# The Problem for 2021

Last year, we introduced our problem of the year with:

*“Fluency with the four operations is important but without flexibility is often simply procedural and limiting.”*

This year, we have chosen a problem for year 5, 6 and 7 classes, that not only relies on fluency and flexibility, but also requires mathematical thinking and reasoning to find solutions (of which there are many).

Before starting, it would be a good idea to ensure that all students have a calculator handy, as that will reduce the stress of having to make the calculations correctly.

**Solution 1: Finding Factors**

Our first mathematical step is to find the factors of 2021, which turn out to be 47 and 43:

2021 = 47 × 43

Which of the digits can we use to make 47? We found that 6 × 5 + 8 + 9 = 47, leaving us with the problem of making 43 with the digits 7, 4, 3, 2 and 1. Using the known fact that 7 × 6 = 42, we can see that a way to make 43 is 7 × 3 × 4 ÷ 2 + 1 = 43. Hence, our first solution is:

(6 × 5 + 8 + 9) × (7 × 3 × 4 ÷ 2 + 1) = 47 × 43 = 2021

We also found another way to make 43 and 47. 43 is made of 40 and 3 by 8 × 5 + 3 = 43 and, okay, a bit harder to see, but 47 is made of 45 and 2 more by 9 × (4 + 1) + 2 × (7 – 6) = 47.

**The Sting**

Rather than change the problem, the sting for this year’s problem is:

“Why stop there? What other ways of making 2021 can you find?”

As with Solution 1, we have suggested a couple of ways in which thinking mathematically about the problem first can set you on the path to a solution.

**Solution 2: Approximating**

Using the calculator, we searched for multiplications that gave a number close to 2021. We found:

9 × 8 × 7 × 4 = 2016

2016 is 5 short of our total and we have 6, 5, 3, 2 and 1 with which to make 5. Easy!

6 + 5 – (3 × 2 × 1) = 5

Our solution is:

(9 × 8 × 7 × 4) + 6 + 5 – (3 × 2 × 1) = 2021

**Solution 3: The Greedy Algorithm – make the largest part first.**

This time we started with making a number that is clearly close:

20 × 100 = 2000

We chose to make the 20 first and used the fact 9 × 11 = 99 (close to 100) to make 2000 as:

(4 × 5) × (9 × (8 + 3) + 1) = 2000

We had 7, 6 and 2 with which to make 21 as 7 × 3 and our solution is:

(4 × 5) × (9 × (8 + 3) + 1) + 7 × 6 ÷ 2 = 2021

Over to you and your students. Feel free to share via our Facebook page results or email them to [email protected].