# This Math Trick Only Works for Everyone at the End of the Year

So it’s the perfect way to usher in the new year

The year 2020 might be the year in which I’ve received the most math problems from friends and seen the most math tricks shared on social media. It turns out that sharing math tricks with one another is one of the things people do most when they’re under lockdown. That fact is tempting me to wish for longer lockdowns.

Recently, I was sent one such math trick from a friend. He sent me a picture which showed an old man who seemed to be teaching mathematics in front of a classroom. He had his back facing a green chalkboard and his left hand pointing towards it as he said the following words to the class:

*“80 bags of cement minus your age plus 40 will give you the last two digits of your birth year. Am I wrong?”*

If you are anything like one of my friends who is very particular about tiny details, you would scream, “Hey! You can’t subtract ‘age’ from ‘bags of cement’. That’s blasphemy!”

But let’s ignore those intense details for a while. The “bags of cement” was, in my opinion, put there for comedic effect. Once we get rid of that, we can realize that this is a pretty neat math trick.

There will be a section of readers who would try this and it will work, while it will fail for others(it worked for me, by the way). This article is an attempt to explain all the possible reasons why the math works or doesn’t work for any particular person. So with the preambles out of the way, let’s start the real math.

WARNING: The math is going to get fairly complex so readers are going to have to explore the math while reading.

# Stripping the Math Trick naked

First of all, we need to ignore the mention of “bags of cement” since the math wouldn’t work otherwise. In pure mathematical form, the math teacher’s claim goes like this:

*80 — Age + 40 = Last Two Digits of Birth Year*

In this form, it’s easy to see what is going on because this is the same as:

*Age + Last Two Digits of Birth Year = 120 — — — —(Equation 1)*

This means that the math teacher’s so-called “math trick” can be reworded in this less impressive way:

“*Your age plus the last two digits of your birth year will give you 120. Am I wrong?*”

Obviously, this gives away the math that is working under the hood and that is why it was coined in an ambiguous manner. However, once we strip the math trick naked, **we can truly separate the math from the trick**.

But is it true that ** Age + Last Two Digits of Birth Year = 120**?

# Relationship Between Age and Birth year

In order to find out whether that relationship between ** Age **and

**was true, I decided to start with what I already knew about two similar quantities:**

*Last Two Digits of Birth Year***and**

*Age***. We all know that in order to calculate our current age, we do the following calculation:**

*Birth Year**Age = Current Year — Birth Year*

In other words, ** Age + Birth Year = Current Year**. Given that the year I received the message from my friend was

**, we then have the identity**

*2020***Age + Birth Year = 2020.**

Let’s explore the above identity some more and see if we find anything interesting. Let ** Age = X₄X₃X₂X₁**,

**, and**

*Birth Year = Y₄Y₃Y₂Y₁***, where**

*Current Year = Z₄Z₃Z₂Z₁***,**

*X₁***,**

*X₂***, and**

*X₃***each represent a digit of the person’s**

*X₄***.**

*Age***,**

*Y₁***,**

*Y₂***,**

*Y₃***,**

*Y₄***,**

*Z₁***,**

*Z₂***, and**

*Z₃***each represent digits in a similar manner.**

*Z₄*Although most people reading this don’t have four-digit ages, I used this approach because the math still works. However, because of how complex the math is going to get if we crossover from A.D. to B.C., what we’re doing here doesn’t cover people who are older than ** 2020 **years at the time of writing this. So I’m sure almost nobody reading this has been left out.

In that case, I need you to convince yourself that if ** X₄X₃X₂X₁+ Y₄Y₃Y₂Y₁ = 2020**, then the following is also true:

*X₄X₃X₂X₁ + Y₂Y₁ = Z₄Z₃20 — — — — (Equation 2)*

This is one of the keys to proving or disproving ** Equation 1**.

Now, let’s try and find the kinds of digits that satisfy * Equation 2*. To do so, we need to go through

*three steps of addition*:

- First of all, in the
*ones*place,must result in either*X₁ + Y₁*or*0*. If*10*then*X₁ + Y₁ = 10*gets carried over into the**1***tens*place. Otherwise, nothing gets carried over. - Next, in the
*tens*place,must result in either*X₂ + Y₂*or*2*(whether or not there is a carryover of*12*from the**1***ones*place). Ifthen*X₂ + Y₂ =12*gets carried over into the**1***hundreds*place. Otherwise, nothing gets carried over. - Finally, there is no digit in the
*hundreds*or*thousands*place ofso we have*Last Two Digits of Birth Year*if there was no carryover from the**X₄X₃ = Z₄Z₃***hundreds*place orif there was a carryover of**X₄X₃ + 1 = Z₄Z₃***1*from the*hundreds*place.

This analysis therefore implies that ** Equation 1**(the math teacher’s claim) is not always true. In fact, his claim is only true if one of the following two things happens:

and*X₂ + Y₂ = 12***X₄ = 0 = X₃**but*X₂ + Y₂ = 2*and*X₄ = 0**X₃ = 1*

Furthermore, with lots of exploration using the *three steps of addition*, you can discover that

*X₄X₃X₂X₁ + Y₂Y₁ = 20+100n — — — —(Equation 3a)*

for any non-negative integer ** n**. In other words,

*Age + Last Two Digits of Birth Year = 20+100n — — — —(Equation 3b)*

So it is easy to see that the math teacher’s claim(** Equation 1**) is the case when

**. However, that isn’t the only possible case. The sum of**

*n = 1***and**

*Age***can be any number in the sequence:**

*Last Two Digits of Birth Year***,**

*20***,**

*120***,**

*220***,…. Let’s call these possible numbers “**

*320***”.**

*Magic Numbers*However, this begs another question. For any particular person or group of people, is it possible to determine what the corresponding ** Magic Number** is?

# Finding the Magic number

Near the beginning of the article, I already established that the math trick essentially degenerates into

*Age + Last Two Digits of Birth Year = 120 — — — — (Equation 1)*

but we’ve realized that we need to change ** 120 **to other

**depending on the situation. It’s important to note that all the**

*Magic Numbers***have their last two digits together always being**

*Magic Numbers***.**

*20*Going back to the *three steps of addition*, it means that the main thing that matters is if there was a carryover of * 1* from step 2(the

*tens*place) to step 3. Whether or not there is a carryover of

**determines what digit(s) should come before**

*1***in the digits of the**

*20***.**

*Magic Number*So what age groups or birth year groups will cause a carryover of ** 1** from step 2 to step 3?

First, we need to consider that there might be carryovers from step 1. In step one, ** X₁ + Y₁** must result in either

**or**

*0***.**

*10*- If
then the only possible*X₁ + Y₁ = 0*pair is*(X₁,Y₁)*(no carryover to the next step).*(0,0)* - If
then the set of all possible*X₁ + Y₁ = 10*pairs is*(X₁,Y₁)*(carryover of*{(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)}*to the next step).*1*

In step 2, ** X₂+Y₂** must result in either

**or**

*2***. After some exploration, we can discover that**

*12*- If
then the set of all possible*X₂ + Y₂ = 2*pairs is*(X₂,Y₂)*if there was a carryover from the previous step or*{(0,1),(1,0)}*otherwise.*{(0,2),(1,1),(2,0)}* - If
then the set of all possible*X₂ + Y₂ = 12*pairs is*(X₂,Y₂)*if there was a carryover from the previous step or*{(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2)}*otherwise.*{(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3)}*

This is all the information we need to answer one of our earlier questions: what age groups or birth year groups will cause a carryover of ** 1** from step 2 to step 3?

There is only a carryover from step 2 to step 3 when ** X₂ + Y₂ = 12**. Based on the possible pairs for

**(X₂,Y₂)**and

**, this happens only when**

*(X₁,Y₁)***is not in the range**

*Last Two Digits of Birth Year***. These are people with birth years in the ranges**

*[0,20]***,**

*[1921,1999]***,**

*[1821,1899]***,…,**

*[1721, 1799]***[21,99]**. (Remember that all these years are in A.D., not B.C.)

On the other hand, there is no carryover if ** Last Two Digits of Birth Year** is in the range

**. These are people with birth years in the ranges**

*[0,20]***,**

*[2000,2020]***,**

*[1900,1920]***,…,**

*[1800,1820]***.**

*[0,20]*So if we modify the math trick, we need to first find out whether or not the last two digits(or the entirety) of the person’s birth year falls in a particular range. Then depending on whether or not there is a carryover of ** 1**, we can decide what the

**is. That will give us**

*Magic Number**Age + Last Two Digits of Birth Year = Magic Number*

But we have to change this into a form that looked like the original math trick:

*P — Age + Q = Last Two Digits of Birth Year*

such that *P + Q = Magic Number*

With all that complex math out of the way, let’s modify the math trick.

# The Modified Math trick

Our new math trick is only going to work under the following conditions:

: The subject(in other words, the person whose age you’re using) should use the age they would have attained if their birthday in 2020 had arrived already.*Assumption 1*: The subject shouldn’t be more than*Assumption 2*years old.*2020*

The new math trick follows this procedure:

- Ask the subject if the last two digits(together) of their birth year is in the range
. If it is go to step 2a, but otherwise go to step 2b.*[0,20]*

For the next step, assume that the person’s age has four digits. If it has less than four digits ask the subject to put as many zeros in front of the age as needed to turn it into four digits. The four-digit form of the age is what you will use as the**actual age**. - (a)Ask the subject what the first two digits of the new
is. If you take those two digits and follow it up with*Age*, the result is the*20*. So if the first two digits of their*Magic Number*is*Age*, then the*AB*is*Magic Number*.*AB20*

(b)Ask the person what the first two digits of the new**Age**is. Addto that number. Take the result and follow it up with*1***20**and what you get is the. So if the first two digits together is*Magic Number***CD**and, then the*AB = CD + 1*is*Magic Number*.*AB20* . So split*Age + Last Two Digits of Birth Year = AB20*into any two summands of your choice and write a new equation like this:*AB20*such that*P — Age + Q = Last Two Digits of Birth Year**P + Q = AB20*- Translate this into a beautiful sentence like: “
”**P**bags of cement minus your age plus**Q**will give you the last two digits of your birth year. Am I wrong? - Smile as the subject stares in awe of your intelligence.

As an example, suppose my subject was a ** 122**-year-old woman which implies that her four-digit age is

**. That would mean that she was born in**

*0122***. The**

*1898***is**

*Last Two Digits of Birth Year***and that isn’t in the range**

*98***so I move to step 2b. The**

*[0,20]***is**

*Magic Number***(or**

*220***) since**

*0220***and AB=**

*CD=01***.**

*01*If I decide to split the ** Magic Number** as

**then**

*220 = 132 + 88***and**

*P = 132***so that I can confidently tell the woman “**

*Q = 88**”*

**132**bags of cement minus your age plus**88**will give you the last two digits of your birth year. Am I wrong?# Conclusion

Try the modified math trick with your friends and let me know how it goes. It is certainly not as simple, short and attractive as the original but it works every time.

However, as I’ve stated before, this isn’t guaranteed to work if your friend’s birthday in 2020 hasn’t arrived yet(that’s the reason for ** Assumption 1**) or if your friend has managed to be more than

**years old. Out of those two issues, the former is more common. That is why this math trick is best used at the end of the year 2020 when everyone’s birthday would have passed.**

*2020*But a math problem that is only guaranteed to work on ** only one day** on

**isn’t fun, is it? There are ways to modify this so that it works for everyone at the end of 2021, 2022, or any year at all. It certainly gets more complex but it still remains fun. I’m unlikely to explore that in an article, however. To all readers, I have to say what all good math textbooks like to say:**

*only*one year“*This is left as an exercise for the reader*”.