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'Ordering Cards' printed from http://nrich.maths.org/
This challenge obviously led to some interesting explorations! Here's a solution sent in by Maicy, Caitlan and Taryn
from Saxmundham Primary School in England
| Calculation |
Method |
Answer |
| 3 x 6 |
mental |
18 |
| 18 - 3 |
mental |
15 |
| 15 ÷ 3 |
mental |
5 |
| 5 x 2 |
mental |
10 |
| 10 x 2 |
mental |
20 |
| 20 + 1 |
mental |
21 |
| 21 ÷ 3 |
mental |
7 |
| 7 x 2 |
mental |
14 |
| 14 - 2 |
mental |
12 |
| 12 ÷ 3 |
mental |
4 |
| 4 x 2 |
mental |
8 |
| 8 - 5 |
mental |
3 |
if you use this card from another set $10 \times 3$ the card that follows is $30$.
if you use this card from another set $12 \div 2$ the card what follows is $6$
Myles, Joshua and William
also from Saxmundham Primary
School from England
also sent in;
| Sum |
Method |
Answer |
| 7 x 2 |
mental |
14 |
| 14 - 2 |
mental |
12 |
| 12 ÷ 3 |
mental |
4 |
| 4 x 2 |
mental |
8 |
| 8 - 5 |
mental |
3 |
| 3 x 6 |
mental |
18 |
| 18 - 3 |
mental |
15 |
| 15 ÷ 3 |
mental |
5 |
| 5 x 2 |
mental |
10 |
| 10 x 2 |
mental |
20 |
| 20 + 1 |
mental |
21 |
| 21 ÷ 3 |
mental |
7 |
If we used the card from a different set $10 \times 3$ it would make $30$.
If we used $12 ÷ 2$ it would make $6$.
Ashton from Raynsford Voluntary Controlled First School in England sent in this solution;
I started with $8 - 5$ at that $3$ then I found the problem starts with $3$
and that was $3 \times 6$ then I work it out and then I did what you have just done
again so it will be $18 - 3$ then $15 ÷ 3$ then $5 \times 2$ then $10 \times 2$ then $20 + 1$ then $21 ÷ 3$
then $7 \times 2$ then $14 - 2$ then $12 ÷ 3$ then $4 \times 2$ then $8 - 5$ and that is what we started
with at the beginning so we have finish the loop game.
Thank you for these well thought out solutions. Thank you also to those who pointed out that we had slipped up on day $1$ with an incorrect card.