Category Archives: Arithmetic

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let x be the number of cats and y be the number of dogs.
Then 0.9x+0.1y think they are cats.
But we also know 20% of the total think they are cats.
0.2(x+y)
Therefore, 0.9x+0.1y=0.2(x+y)
0.9x+0.1y=0.2x+0.2y
0.7x=0.1y
7x=y
Percentage of cats is \frac{x}{x+y}\times100
Substitute 7x for y
\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%
\therefore 12.5% of the animals are cats

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Filed under Algebra, Arithmetic, Percentages, Simplifying fractions, UK Mathematics Challenge

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
7^01
7^17
7^249
7^3343
7^42401
7^516807
As numbersAs Powers of 7
3438=1\times2401+10373438=1\times7^4+1037
1037=3\times343+81037=3\times7^3+8
8=1\times7+18=1\times7^1+1
1=1\times11=1\times7^0

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.

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Filed under Number Bases, Polynomials

Fractions to decimals

People usually know some fractions as decimals, for example

    \[\frac{1}{4}=0.25\ \textnormal{or }\frac{4}{5}=0.8\]

And denominators that are powers of ten are also easy,

    \[\frac{47}{100}=0.47\ \textnormal{or }\frac{256}{1000}=0.256\]

But what if it is something else? One that you don’t know. For example,

    \[\frac{5}{12}\ \textnormal{or }\frac{15}{37}\]

I like to do these as a long division

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Filed under Arithmetic, Decimals, Fractions

Divisibility Rules – Integers 1 to 10

I think it’s useful to knot the divisibility rules, at least up to ten. Although, let’s face it, 7 is a bit tricky.

Divisibility Rules pdf

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Filed under Arithmetic, Divisibility