Number of digits of the number of digits of the number of digits of $2014^{2014}$

Question

How would you solve that problem :

What is the number of digits of the number of digits of the number of digits of $2014^{2014}$ ?

(for instance the number of digits of $12345678901234567890$ is $20$, and the numbers of digits of $20$ is $2$, and the numbers of digits of $2$ is $1$, so the number of digits of the number of digits of the number of digits of $12345678901234567890$ is $1$)

@user153841 To be honest, i have no idea how to do it - there is obviously some trick here, but i cannot find it — Hippalectryon, Jun 12 '14 at 17:51
Well let's try to start somewhere. First, is the answer dependent upon the base of the number system? As in would it be different if we used base 2 or 8 or 2014, for example? — Kyle, Jun 12 '14 at 17:52
It clearly isn't a silly problem, and requires using some mathematical imagination. How many "numbers" (most of us would say "digits") does $10^n$ have? — Mark Bennet, Jun 12 '14 at 17:52
@MarkBennet Well of course $10^n$ has (in base $10$) $n+1$ digits. How can I use this ? — Hippalectryon, Jun 12 '14 at 17:56
@user153841 I really don't see how to answer that. Which way should I try ? — Hippalectryon, Jun 12 '14 at 17:58
I believe you won't; a counterexample should be $n = 10^{15}$. — Avraham, Jun 12 '14 at 18:30
@A.E $log_{10}(log_{10}(log_{10}((10^{10})^{10^{10}}))+1\geq 2$ so i believe it's false — Hippalectryon, Jun 12 '14 at 18:31
Every positive integer is between two powers of 10. How do you find out which they are? — Mark Bennet, Jun 12 '14 at 18:32
Indeed - and you've accepted an answer which uses that approach. Note that iterating log several times makes large numbers quite small. — Mark Bennet, Jun 12 '14 at 18:38
A little note.This is a past IMO problem with the numbers changed. — rah4927, Jun 13 '14 at 14:25

Avraham · Accepted Answer · 2014-06-12T18:34:05.657

6

We know that the number of digits in base $b$ of a given integer $N$ is $\lfloor\log_b(N) + 1\rfloor$. We also know that $\log_b(n^x) = x \log_b(n)$. Here, $n^x = 2014^{2014}$ and $b = 10$. So we can turn the ridiculously large number $2014^{2014}$ into something more manageable using logs of base $10$. That will give you the number of digits in $2014^{2014}$; the rest should follow.

Update: To get the length of $2014^{2014}$ in base $10$, we need $\log_{10}(2014^{2014})$. Well, that is equal to $2014\cdot\underbrace{\log_{10}(2014)}$. That second term is easily handled on a computer or calculator. Now multiply the underbraced term by $2014$, add $1$, and take the integer part of that final number; that is the number of digits in $2014^{2014}$. Apply your "number of digits" reduction twice more, and you have your answer.

Update 2: OK, you cannot use a calculator, right? but you do know how many digits are in $2014$ itself, 4, of course. That means that $3 \leq \log_{10}(2014) < 4$. Now $2014\cdot 3 = 6042$ and $2014\cdot 4 = 8056$, so no matter which way you slice it, the "number of digits of the number of digits" of $2014^{2014}$ is 4. Now you have one more reduction, I believe, right?

edited Jun 12 '14 at 18:34

answered Jun 12 '14 at 18:08

Avraham

3,237

Same comment as for @Joel, 'Well i could do #$2014^{2014}=⌊2014(log_{10}(2)+log_{10}(19)+log_{10}(53))⌋+1$, but how can I solve that without a calculator ? ' – Hippalectryon Jun 12 '14 at 18:11
All you need is a calculator with $log_{10}$ on it. I'll expand the answer. – Avraham Jun 12 '14 at 18:12
I don't think i'm allowed to use a calculator here – Hippalectryon Jun 12 '14 at 18:14
1

OK, you actually don't need to with the extra reductions, let me expand again. – Avraham Jun 12 '14 at 18:20
2

@Hippalectryon $\log_{10} 2014 = 3+\log_{10}2.014$ and knowing that $2^{10}\approx 10^3$ or otherwise, $\log 2$ is about $0.3$ [the value I remember is $0.30103$] use that and see if you need a closer approximation once the calculation is complete. – Mark Bennet Jun 12 '14 at 18:36

score 2 · Answer 2 · answered Jun 12 '14 at 18:00

Hint: write it as number of digits(number of digits(number of digits($2014^{2014}$))) As usual evaluate parentheses first, so start from the inner set. Given your comment the number of digits of $10^n$ being $n+1$, which is correct, you need to estimate number of digits($2014^{2014}$) first.

score 2 · Answer 3 · edited Jun 12 '14 at 18:05

2

If you wish to determine the number of digits a given integer has in base ten, then you would use a logarithm. $$ \# \text{ of digits of } n = \lfloor \log_{10}(n) \rfloor + 1$$

Here $\lfloor x\rfloor $ is the largest integer smaller than $x$. For instance $5$ has one digit, and $0< \log_{10}(5) < 1$ so $\lfloor\log_{10}(5)\rfloor + 1= 1$.

You can iteratively repeat this process to find the answer you are looking for.

edited Jun 12 '14 at 18:05

gebruiker

6,154

answered Jun 12 '14 at 18:00

Joel

16,256

is [] the floor ? – Hippalectryon Jun 12 '14 at 18:01
Yes it is the floor function. – Joel Jun 12 '14 at 18:02
Well i could do #$2014^{2014}=⌊2014(log_{10}(2)+log_{10}(19)+log_{10}(53))⌋+1$, but how can I solve that without a calculator ? – Hippalectryon Jun 12 '14 at 18:07

score 2 · Answer 4 · answered Jun 12 '14 at 18:46

We know the answer is at least $1$ (You can't have a number with zero digits. If we define zero as having zero digits, then that would require $2014^{2014} = 0$).

Suppose the answer is at least $2$. Then the number of digits of the number of digits of $2014^{2014}$ is at least $10$. So the number of digits of $2014^{2014}$ is at least $1,000,000,000$. But, $2014^{2014} < 10000^{2014}$, the latter of which has only $8053$ digits, a contradiction.

Thus the answer is $1$.

Number of digits of the number of digits of the number of digits of $2014^{2014}$

4 Answers4