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Number Theory Level 2

If $x$ in an integer such that $2\leq x \leq 10,$ how many digits will $x^x$ have when written in base- $x?$

10 x x-1 x+1

2 solutions

Christopher Boo
Nov 14, 2014

$1\cdot x^x + 0\cdot x^{x-1}+0\cdot x^{x-2}+\dots+ 0\cdot x^0=x^x$

In base $x$ , the value of $x^x$ is $1$ followed by $x$ $0$ 's. So, the number of digit will be $x+1$ .

I wasn't sure with the case of base $1$ . I've searched for information on MSE and the answer stated that

Such a system is known as a Unary Numeral System (Wikipedia Entry):

The unary numeral system is the bijective base-1 numeral system. It is the simplest numeral system to represent natural numbers: in order to represent a number N, an arbitrarily chosen symbol representing 1 is repeated N times. This system is used in tallying. For example, using the tally mark |, the number 6 is represented as ||||||.

So, in the case of base $1$ , even if we use the digit $1$ instead of the tally mark, the number of digit of $1^1$ is not $2$ .

Christopher Boo - 6 years, 7 months ago

Nice research! I've updated the question so that $x \neq 1$ .

Calvin Lin Staff - 6 years, 6 months ago

Parth Lohomi
Nov 13, 2014

I took an example let $x$ =2

So $x^{x}$ =4

So $4_{2}$ = $100$

Three digits $3$ = $2$ + $1$

$\boxed{x+1}$

what if 4^4=256 hence, x+1 isnt quite appropriate!!! it has many solutions (this problem)

anushka janawlekar - 6 years, 6 months ago

This is not the appropriate solution.

Shubhendra Singh - 6 years, 7 months ago

this is the quick solution

math man - 6 years, 7 months ago

There is more than one solution to a problem.

Trevor Arashiro - 6 years, 6 months ago

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