Starting tomorrow, 2024 will be the most viewed, most written and most mentioned number.

But what do you know about him?

And no, I don’t have a crystal ball to predict the future.

I am talking, of course, about the curiosities, proposals or mathematical reflections that can be extracted from the number that will bring together the next 365 days of your life.

Here are 15 questions that you can use as suggestions to ask with your children or your students. Let’s see what you think:

- How many figures does 2022 have?
- How many decimal places are needed to write it?
- How do you write with Roman numerals?
- Is it an even or odd number?
- Is it a prime or composite number?
- What is the sum of its digits?
- Do you know how many four-digit numbers can be written with exactly three 2’s?
- How many of those numbers could be years past?
- What are its prime divisors?
- What is the sum of the prime divisors?
- What are ALL your divisors?
- What is the sum of all its divisors?
- Did you know that 2022 is an abundant number?
- And that we are looking at a Harshad number?
- Could you say the square of 2022 without (almost) doing calculations?

And here are the answers:

## 1. How many figures does 2024 have?

2022 has 4 figures: 2, 0, 2 and 4.

## 2. How many decimal figures are necessary to write it?

It is true that, generally, when we talk about “decimal figures” we are referring to the figures to the right of the comma, and which represent the *non-integer* part (that is, less than 1) of a number. They are tenths, hundredths, thousandths, etc.

However, the term “decimal figures” can have another interpretation: the figures used in the decimal system, which are: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. To write the 2022 only two of them are needed: 0 and 2.

**Observation** : I will not go into details now, but, if you used the binary system (which only has two digits, 0 and 1), 2022 would be written like this: 11111100110. As you can see, it is a very long number.

## 3. How do you write with Roman numerals?

2024 in Roman numerals is written like this: **MMXXII** . With only three letters and, furthermore, repeated two by two.

## 4. Is it an even or odd number?

**Even numbers** are those that are divisible by 2, that is, they have an integer half. Remember that to know if a number is even or odd you only have to look at the last figure: if it is 0 or an even number (2, 4, 6, 8), the number will be even and if it is odd (1, 3 , 5, 7, 9), the number will be odd.

Clearly, 2022 is an even number, since its last digit is 2.

## 5. Is it a prime or composite number?

Since 2022 is an even number, it cannot be a prime number, since it will at least be divisible by 2. Remember that prime numbers are those that are only divisible by 1 and by themselves. Thus, 2022 is a composite number.

## 6. What is the sum of its digits?

The sum of its digits is 2 + 0 + 2 + 2 = **6** .

## 7. Do you know how many four-digit numbers can be written with exactly three 2's?

Using three 2’s you can write no more and no less than 35 four-digit numbers.

If there are four digits and three of them are a 2, there is only one digit that is not a 2. This digit can occupy four different positions.

If it occupies the units position, we obtain a number of the type 2 2 2 A, in which A can be replaced by 0, 1, 3, 4, 5, 6, 7, 8 and 9. Thus, there are **9** different possibilities.

The same thing happens if said figure occupies the tens positions (2 2 A 2) or the hundreds positions (2 A 2 2), which adds **18** more possibilities.

Finally, if the figure occupies the thousands position (A 2 2 2) there are only **8 possibilities** , since the number 0 2 2 2 is not a four-digit number.

Thus, there are 9 + 9 + 9 + 8 = **35** different four-digit numbers with three 2s.

## 8. How many of those numbers could be years past?

The numbers of the type 2 2 2 A, 2 2 A 2 and 2 A 2 2 do not correspond to past years, since the smallest of all of them is 2022.

Of the numbers of type A 2 2 2, only **1222** is a year that has already passed. I have already said before that 0 2 2 2 cannot be considered a four-digit number, and the next one would be 3222, which is obviously still to come.

## 9. What are its prime divisors?

A **divisor** of 2022 is a number that multiplied by another gives 2022 or, in other words, a number that divides, exactly, 2022.

As you already know, a **prime number** is one that can only be divided exactly by 1 and by itself. Thus, the prime divisors of 2022 will be those that divide it exactly and that, in turn, can only be divided exactly by 1 and by themselves.

To begin with, you already know that 2022 is an even number and, therefore, that it is divisible by 2.

2024 : 2 = 1012

1011 is also a divisor of 2022 (2022 : 1011 = 2). However, he is not a cousin, so I cannot add him to the list. Even so, it is useful to know it, since all its divisors are at the same time 2022. I can continue the search, then, starting from 1011.

1011 is not divisible by 2, because it is an odd number. On the other hand, it is divisible by 3. To check it, there are two options: do the division and see if it is exact or remember the rule that says: “A number is divisible by 3 if the sum of its digits is a multiple of 3.” .

The sum of the figures of 1011 is 1 + 0 + 1 + 1 = 3, which is obviously a multiple of 3 (3 x 1 = 3), so 1011 is divisible by 3:

1011 : 3 = 337

Thus, 3 is also a divisor of 2022 (2022 : 3 = 674).

Now I can start from 337, since all the divisors of 337 are at the same time divisors of 1011 and, therefore, of 2022.

I already told you that 337 is a prime number.

But how do I know?

Have I tried dividing it by all the numbers between 2 and 336? The truth is, no.

To begin with, because it is not necessary.

To check if a number is prime, it is enough to check if it is **divisible by some other prime number less than its square root** . I explain.

Firstly , when a number is divisible by a composite number (for example, 6) this implies that it is also divisible by its prime factors (in **this** case, 2 and 3, since 6 = 2 x 3). Look at 36, for example: it is divisible by 6 and, therefore, it is also divisible by 2 and 3. Or, put another way: it is divisible by 2 and 3 and, therefore, it is also divisible by 6. So, if I check that a number is not divisible by 2 and 3, which are prime numbers, I already know that it will not be divisible by 6, which is a composite number, so I will not need to check it.

Secondly , when a number is divided by any of its divisors, the result obtained is also a divisor of said number **. **36 : 2 = 18, so 18 is also a divisor of 36 (36 : 18 = 2). Likewise, 36 : 3 = 12, so 12 is also a divisor of 36 (36 : 12 = 3). In this way, every time a divisor of 36 is found, two are actually found: each divisor corresponds to another divisor. 4 and 5 are not divisors of 6. But what about 6? 6 is the square root of 36 (36 : 6 = 6). Thus, all numbers greater than 6 will give quotients less than 6, and all divisors less than 6 I have already found, so I can end my search.

The square root of 337 is approximately 18’36. Prime numbers less than 18 are: 2, 3, 5, 7, 11, 13 and 17.

**337 is not divisible by 2** because it is not an even number.

**It is also not divisible by 3** because the sum of its figures (3 + 3 + 7 = 13 and 1 + 3 = 4) is not a multiple of 3.

**It is not divisible by 5** because its last digit is neither 0 nor 5.

**It is also not divisible by 7** . There is a little-known rule to know if a number is divisible by 7. It goes like this: “A number is divisible by 7 if the result of removing the units figure and then subtracting twice that figure is a multiple of 7.”

If I take away the units number (7) from 337, I have 33 left. Now I subtract double that number (that is, 7 x 2 = 14) and I get 33 – 14 = 19.

19 is not a multiple of 7, so neither is 337.

So that the rule is understood, I am going to verify that, for example, 336 is divisible by 7:

- If I remove the number of units from 336 (6), there remains 33
- If I subtract double 6 from 33 (that is, 12), it gives 33 – 12 = 21
- 21 is a multiple of 7 (3 x 7 = 21), so 336 is divisible by 7 (336 : 7 = 48)

I follow.

**337 is not divisible by 11** . For a number to be divisible by 11, “the sum of the numbers occupying the even position minus the sum of the numbers occupying the odd position must be 0 or a multiple of 11 (11, 22, 33…)”.

337 has one even position, occupied by the number 3, and two odd positions, occupied by 3 and 7.

7 – (3 + 7) = -3,

so 337 is not divisible by 11.

Look at 341, for example. In this case, 4 – (3 + 1) = 0, so it is divisible by 11 (341 : 11 = 31).

There are rules to check if a number is divisible by 13 and 17, but they are more difficult to remember, so I recommend that you simply do the division and check if it is exact.

337 is not divisible by 13 or 17.

In conclusion: the prime divisors of 2022 are **1** , **2** , **3** and **337** and therefore:

2022 = 1 x 2 x 3 x 337

## 10. What is the sum of the prime divisors of the number 2022?

The sum of the prime divisors of 2022 is 1 + 2 + 3 + 337 = **343** .

## 11. What are ALL your divisors?

As you have seen before: 2022 = 2 x 3 x 337 (I have ignored the 1, since the result does not vary).

To determine all the divisors of the number 2022, I am going to use two properties of multiplication:

- the
**commutative property**, according to which the order of the factors does not alter the product - the
**associative property**, according to which the way of grouping the factors does not alter the product either

Thus, the previous product can also be expressed as:

2022 = (2 x 3) x 337 = **6** x 337

2022 = 2 x (3 x 337) = 2 x **1011**

2022 = 3 x (2 x 337) = 3 x **674**

The divisors of 2022 are: **1** , **2** , **3** , **6** , **337** , **674** , **1012** and ** 2024** .

## 12. What is the sum of all its divisors?

The sum of all divisors of 2022 is: 1 + 2 + 3 + 6 + 337 + 674 + 1012 + 2024 = **4056** .

## 13. Did you know that it is an abundant number?

An **abundant number** or **excessive number** is a number such that the sum of its **proper divisors** (that is, all its divisors except itself) is greater than it. The difference between the sum of the proper divisors and the number in question is known as the **abundance** of said number.

In the case of the number 2022, the sum of its proper divisors is 1 + 2 + 3 + 6 + 337 + 674 + 1011 = 2034. Thus, the abundance of 2024 is 2034 – 2024 = 12.

## 14. And that we are looking at a Harshad number?

A **Harshad number** (from the union of the Sanskrit terms *harsa* , meaning joy, and *da* , meaning to give) is a number divisible by the sum of its digits.

The number 2022 is a Harshad number, since the sum of its digits is 2 + 0 + 2 + 2 = 6 and 2022 is divisible by 6.

## 15. Could you say the square of 2022 without (almost) doing calculations?

The square of 2,022 is 4,088,484 (I calculated it with the calculator ;-)). When I saw it for the first time, it immediately caught my attention. Just like the number 2022 only has *twos* and *zeros* , 4,088,484 only has *fours* , *eights* and *zeros* , with 4 = 2 ^{2} and 8 = 2 ^{3} .

Notice:

Seeing this, I asked myself: given a number 20AB (that is, a number between 2000 and 2099, where A and B represent two figures between 0 and 9), will its square always comply with this regularity: (2 x 20) (4 x AB) (AB) ^{2} ?

What I am going to do next is not a demonstration. It is a logical reasoning that will allow me to deduce whether or not it is possible for said regularity to be met and why.

I’m going to start by polynomially decomposing a number like 20AB:

20AB = 2 x 10^{3} + 0 x 10^{2} + A x 10^{1} + B x 10^{0}

Thus, the square of 2022 is equivalent to multiplying:

(2 x 10^{3} + 0 x 10^{2} + A x 10^{1} + B x 10^{0 }) x (2 x 10^{3} + 0 x 10^{2} + A x 10^{1} + B x 10^{0 })

Its result corresponds to multiplying each of the terms in the first parenthesis by each of the terms in the second parenthesis. So that you can visualize it better, I am going to make a list of all these products and their results. Remember that when you multiply two powers with the same base, you keep the base and add the exponents.

(2 x 10^{3}) x (2 x 10^{3}) = 4 x 10^{6}

(2 x 10^{3}) x (0 x 10^{2}) = 0

(2 x 10^{3}) x (A x 10^{1}) = 2A x 10^{4}

(2 x 10^{3}) x (B x 10^{0}) = 2B x 10^{3}

(0 x 10^{2}) x (2 x 10^{3}) = 0

(0 x 10^{2}) x (0 x 10^{2}) = 0

(0 x 10^{2}) x (A x 10^{1}) = 0

(0 x 10^{2}) x (B x 10^{0}) = 0

(A x 10^{1}) x (2 x 10^{3}) = 2A x 10^{4}

(A x 10^{1}) x (0 x 10^{2}) = 0

(A x 10^{1}) x (A x 10^{1}) = A^{2} x 10^{2}

(A x 10^{1}) x (B x 10^{0}) = AB x 10^{1}

(B x 10^{0}) x (2 x 10^{3}) = 2B x 10^{3}

(B x 10^{0}) x (0 x 10^{2}) = 0

(B x 10^{0}) x (A x 10^{1}) = AB x 10^{1}

(B x 10^{0}) x (B x 10^{0}) = B^{2} x 10^{0}

Adding all the results we obtain:

(20AB)^{2} = 4 x 10^{6 }+ 2 x (2A x 10^{4}) + 2 x (2B x 10^{3}) + A^{2} x 10^{2} + 2 x (AB x 10^{1}) + B^{2} x 10^{0} =

= 4 x 10^{6 }+ 4A x 10^{4} + 4B x 10^{3} + A^{2} x 10^{2} + 2AB x 10^{1} + B^{2} x 10^{0}

Which corresponds to the polynomial expression of the number:

4 0 4A 4B A ^{2} 2AB B ^{2}

Notice that:

- the tens of thousands digits and the thousands units (i.e., the 3rd and 4th digits from the left) correspond respectively to 4 x A and 4 x B, the last two digits of the number 20AB multiplied by 4
- The figures of hundreds, tens and units correspond to the square of AB, since:
- The polynomial expression of the number AB is A x 10
^{1}+ B x 10^{0}= A x 10 + B - according to the notable identity (a + b)
^{2}= a^{2}+ 2ab + b^{2}, we obtain that:

- The polynomial expression of the number AB is A x 10

(A x 10 + B)^{2} = (A x 10)^{2} + 2 x (A x 10) x B + B^{2} = A^{2} x 10^{2} + 2AB x 10 + B,

which is the polynomial expression of the number A ^{2} 2AB B ^{2}

So, is the result of (20AB) ^{2} a 7-digit number in which the first two digits are a 4 and a 0, the next three are the result of multiplying 4 x AB and the last three are (AB) ^{2} ?

I’m going to try number 2,013…

2.0132 = 40(4 x 13)(13)^{2} = 4.052.169

Eureka! It seems that everything fits. And, in a way, it is so.

But… what happens when the result of 4 x AB has more than two figures? Or when (AB) ^{2} has more than three figures?

I’m going to leave it here, because this is enough for another article 😅 Anyway, what’s really interesting is the entire analysis process, the logical reasoning and the deductions made so far.