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 Last Two Digits of Birth Year was true, I decided to start with what I already knew about two similar quantities: Age and Birth Year. We all know that in order to calculate our current age, we do the following calculation:

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 2020, we then have the identity

Age + Birth Year = 2020.

Let’s explore the above identity some more and see if we find anything interesting. Let Age = X₄X₃X₂X₁, Birth Year = Y₄Y₃Y₂Y₁, and Current Year = Z₄Z₃Z₂Z₁, where X₁, X₂, X₃, and X₄ each represent a digit of the person’s Age. Y₁, Y₂, Y₃, Y₄, Z₁, Z₂, Z₃, and Z₄ each represent digits in a similar manner.

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:

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

• X₂ + Y₂ = 12 and X₄ = 0 = X₃
• X₂ + Y₂ = 2 but X₄ = 0 and 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 n = 1. However, that isn’t the only possible case. The sum of Age and Last Two Digits of Birth Year can be any number in the sequence: 20, 120, 220, 320,…. Let’s call these possible numbers “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 Magic Numbers depending on the situation. It’s important to note that all the Magic Numbershave their last two digits together always being 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 1 determines what digit(s) should come before 20 in the digits of the 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 0 or 10.

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

In step 2, X₂+Y₂ must result in either 2 or 12. After some exploration, we can discover that

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

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 (X₁,Y₁), this happens only when Last Two Digits of Birth Year is not in the range [0,20]. These are people with birth years in the ranges [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 [0,20]. These are people with birth years in the ranges [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 Magic Number is. That will give us

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:

• Assumption 1: 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 2: The subject shouldn’t be more than 2020 years old.

The new math trick follows this procedure:

1. Ask the subject if the last two digits(together) of their birth year is in the range [0,20]. If it is go to step 2a, but otherwise go to step 2b. 
   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.
2. (a)Ask the subject what the first two digits of the new Age is. If you take those two digits and follow it up with 20, the result is the Magic Number. So if the first two digits of their Age is AB, then the Magic Number is AB20.
   (b)Ask the person what the first two digits of the new Age is. Add 1 to that number. Take the result and follow it up with 20 and what you get is the Magic Number. So if the first two digits together is CD and AB = CD + 1, then the Magic Number is AB20.
3. Age + Last Two Digits of Birth Year = AB20. So split AB20 into any two summands of your choice and write a new equation like this:
   P — Age + Q = Last Two Digits of Birth Year such that P + Q = AB20
4. 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?”
5. 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 0122. That would mean that she was born in 1898. The Last Two Digits of Birth Year is 98 and that isn’t in the range [0,20] so I move to step 2b. The Magic Number is 220(or 0220) since CD=01 and AB=01.

If I decide to split the Magic Number as 220 = 132 + 88 then P = 132 and Q = 88 so that I can confidently tell the woman “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 2020 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.

But a math problem that is only guaranteed to work on only one day on onlyone year 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:

“This is left as an exercise for the reader”.