base 7 | Racquel Sanderson

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July 16, 2024 · 7:35 am

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
$3438=1\times2401+1037$	$3438=1\times7^4+1037$
$1037=3\times343+8$	$1037=3\times7^3+8$
$8=1\times7+1$	$8=1\times7^1+1$
$1=1\times1$	$1=1\times7^0$

Tag Archives: base 7

Australian Mathematics Competition – Polynomial Question

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