Standard index form matching
Problem
When working with very large or very small numbers, it's useful to be able to convert them into standard index form, also known as scientific notation. Numbers in standard form are expressed as $a \times 10^{b}$, where $a$ is a number between 1 and 10.
The aim of this game is to match pairs of cards.
Click on a card in the interactivity below to select it. Then click on another one. If the two cards match, they will change colour. If the two cards do not match, select another pair.
The game ends when all the cards have been matched in pairs.
Click on the links below if you would like to try some alternative versions of the game
Play Level 1 against the clock
Play Level 2 against the clock
Alternatively, you can print a set of cards out to play away from the computer. There are two sets to choose from, Set 1 and Set 2 (which is more challenging). Here are some activities you could try on your own or with a partner:
- Shuffle the cards and deal them out face up. Take it in turns to find a matching pair. If your partner chooses two cards and you don't think they form a matching pair, challenge them to explain their reasoning.
- Shuffle the cards and deal them out face up. Time yourself find all the matching pairs. Then try it again - can you beat your time?
- Shuffle the cards and deal them out face down. Turn two cards over at a time. If they match, keep them. If they don't, put them back.
Once you are confident at matching the pairs, here are some questions you might like to consider:
- How do you know that two cards match?
- Are some pairs easier to match than others?
- Can you create some more cards to add to the sets?
Send us your strategies and your thoughts about these questions.
Getting Started
$7.5 \times 10^6$ multiplied by $3 \times 10^2$ is the same as $7.5 \times 3 \times 10^6 \times 10^2$
$7.5 \times 10^6$ divided by $3 \times 10^2$ is the same as $\frac{7.5 \times 10^6}{3 \times 10^2}$ which is the same as $\frac{7.5}{3} \times \frac{10^6}{10^2}$
Student Solutions
Hugo from Bangkok Patana School in Thailand, Freya and Jonty from Long Field Academy in the UK, Kota from Latymer School in England and Arsh from Jamnabai Narsee International School in India explained how you can tell that two cards match. This is some of Jonty's work:
This is some of Arsh's work:
Kota wrote all of that algebraically:
When you multiply two standard forms together, the answer will become the product of the coefficients, and the index will be the sum of the two indices.
So $$a\times10^x \hspace{2mm}\times\hspace{2mm} b\times10^y = ab\times10^{x+y}$$
Similarly $$\frac{a\times10^x }{ b\times10^y} = \dfrac{a}{b}\times10^{x-y}$$
Deeya from St Mary's Catholic High School Astley Manchester in England used this strategy to match the cards quickly:
Ariel started from the second part of Deeya's strategy:
For level 1, modifying of index is not needed. Therefore you can match some unique index ($b$) first, then multiply the $a$ to identify the rest. Example:
Four of the cards are $\left(2.4\times10^{-5}\right)\times\left(2\times10^4\right),$ $2.4\times10^{-2},$ $\dfrac{7.2\times10^1}{3\times10^3}$ and $4.8\times10^{-1}.$
$-5+4=-1, 1-3=-2,$ so the first should match the fourth and the second should match the third.
For level 2, it is similar. Except you need to divide the $a$ by $10$ if $a$ is larger than $10$ (or less than $1$)
Example (level 2):
Four of the cards are $\frac{1.5\times10^{-4}}{7.5\times10^{-6}},$ $2\times10^1,$ $\frac{4.5\times10^{-3}}{1.5\times10^{-5}}$ and $3\times10^2.$
Both pairs have a original index of $2,$ (since $-4--6=2$ and $-3--5=2$) so comparison of $a$ is needed.
$1.5\div7.5$ is $0.2,$ so the actual index of the first card is $2-1=1$ (because $0.2\times10^2 = 2\times10^1$)
Therefore the first matches the second and the third matches the fourth.
Mahdi from Mahatma Gandhi International School in India described a similar strategy, but startest with the hardest cards instead of the easiest ones:
The key strategy here I used was to boil down the possibilities every time two cards matched. Throughout the games, I observed that half of the card is simplified and half weren't. Thus over time, I started to take the hardest one (not- simplified) and simplify that using the powers of 10. Mainly I did this but also started to find some connection between the numbers in the numerator and the denominator. So now I had simplified one card and approximated it in less than 3-4 seconds. Next, I looked for the simplified cards and tried to match the powers. There were most of the times two cards that had matching powers of 10, so I used the simplified numbers in division or multiplication (numerator and the denominator) to boil down that as well. So now I had a matching pair in less than 6-7 seconds. I continued doing this (starting to find hardest and matching it with one of the simplified ones) and quickly the possibilities of matching 10 powers reduce. Then I could easily find the matching pairs by just approximating powers of 10. I did not have to be precise with division or multiplication. One key thing I had to remember was that sometimes the numbers in the numerator and denominator also changed the powers of 10, so I had to keep that in mind.
Teachers' Resources
Why do this problem?
This problem offers an excellent opportunity for students to practise working with numbers and calculations expressed in standard index form.
Possible approach
The game can be played online using the interactive version. Alternatively, print off enough copies of the cards (Set A and Set B) so that students can work in pairs. The top 12 cards and the bottom 12 cards on each sheet form a set of six pairs. The cards becomes progressively more difficult. This means that students can work with a set of 6, 12, 18 or 24 pairs.
Here are some suggestions for games and activities that they could do:
- Shuffle the cards and deal them out face up. Take it in turns to find a matching pair. If your partner chooses two cards and you don't think they form a matching pair, challenge them to explain their reasoning.
- Shuffle the cards and deal them out face up. Time how long it takes your partner to find all the matching pairs. If they match any incorrectly, add a twenty second time penalty. Then swap over. Who can get the fastest time?
- Shuffle the cards and deal them out face down. Turn two cards over at a time. If they match, keep them. If they don't, put them back. Take it in turns until all the cards have been chosen. Who can collect the most pairs?
Whether students have played online or with the cards, set aside some time at the end of the lesson to discuss the strategies they used to work out the calculations, and which pairs they found easiest/most difficult to match.
Key questions
How do you know that two cards match?
Are some pairs easier to match than others?
Possible support
Students could start by just matching the cards in the top half of Set A.
Possible extension
Invite students to create their own sets of cards with tricky examples, which they can then swap with another pair of students.