Category Archives: Number Bases

October 4, 2025 · 8:36 am

Converting base 10 numbers to base 2

Converting integers to base 2 is reasonably easy.

For example, what is 82 in base 2?

Think about powers of 2

$n$	$2^n$
$0$	$1$ (‘ones’)
$1$	$2$ (‘tens’)
$2$	$4$ (‘hundreds)
$3$	$8$ (‘thousands’)

Make $82$ the sum of powers of $2$ .

$82=64+16+2=1\times 2^6+0\times 2^5+ 1\times 2^4+0\times 2^3+0\times 2^2+1 \times 2^1+0\times 2^0=1010010$

We follow the same approach for real numbers

$n$	$2^n$
$-3$	$\frac{1}{8}=0.125$
$-2$	$\frac{1}{4}=0.25$
$-1$	$\frac{1}{2}=0.5$
$0$	$1$ (‘ones’)
$1$	$2$ (‘tens’)
$2$	$4$ (‘hundreds)
$3$	$8$ (‘thousands’)

Convert $0.765625$ to base $2$

$0.765625\times 2=1.53125$ the first number is 1

$0.53125\times 2=1.0625$ the second number is 1

$0.0625\times 2=0.125$ the third number is 0

$0.125\times 2=0.25$ the fourth number is 0

$0.25\times 2=0.5$ the fifth number is 0

$0.5\times 2=1$ the sixth number is 1 and we have finished

$0.765625=0.110001_2$

What about something like $2.\overline{4}$ ?

The non-decimal part $2=10$

$0.\overline{4}\times 2=0.\overline{8}$ first number is zero

$0\overline{8}\times 2=1.\overline{7}$ second number is 1

$0.\overline{7}\times 2=1.\overline{5}$ third number is 1

$0.\overline{5}\times 2=1.\overline{1}$ fourth number is 1

$0.\overline{1}\times 2=0.\overline{2}$ fifth number is 0

$0.\overline{2}\times 2=0.\overline{4}$ sixth number is 0

We are back to where we started, so $2.\overline{4}=10.0111000111000..._2=10.\overline{0.11100}_2$

Divisibility Rule for 11

I was working on a question and involved 11 and I wondered what the divisibility rule was?

So then I had a bit of a think about it.

Let $N$ be a number divisible by $11$ . The $N (mod11)=0$

$\begin{equation*}N=a_n\times10^n+a_{n-1}\times10^{n-1}+a_{n-2}\times10^{n-2}+...+a_0\times10^0\end{equation}$

$\begin{equation*}0=a_n\times10^n (mod11)+a_{n-1}\times10^{n-1}(mod11)+a_{n-2}\times10^{n-2}(mod11)+...+a_0\times10^0(mod11)\end{equation}$

Now $10 (mod11)=10$ which is congruent to $-1$ because $10-(-1)=11$ , which is a multiple of 11.

Thus

$\begin{equation*}0=a_n(-1)^n+a_{n-1}(-1)^{n-1}+a_{n-2}(-1)^{n-2}+...+a_0(-1)^0\end{equation}$

Odd powers will be negative and even positive.

So if we start at one end of the number and add every second digit (i.e. first digit plus third digit plus fifth digit etc.) and then subtract the other digits (i.e. second digit, fourth digit, six digit, etc.), if that equals zero then the number is divisible by 11.

For example, is $1756238$ divisible by $11$ ?

$1+5+2+8-7-6-3=16-16=0$

Hence $1756238$ is divisible by $11$

Australian Mathematics Competition – Polynomial Question

I came across this question from the 2010 Senior Australian Mathematics Competition:

A polynomial $f$ is given. All we know about it is that all its coefficients are non-negative integers, $f(1)=6$ and $f(7)=3438$ . What is the value of $f(3)$
Australian Mathematics Competition 2006-2012

I thought ‘excellent, a somewhat hard polynomial question for my students’ and then I tried it. Now I know why only 1% of students got it correct.

As we don’t know the order of the polynomial, let

$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0$

We know all of the coefficients are greater than or equal to zero. We also know

$f(1)=a_n+a_{n-1}+...+a+a_0=6$

Which means that all of the coefficients are between zero and six

$0\le{a_n}\le{6}$

We have also been given $f(7)$

$f(7)=7^na_n+7^{n-1}a_{n-1}+ ... +7a+1=3438$

As all of the coefficients are between zero and six, this is $3438$ written in base 7.

Let’s calculate a few powers of 7

	Powers of 7
$7^0$	1
$7^1$	7
$7^2$	49
$7^3$	343
$7^4$	2401
$7^5$	16807

Hence $3438$ written in base $7$ is $13011$

Therefore $f(x)=x^4+3x^3+x+1$

$f(3)=3^4+3\times3^3+3+1$

$f(3)=81+81+4$

$f(3)=166$

I really like this question. I think it could work well as a class extension activity with a bit of scaffolding.

As numbers	As Powers of 7
$3438=1\times2401+1037$	$3438=1\times7^4+1037$
$1037=3\times343+8$	$1037=3\times7^3+8$
$8=1\times7+1$	$8=1\times7^1+1$
$1=1\times1$	$1=1\times7^0$

Category Archives: Number Bases

Converting base 10 numbers to base 2

Divisibility Rule for 11

Australian Mathematics Competition – Polynomial Question

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