +d_1^{d_1}+d_0^{d_0} leq (k+1) 9^9 leq 9^9(log_{10}(n)+1)$

This says that $n^*$ cannot grow too big in terms of the number of digits of $n$.

#### An inequality for large $n$

We look at the behavior of the function $G(x):=frac{x}{log_{10}(x)}$:

Graph of x/log10(x) for x>1

The derivative of $G(x)$ is $frac{dG(x)}{dx}=log(10)frac{log(x)-1}{log^2(x)}$ and this is positive for $x>e$.

So, for $n>2 imes 10^{10} =20000000000 >e$ we have:

$frac{n}{log_{10}(n)+1}>frac{2 imes 10^{10}}{10log_{10}(10)+log_{10}(2)+1}$

$>frac{2 imes 10^{10}}{2 imes 10 log_{10}(10)}=10^9>9^9$

x

In other words, for $n> 2 imes 10^{10} = 20000000000$ we have $n> 9^9 (log_{10}(n)+1)$

#### Positive integers $n> 2 imes 10^{10}=20000000000$Â  cannot be equal to $n^*$

x

If $n$ is too big – for instance, $n>2 imes 10^{10} = 20000000000$ – then $n$ cannot be equal to $n^*$.

The reason is that for $n>2 imes 10^{10}$ we have $n> 9^9(log_{10}(n)+1)$ while $n^*leq 9^9(log_{10}(n)+1)$

So we search through all positive integers up to $20000000000$ and find no integers $n$ with $n=n^*$ other than $n=1, n=3435$.

#### Variation on this theme

Suppose instead of forming $n^*$ as we did, we move the digits forward, with the first digit $d_0$ moving around to be last: in other words, we form a new number $n^{**}:= d_k^{d_0}+d_{k-1}^{d_k}+ldots +d_1^{d_2}+d_0^{d_1}$.

Do we ever have $n=n^{**}$ ?

Well, yes: we do for $n=1, 4155, 4355$ because $4^5+1^4+5^1+5^5 = 4155$ and $4^5+3^4+5^3+5^5=4355$

Are there any others?

What if we move each digit forward two (with the first two cycling around to be be last and second last)?

#### References

van Berkel, D. (2009) On a curious property of 3435. Retrieved from arxiv.org: On_a_curious_property_of_3435 [This article provides the argument I have described in this post]

Perfect digit-to-digit invariant Wikipedia.org [In this reference another number is counted as a Munchausen number due to their using the convention 0^0=0]

### 8 Responses to "The curious case of the number 3435"

You link to Wikipedia, which gives 438579088 as another Munchausen number. Did you even read it?

You will see that there they choose to define 0^0=0, which is how this other number comes to be counted as a Munchausen number.

I explained how for the proof in the post I was taking 0^0=1 and why I was doing that.

You have a small typo in your last sentence: “… due to their using the convention 0^0=1”. Their convention os “0^0=0”, not “0^0=1”.

Thanks – fixed.

In the Wikipedia case, the 0^0 = 0 is what qualifies solely ‘0’ to be a Munchausen number. It in no way qualifies that 438579088 to be a Munchausen number. At the same time, the definition merely asks for a number to satisfy ‘n = n*’ to be a valid Munchausen number, in which case, if you suggest that a method of proof does not say so, then the method of proof is obviously flawed (again, which is the not the case). And I say so, because the only conclusion your proof arrives at is that, any n, such that n > 2×10^10 cannot be a Munchausen number and then you go ahead to make an argument that the only number between 1 and 2×10^10 which is a Munchausen number is 3435. The post-conclusion is wrong and is not in any ways affected by how you chose to define your function at a particular discontinuity.

Rohan,

the argument shows the Munchausen numbers are bounded by a – relatively – small number. An exhaustive search up to that number shows there are none, other than 1 and 3435.

This is so amazing. I wonder what other properties 3435 has?

Wow, that’s fantastic :) While I’m sure this is a relatively simple example, it’s still cool to see a combination of mathematical logic and computing power working in symphony.