Problem of the Year for 2022
In keeping with the Problem of the Year for 2020 and 2021, we have again focused on fluency and flexibility with number, place value and multiplicative thinking.
The problem should be within the reach of Year 4, 5 and 6 students especially if estimation, risk taking and calculator use are encouraged. So, here is your Problem of the Year for 2022.
As students prepare to work on the problem remind them that strategies such as
- Guess, Check and Improve
- What if I tried….
- Use the greedy algorithm to make 2000 first and then worry about 22 later
- Keeping a clear record of what is being tried and changed
are all part of the messy process of solving the problem. Possible pathways (of which there are many) could include:
2 multiplied by itself 11 times gets us to 2048 which is 26 too many. It’s remains to find a way to make 28 using eleven 2s and that is possible.
Working with 10s
10 can be made with four 2s as 2 × 2× 2× 2 + 2. It will take twelve 2s to multiply 10 by itself three times to make 1 000 and one more × 2 to make 2 000. Can we make the remaining 22 with the nine 2s? Yes, add 10 + 10 + 2 and the total is made!
2022 is even and the digits add to 6 which means that 2022 is a multiple of 3.
2002 = 6 × 337 and 6 = 2 × (2 + 2 ÷ 2). Not promising! But we can show that 337 = 256 + 81 and 256 is 2 multiplied by itself 8 times. With a possible twelve 2s used so far, it means that the remaining 81 needs to be made with nine 2s. That can be done if we use 1 = 2 ÷ 2 again.
Here is another form of the problem for your early finishers, based on last year’s challenge.