Logic Puzzle - Age equals sum of digits

[@MadisonM1996]

Based on personal experience, i turned 25 in 2021 and I was born in 1996. 1+9+9+6=25. So I'm fairly certain the answer is 1996

[@user-jc2lz6jb2e]

AN* answer

[@fadhil4008]

Lmao, this is kinda funny. Sometimes you don't have to be smart to figure out a problem, you just have to experience it

[@couch5717]

I thought 2014 to make 7

[@bharathram7245]

​@@couch5717wow 👏👏

[@jongyon7192p]

@couch5717 banned from KZbin cuz tos 13 year restriction /s

[@morrispearl9981]

I figured that the maximum sum of digits can be 28 (1+9+9+9), and from that I can check the 28 cases in less time than the duration of this video.

[@tonyhasler985]

A nice and quick idea. Mind you if you don't have to stop and explain the approach, you can follow Presh's method and solve it in 2 minutes anyway!

[@maverick627uk]

Exactly the same method I used 👍

[@beauthestdane]

Took me about 30 seconds to do it in my head.

[@dawnscript1]

Same

[@j.r.1210]

Yes, but you don't even have to check all 28 cases. Just start with the max number for each of the two centuries (1999 and 2019), and then move backward until you get a match. E.g., 1999: 28/22; 1998: 27/23; 1997: 26/24; 1996: 25/25. That's only four cases. Then do the same thing starting with 2019. Only five more cases.

[@peterkwan1448]

There are two possible solutions: Consider the first case when the year is between 1900 and 2000 (exclusive), in this case, let the year is 19xy. The sum of the digits is 1 + 9 + x + y = 10 + x + y The age is 2021 - 19xy = 2021 - 1900 - 10x - y = 121 - 10x - y. Since the sum of the digits is equal to the age: 121 - 10x - y = 10 + x + y => 111 - 2y = 11x By searching for y, it gets y=6 such that 111 - 2y is divisible by 11. In this case, the value of x is 9. Therefore, the age is 10+9+6=25, and the year is 1996. Consider the first case when the year is between 2000 and 2021 (exclusive), in this case, let the year is 20xy. The sum of the digits is 2 + 0 + x + y = 2 + x + y The age is 2021 - 20xy = 2021 - 2000 - 10x - y = 21 - 10x - y Since the sum of the digits is equal to the age: 21 - 10x - y = 2 + x + y => 19 - 2y = 11x By searching for y, it gets y=4 such that 19 - 2y is divisible by 11. In this case, the value of x is 1. Therefore, the age is 1 + 4 + 2 = 7, and the year is 2014.

[@Embracing_Impermanance]

Divisibility merhod is really good

[@05arjunagarwal59]

did exactly the same thing. quite simple imo

[@macphearsome]

I did the same thing except I completely overlooked that some people were born after 1999 😅

[@doublej1076]

@@macphearsomeIt's OK, no nine-year-old (in 2023) is going to go around giving people word problems. A 25-year-old might. So we can eliminate one potential answer and say that the person had to be born in '96 :)

[@user-ol7dl3mk3k]

Another possible solution is to see that both year and age have the same remainder from division by 9 (7), and then see the upper limit of age as the maximum possible sum of digits in a year - 1+9+9+9 which left us with only three possible values for age: 7, 16 and 25 years old, which can be easily checked (and option 16 therefore discarded).

[@RGP_Maths]

That's what I did. Sum of digits immediately indicates using mod 9 arithmetic. Year + age = 2021, where year (of birth) and age are congruent mod 9. So 2x ≡ 5 and x ≡ 7(mod 9). Working back, 2014 works, 2005 doesn't, 1996 does, and it's evident that going further will increase the age but won't increase the digital sum. So there are no more solutions.

[@yurenchu]

@@RGP_Maths Well, it will increase the sum of digits if you go more than 4000 years back (i.e. further back than 1999 _BCE_ ), but by that time the sum of digits cannot catch up with the age, because the age increases at a faster rate. For any positive integer n, the number of digits of n equals floor(1+log₁₀(n)) ≤ 1+log₁₀(n) with the = case only in the event that n is a power of 10 , so we know {sum of digits of n} < 9*(1+log₁₀(n)) . But for years BCE, the age will be (n+2021), which is strictly greater than 9*(1+log₁₀(n)) and hence also strictly greater than the sum of digits (this can be easily proven by comparing the rate of growth of (x+2021) and 9*(1+log₁₀(x)) with each other).

[@vitriolicAmaranth]

Wow, this is even more elegant than my solution, which I already felt was much more elegant than the one in the video. Lots of ways to solve this problem.

[@OfficialScottR]

Intuitively thought of 2014 immediately, but wasn't sure how to make a full method. Interesting video!

[@Мопс_001]

Hah, I intuitively thought of something about 1990-s ages. Quickly figured it out by just calculating 1999 and lower it to 1996 to have a match

[@gafjr]

Not quite intuitively, but very close. I thought there was only one answer, so I didn't go any further.

[@TheEulerID]

You have to question the existence of a 7 year old capable of presenting a problem in such an erudite manner (and the question is posed in the present tense, so it really would be a 7 year old).

[@yurenchu]

@@TheEulerID What about a 7-year old AI? Or maybe the question is posed in the year 2113 by a not-so-erudite 99 year old who doesn't know the difference between present tense and past tense.

[@robertlewis5439]

I used Excel - nesting numbervalue(), left(), and right() functions to extract and sum the individual year digits. Not as elegant as Mr. Talwalkar's solution, but got the job done.

[@luketurner314]

Matt Parker would be proud

[@JohnSmith-gq3xy]

I did this too. I noticed that the “less than 100 years old” is irrelevant since the maximum sum of the positive years less than 2021 is 28. Going further back gets you further from the answer. It was also good to see which target years have 2 solutions (like 2021), 1 solution (like 2020) or none (like 2007)

[@blasterTC]

I did some nested monstrosity in excel too. Put 2021 as bday, listed years from 2020 downward in column B, put age (2021-whatever row B is) in column C, and then this: "=VALUE(MID(TEXT(B2;"####");1;1)+MID(TEXT(B2;"####");2;1)+MID(TEXT(B2;"####");3;1)+MID(TEXT(B2;"####");4;1))" i'm positive there must be a better way to sum digits in a number than this but yeah...

[@stevenz933]

@@blasterTC You are correct, here is a much simpler formula for adding up the digits of a number. Assuming the number is in cell A1, then the formula is =SUM(MID(TEXT(A1,"0"),{1,2,3,4},1)*1). This basically converts each digit to "text" then sums it all up!

[@blasterTC]

@@stevenz933 Thank You very much. I had NO idea whatsoever that you can use array in sum like that... It's like a whole new world now. (I also had trouble what to put in TEXT function as format text)

[@ParadoxDev_]

I picked a year in the late 20th century and calculated the age and the sum. My first guess was 1990, whose sum is 19 and the age would be 31. As the year increases, the age decreases but the sum increases, so i then checked 1991 which got me 20 as the age and 30 as the sum, noticed that 5 years further would get a 25 and a 25, so the answer would be 1996.

[@kyleward3914]

I did something similar. I figured 1999 would give the highest sum, thus the highest possible age. Substracting 28 from 2021, I realized the person couldn't have been born before 1993. I started figuring from there but neglected to realize there were two solutions.

[@The14Some1]

The calculation of the first case might be greatly simplified if we consider, that the maximum possible age is in fact 1 + 9 + 9 + 9 = 28 (there are no combinations greater than that), thus the minimum considered year is 1993, therefore, a is 9, and b in [3..9].

[@brianvernaglia9449]

That is a great bit of insight that I missed totally.

[@buffuniballer]

I did something similar. I also determined the max age would be 28, but 28 years prior to 2021 would be 1993. The digits for 1993 summed to 22. The difference was six years. As the year increased, the age also decreased, meaning I would have to go three years later than 1993, or 1996. I checked and 1996 satisfied the conditions.

[@Sonicfan138]

Exactly what I did. I like to do the least amount of work possible, and this got me there much faster than running through all the formulas.

[@Wildcard71]

@@buffuniballer arithmetical average: (28+22)/2 respectively (1999+1993)/2 simplified: 1990+(9+3)/2

[@rbowler88]

This is an incredibly complicated way to get to the answer. Also slow and time consuming. The simple way is take 1999. 2021-1999=22. 1+9+9+9=28. (28-22)/2=3. 1999-3=1996. Simple and quick. No need to pretend to be so clever. Not everything needs to be solved by algebra. A coder will just use a binary split algorithm to find the number.

[@tonyhasler985]

A bonus question A draft version of the video omitted the words "On my birthday." This leads to one more possible answer. Left as an exercise for the reader!

[@idk_6385]

Is this some joke or something? You cant have integers add into rationals so whats the catch here

[@wyattstevens8574]

So this turns into, "If my age is less than 100 and is equal to the digit sum of my birth year, what could that year be?" Tony seems like he's saying that this is less constrained than the original: 1996 and 2014 are answers, sure, but there's another answer!

[@The14Some1]

I guess is for the case of age 9½ ?

[@carultch]

@@idk_6385 Your age is usually considered an integer, even if the current date is half-way between your consecutive birthdays. If you hadn't had your birthday yet in 2021, then your (2021 - your age) is one less than your birthyear. By specifying that the narrator has had a birthday in 2021, we know that we don't need to consider this problem for the case of a date in 2021 prior to the narrator's birthday. As a loophole for this to still be an applicable condition, one possibility is if the current time is earlier than the time of your birth, and you don't consider your age to advance until the time of your birth, then this possibility could still apply. Even though legally, your age starts at midnight in the current time zone.

[@tonyhasler985]

If you were born late in 2009 you would still be 11 early in 2021!

[@VidaV]

I'm typing this up before watching the video. EDIT: looks like we used the same method :) I expressed the problem as the equation 2021-a||b||c||d= a+b+c+d (eq 1) a= 1000w, b= 100y, c= 10x, d= z, so we have the following: 2021 - 1000w - 100y - 10x - z = w+x+y+z Since their age is less than 100, w= 1 or 2. Case 1: If w= 1 then x= 9 Case 2: If w= 2 then y=0 and x= 0 or 1 Solving eq 1 for case 1 yields 2021 - 1*1000 - 9*100 - 9*10 + z = 1+9+9+z. z = 6, age = 25 (born 1996). For Case 2, x cannot equal 0 because solving the original equation for x= 0 yields 19-11(0)-2z=0. z=9.5 which is not possible, so x = 1 for case 2. Solving eq 1 for case 2 yields 2021 - 2*1000 - 0*100 - 1*10 + z = 2+0+1+z. z = 4, age = 7 (born 2014). 1996 and 2014 are the correct answers.

[@Dan-vt9vk]

Faster method from 1:57 - Since b=93/11>8, so a=9 is the only possibility. You can use the same method to show that there are no solutions if you were born in the 1800s (or earlier), so the statement "I am less than 100 years old" is redundant. Thanks for the problem, enjoyed this one!

[@alexsinclair5358]

Interesting approach, I took the cheap option and just used Python to brute force it since there was only 100 options lol for age in range(0,101): year = str(2021-age) if age == int(year[0])+int(year[1])+int(year[2])+int(year[3]): print(age, year)

[@mrityunjaykarmankar9239]

How did you learn this

[@gearrotation]

i also tried using python despite knowing baely anything about it lol i used the year number instead of age for the range, and tried to individually assign each number of the year with a letter it looks wonky and lengthy but i got the correct answer in the end: import math years=[] ages=[] for i in range(1921,2022): w=math.floor(i/1000) x=math.floor(i/100)-10*w y=math.floor(i/10)-(100*w+10*x) z=i-(1000*w+100*x+10*y) age=w+x+y+z if w+x+y+z==2021-i: years.append(i) ages.append(age) print(years) print(ages)

[@Lovuschka]

Since the maximum age is 28 with those parameters (in 1999), it is very easy to brute force this by mind. You didn't need to check any year below 1990 really.

[@mgancarzjr]

You can get the least significant digit of an integer (the ones' place) by getting the remainder from a division by 10. sum = 0 while (number > 0) { sum += number % 10 number = number / 10 } This isn't python code, but you should be able to understand it.

[@alexsinclair5358]

@@mgancarzjr that doesn't work actually cause doing num/10 would either turn the variable into a float or cause an error if num isn't a multiple of 10 (eg 2021/10 = 202.1) This also causes the sum to be incorrect if you let num be a float since modulo gives the remainder specifically and not just the 1s column (eg 202.1 % 10 = 2.1) Either way tho my solution is fine anyways since Python still got it in under a second and that's a slow language lol

[@emtheslav2295]

Since you’re less than 100 years old, that limits the range of years to 1922 or after. Then, with some trial and error, I stumbled upon 1996.

[@koibubbles3302]

I used a different approach that was friendlier for head maths since I didn’t feel like taking out a pen and paper. I thought about the problem for a little bit and realized that the maximum sum of digits for the 2010s would be 12, because 2019 is the biggest. 2019 is also 2 years away from 2021. Because each year decreases the sum value by one, that meant I could take the average of the two numbers to find a middle value that had a sum value which was the same as its distance from 2021. 12 + 2 is 14, 14/2 is 7, 2021 - 7 is 2014, 2 + 0 + 1 + 4 is 7. I thought there might be multiple answers but i was too lazy to look for a second one. The 1996 answer can be proved with the same method, however. The max sum value of 1990s would be 28, and 1999 is 22 years away from 2021. 28 + 22 = 50, 50/2 = 25, 2021-25 = 1996, 1 + 9 + 9 + 6 = 25.

[@OneWeirdDude]

Not bad! My only gripe is that we have no need to know that he is less than 100, because the number with the largest sum-of-digits is 1999, with a sum of 28. Therefore, he is less already than 30.

[@The14Some1]

true

[@scienceaddict2589]

I think you forgot negative birth years whose sum-of-digits can be as big as you want. But it's an easy case knowing that to increase the number of digits in the negatives you have to subtract to the year number at least 9 so it won't add any more solutions (knowing that -1 obviously doesn't work).

[@RGP_Maths]

​@@scienceaddict2589Which calendar system uses negative year numbers?

[@scienceaddict2589]

I don't know (if I search on Google there is the Astronomical year numbering for example). But it's not really what I wanted to point out. In fact no matter how you call these years, -9999, 9999 BC, or I don't know what else, you can have years whose sum-of-digits is greater than 28 and still be in the past of 2021 and so keeping a logical sense.

[@yurenchu]

@@scienceaddict2589 One reason why "negative" birth years (or birth years before Year 0 / Year 1) can't work is because the age grows at a faster rate than the sum-of-digits, so the sum-of-digits can never catch up with the age's 2021 headstart anyway. (This is basically the reason you already gave.) Another reason why it can't work, at least not if the current year is 2021, is because of the following: We're given that age = {sum of birth year digits} , and since {sum of birth year digits} ≡ |birth year| (mod 9) , we can derive age ≡ |birth year| (mod 9) But for "negative" birth years, we would also have age = |birth year| + 2021 . For example, someone who's born in the year "-5" would be 2026 years old in 2021. So substituting age into the previous congruence equation yields |birth year| + 2021 ≡ |birth year| (mod 9) which cannot work since 2021 is not a multiple of 9 . The interesting part is that for the same reason, it also cannot work if the current year was "-2021" (or 2021 BC , or any year BC unless that year is divisible by 9).

[@BangkokBubonaglia]

Where this video and I diverged in our solution is at 2:17. I used x,y instead of a,b, but they are exactly the same. I went to 111 - 11x = 2y, then, since y can't be greater than 9, I turned it into the inequality 111 - 11x

[@chevand8]

I was able to get both correct answers, but through a completely different route. I did start with the realization that someone celebrating a birthday in 2021 would either be born in the 20th or 21st Century, so a year starting with 19 (the sum of which is 10) or 20 (the sum of which is 2). I quickly realized that the year with the highest possible sum in that range, 1999, would yield a sum of only 28-- and since someone born in 1990 would be 31 in 2021, it immediately ruled out anything before the 1990s. 2020 could also be ruled out, since the sum of the digits was 4, which is greater than than 1. That meant the person in question was born in the 1990s, 2000s, or 2010s. Next, I calculated the ranges of both the sums of the digits and the possible ages for each decade, based on 2021 being the given year in the question, in order to see if there were any overlaps in the two sets of numbers that could potentially indicate an answer. For the 1990s, the sum range was 19-28, and the age range was 31-22 (going from 1990 to 1999); there was an overlap here. For the 2000s, the sum range was 2-11, and the age range was 21-12; no overlap at all, so the entire decade could be discarded as a possibility. For the 2010s, the sum range was 3-12, and the age range was 11-2; the ranges overlapped here too, so a second date could be possible. So now I knew I had one possible answer in the 1990s, and one in the 2010s. From there, it was just a simple matter of figuring out how the two ranges fell in each decade, and where they lined up in the middle. By listing the two values for each year of the 1990s and 2010s, I was able to find both matches pretty rapidly-- 1996 and 2014. I double-checked each to make sure the math worked out, and had my answers. Not a particularly elegant or efficient way to arrive at the solution, but it worked. :p

[@TheHolySC]

I found the highest possible number to add up to which is 28 (1999). Someone born in 1999 would be 22, meaning that I was off by 6. Each year I move back, the person gets 1 year older and the year gets 1 number smaller, meaning I only need to go back 3 years.

[@dujas2]

Pretty easy with modular arithmetic. A number is equivalent to the sum of its digits modulo 9. So if the birth year and age are equivalent to x mod 9, 2021 is equivalent to 2x. 2021 is equivalent to the sum of its digits, 5. If 2x is equivalent to 5 modulo 9, x is equivalent to 7. So we try ages of 7, 16, and 25. 16 doesn't work as an age, but the other 2 check out.

[@benjamincruz6633]

I did this in my head. My first guess was 2005, and the sum was way too low and I recognized that all of the 2000s would be too low. Then I went up to the 2010s, where I found the 2015 had a sum that was too large, and saw that 2014 was correct. Then I recognized that all of the other 2010s wouldn't work since the sum and the age went in the opposite direction. Lastly, I checked the 90s, and 1995 had a too small sum, so I found that 1996 was right. I saw that no other 90s would work for the same reason as 2010s, and no others would work since the90s have the highest sums

[@rohangeorge712]

exactly the answers just pop up in ur head, i miss the difficult problems. i liked his tricky algebra or geometry problems

[@timetraveller6643]

I used a short progression of columns. Sum on the left diminishing, age on the right advancing. They converged very quickly.

[@yogisaputro3410]

Practical, just like most of real world problems. Math teachers dislike that, though.

[@timetraveller6643]

@@yogisaputro3410 - It reminded me of Gauss' solution to the sum of the numbers 1 to 100. (stack the columns)

[@maraann330]

Somehow he made the solution prozess more complicated than it had to be. 2021 - abcd = x a+b+c+d = x x < 100 case 1 a= 1 , b= 9 assume c=9 ,d=9 => a+b+c+d = 28 , 22 = 2021 - abcd (difference 6, half 3) assume c=9, d=6 => a+b+c+d = 25 , 25 = 2021 - abcd => 1996 case 2 a= 2 , b = 0 assume c=1, d=9 => a+b+c+d = 12 , 2 = 2021 - abcd (difference 10, half 5) assume c=1, d=4 => a+b+c+d = 7 , 7 = 2021 - abcd => 2014

[@rohangeorge712]

i miss your tricky algebra or geometry problems, can we get some more of that please?

[@saraflint2982]

This person was born in 1996 and was 25. A 7-year-old couldn't tell you this. ;D

[@kennethsizer6217]

🤣

[@halwa9528]

Maybe American 7 year olds, Indian, Korean and Chinese 7 year olds out here doing simple algebra and surds bro they smart as hell.

[@saraflint2982]

@@halwa9528 The "average" 7-year-old can't do this. I sure couldn't have done this when I was 7 years old! Of course, there are exceptions.

[@halwa9528]

@@saraflint2982 tbh it was semi tongue in cheek, but I'm pretty sure those kids low-key could do it lmao

[@saraflint2982]

@@halwa9528 Maybe some. But most likely not a 7-year-old of average intelligence. Like I said, I couldn't do it, and I'm 4 times that age!

[@faaiz6672]

for i in range(1921,2021): age=0 for a in str(i): age+=int(a) if i+age==2021: print(i)

[@platosbeard3476]

Split into 2 cases (x is the 10s, y the unit) Case 1: 2000-2021 => 2y+11x-19=0 By inpection, integer solutions only exist when x is odd, hence 2014 is the only answer Case 2: 1922-2021 => 2y+11x-111=0 By inspection, x needs to be odd and greater than 8 (111-88=23 >10 when divided by 2), hence 1996 is the only solution in the range

[@simonkampfhenkel2117]

I really like your videos - they really encourage me to do some math problems just for the fun of math! Here is my way to the solution: Narrowing down the possible birth year to - between 1990 and 1999 - sums between 19 and 28 - ages between 31 and 40 - first guess was 1996 - between 2010 and 2019 - sums between 3 and 12 - ages between 2 and 11 - first guessed 2015 and arrived with the second at 2014 I should proof that there are no more possible solution, but I leave that as "trivial"!

[@mikefochtman7164]

Always interesting problems when you have to break the number apart by digits (e.g. 10*a+1*b). You immediately know that a and b are in the range 0-9 which also helps. Nice solution.

[@coryshepherd3495]

That has to be the hardest way of solving the easiest problem in the history of this channel.

[@blacktigershearthstoneadve6905]

Solved in Excel in 2 minutes. First coulm - years, second column - sum of digits, third column - age difference. You only need 28 rows, since 28 it the max possible sum (in 1999). All values can be obtained by copypasting simple formulas, although sum of digits is probably easier to fill in manually for most people. Then just look for matching values in the 2nd and 3rd columns.

[@StarTheTripleDevil]

So basically what I did: Since 1+9+9+9=28, that's the largest possible sum. 2021-28=1993 so only years after that can be answers. There can be max. one year per decade since increasing the last digit decreases the age (and vice versa). 2021 is odd, which means the year of birth and age must have opposite parities, so the first 3 digits of the year of birth must add up to an odd number. This leaves only the 1990s and 2010s as options. And checking years from those decades gives us 1+9+9+6=25 and 2+0+1+4=7.

[@piman9280]

Me too - a much quicker way, which requires no pen and paper. However, it's still nice to see the algebraic approach.

[@vacuumcarexpo]

The numbers of the solutions are fewer than I expected.

[@carultch]

I did it by setting up a system of equations, with one equation missing. Wolfram Alpha can then solve for expressions that find integer solutions, when you have one degree of freedom. Then we search for the first one that without a negative, and with single digits. Usually the positive range starts on n=0, where n is the integer parameter to choose a solution. Define M, C, D, and Y, to be the millennia, centuries, decades, and years respectively. Your age A, is therefore: A = 2021 - (1000*M + 100*C + 10*D + Y) as well as: A = M + C + D + Y For a birthyear in the 20th century, C=9 and M = 1. The integer solutions for D and Y are: A = 9*n + 7 D = 13 - 2*n Y = 11*n - 16 n = 2 gives us the first case of all positive solutions: A = 25, D = 9, Y = 6 Year = 1996 For a birthyear in the 21th century, C=0 and M = 2. The integer solutions for D and Y are: A = 9*n + 7 D = 1- 2*n Y = 11*n + 4 n = 0 gives us the first case of all positive solutions: A = 7, D = 1, Y = 4 Year = 2014

[@tigerinthejungle_14]

Here's how I did it in very few seconds :- The maximum sum of digits in a birthyear would be in the year 1999 i.e 28, then as we go back sum of the digits will reduce to 25 and the age will increase to 25 from 22(the age if you were born in 1999) i.e birthyear would be 1996. And minimum sum is 2 at the year 2000 where you will be 21 and then max sum in 2000s is 2009 i.e 11 but your would be 12 so then check in 2010's, you will get 2014 as another answer

[@ClarinoI]

Hey Presh, much as I'm impressed with the mathematical method of working this out, I got to the answers quicker by just picking a year and adding the digits together and seeing how far off I was and then working my way to the answers by doing the same for the years closer to the correct answers.

[@irhamma1944]

Using (mod 9) is easier I think. Assume the age as x, and birthyear is abcd, 0

[@tannercarawan6756]

Highest possible sum of birth year digits is 28 corresponding to 1999. If born in 1999 the person would be 22 years old. For each of the first 9 years earlier the person is born the sum of digits of their birth year decreases by 1 while their age increase by one. Thus a number of years before 1999 giving a solution is k < 10 such that 22 + k = 28 - k. K = 3 so 1996. If the person was born on 1989 or earlier the sum of digits of their birth year would be 27 or fewer while their age would be 32 or greater. The 2000s case is a little different

[@tannercarawan6756]

Can check endpoints of each decade. 2000 - 2010. Age is 11 - 21 while sum of digits is at most 11. 11 doesn’t work. 2010 - 2020 is done like the message this is a comment to

[@asphics7108]

I used a different approach for the solution: Let us take the example of the 1900s. The max possible digit sum of the years in 1900s will be for 1999 and it will be 28 (1+9+9+9). So the person can't be more than 28 years old. But if the person was born in 1999, his age would have been 22. With this knowledge, we can do the following iteration: 1. Subtract one from the digit sum of the years (Starting from 28 for 1999, represents the birth year shifting back one year) 2. Add one to the age (Since the brith year got shifted by a year) 3. Repeat the process till the digit sum and Age are equal. This iteration needs to be done 3 times to get both of the numbers to be equal (22+3 = 25 = 28-3). This means that the Birth year which satisfies this criteria is 1999-3 = 1996 and the person's age is 25. Similarly for the 2000s, the max digit sum would be 12 (for 2019) and the age if the person was born in 2019 would be 2 years old. Using the same method as above, with 5 iterations we get 12-5=7=2+5. This means for the 2000s, the birth date would be 2019-5 = 2014. Note: Although this method works, but if you're trying to code it as a program then keep in mind that the max digit sum needs to be refreshed for each decade (i.e. after 10 iterations).

[@julienteychene6022]

Did that too, once you have age

[@yurenchu]

What if the year were 2020 (instead of 2021)?

[@beefjerky2372]

I'm just trying to have fun with Excel Top row headers for columns A through E, Potential Birthyear, 2021 year, Sum of digits, Age, Sum = Age Data starting from row 2 Column A: All years from 1922-2021 Column B: 2021 for the entire column Column C: =LEFT(A2,1)+MID(A2,2,1)+MID(A2,3,1)+RIGHT(A2,1) Column D: =B2-A2 Column E: =IF(C2=D2,"Yes","No")

[@TxHornyToad]

If you assume the year is 2000 or later, then the possible sums of digits is 2 to 12. The sum goes up from 2000 and the age goes down as the year goes up. By inspection you get 2014. The sum of 1999 is 28. The age in 1999 is 22. As year decreases the sum goes down and age goes up so by inspection year is 1996.

[@Kavyatej]

some mental calculations and odd even tricks got me there

[@JohnJones-pu4gi]

Solved much the same way. SInce there are only 99 possible solutions this is clearly solvable by brute force, as in the following V program object y, aSum:integer for y {1922..2020} [ digitizer(y,10) → sum → aSum assert aSum = 2021 - y fprint("",{y}) ]

[@crowbar_the_rogue]

Neat problem. 1996 and 2014 are both suitable solutions.

[@amoghverma3108]

Good teaser for CAT Exam 😊

[@hvnterblack]

Only one hundred possibilities? Bruteforce method!

[@Sven_vh]

I'm a coder so I wrote a quick little piece of code for it and found the same answers. You can run it too by pressing ctrl+shift+i and pasting the code. for(let i = 1921; i < 2021; i++){ //Convert the year to an array, for example 1975 => [1,9,7,5] let numbers = numberToArray(i); //Calculate the sum by looping through the array let age = numbers.reduce((a, b) => a + b, 0); // check if its correct if(age == 2021-i){ console.log(i); } } function numberToArray(number) { let array = number.toString().split("");//stringify the number, then make each digit an item in an array return array.map(x => parseInt(x));//convert all the items back into numbers }

[@Codraroll]

I considered the upper bounds of the answer. The birth year with the highest possible sum of its digits is 1999, which means the person could at most be 28 years old. That means the birth year is 1993 or later. The sum of the digits in 1993 is 22. Likewise, if the person was born in the 1990's, their age in 2021 must at least be 22 years old. For each possible birth year after 1993, the sum of the digits increases by 1, while the age decreases by 1. That made it a simple matter to guess. Not very elegant, but at least I narrowed the possibility space quite a bit before brute-forcing the answer. Likewise, a person born in the 2000's would be at least 12 years old in 2021, but the highest possible sum of digits in one of the possible birth years (2009) is 11, so no solution exists. A person born in the 2010's would have the sum of their birth year somewhere between 3 (2010) and 12 (2019), but can't be older than 11 or younger than 2, which eliminates a lot of the possibilities early. Brute-forcing takes you to the correct answer quickly enough here too.

[@thomaschonps]

Just counted the digits of each year starting from 2021 down to the answer: 2+0+2+1 = 5 =/= 0 2+0+2+0 = 3 =/= 1 2+0+1+9 = 12 =/= 2 2+0+1+8 = 11 =/= 3 ... Using this "going blind" method, we can already see right here that we'll get an answer before getting to 2010: the numbers on each side having the same parity, as we're substracting 1 to the left side and adding 1 to the right side at each step, we are sure they will coincide at some point. That's how we can find 2014. We can similarly find 1996 by having the intuition of starting at 1999 (being the year that gives the biggest age) and doing the same method.

[@EricScheid]

If born in the 1900s, the max age is 1+9+9+9 = 28. If born in the 2000s, the max age is 2+0+1+9 = 12. This greatly simplifies the logic.

[@tristantheoofer2]

1996. took me a couple minutes to figure out in my head but i got it. oh amd 2014 would work too. 2 + 0 + 1 + 4 is 7. 2021 - 2014 is 7

[@RyanTradesFutures]

As years decrease before 2000, the sum decreases while age increases. Need to observe each decade until age exceeds max of sum. After 2000, 2014 at 8yrs is possible. Prior to 2000, 1996 is possible. By 1989, age 32 already exceeds the sum of 27. Any years before this will be less than age.

[@mattt961]

Another way to solve is treating it as an optimization problem. Start your search at a maximum, looking at the sum of the year digits, 1999. Born in 1999 you get an age of 22 and a sum of 28, a difference of 6. Since you need to go up in age and down in sum, changing the delta by 2 for each year you subtract from 99, you can see that you need to move 6/3 years which puts you at 1996. Before 1996 the sum is less than the age so no use searching there. Since the sum of the digits has a discontinuity in it every decade, you need to search in 10 years blocks. The 2020 block is trivial. The 2000 block results in a false start because the delta between age and sum is always off and you can only adjust it by 2. But the 2010 block starts with you being 11 and a sum of 3, a difference of 8. So you need to reduce the age by 4 and increase the year by 4 to get 2014. Just a comp-sci approach using search and understanding that the sum of the digits function has a repeating discontinuity every 10 years, so you need to check every 10 years span for a solution.

[@HoSza1]

answer = list(year for year in range(1921, 2021) if 2021-year == sum(int(digit) for digit in str(year))) [1996, 2014]

[@doq]

I got the cheeky third answer of 2009, but it rides on a condition that the person's birthday had not yet happened, which is not actually a solution, but an interesting observation.

[@MasterCivilEngineering]

Master the concepts of engineering here

[@renderthepi]

I used the true programmer solution: Spend an hour writing a program that does something I could have done logically in 5-10 minutes

[@tabletoparcade4203]

Why does it include the clause "I am less than 100 years old"? Aside from using BC, it isn't possible to get an answer past 28yo (1999; 1+9+9+9). Or am I missing something?

[@carultch]

I think it is just to narrow down the search, so people aren't perpetually searching unrealistic possibilities. But even if you do consider a historical figure born in a BCE year, the maximum possible 4-digit year's digits will add up to 36, which is obviously not a BCE birth year. Adding more digits to the person's birth year, will just exacerbate the difference between their age and the sum of the digits in their birth year.

[@tabletoparcade4203]

@@carultch Of course! I hadn't even considered the math with BC, I just assumed it would work. Now I'm really confused as to why that clause was added.

[@i.setyawan]

The way you said "... born in the 1900s..." makes me feel sooooo old, nay, it made me feel ancient...

[@fryingpan94]

I got 1996 for 25 really fast, because I knew they had to be younger than 1921, but also mathematically 1999 is the largest sum possible. 28. A person born in 1999 would be 22 in 2021. Took the difference and divided by two. 28-22 = 6/2 = 3. Then took 3 off 1999 for 1996.

[@fryingpan94]

Additionally, this method works for 2014 as well. The largest sum possible between 2000 and 2021 is year 2019. Which is 12. The person would be 2. 12-2 = 10/2 = 5. Then subtract 5 from 2019. 2014.

[@yurenchu]

What if the current year had been 2005 (instead of 2021)? Or six years from now, in the year 2029? How would this method work then?

[@fryingpan94]

@@yurenchu You know what. I just tried it, and it didn't work. There may just be another rule in my method that I just HAPPEN to skip by naturally with this year combination. The method as I explained does not work for those years. If I were to guess, I think it would have something to do with the decades 'resetting' the SUM progression as you move along the 100 years.

[@yurenchu]

​@@fryingpan94 It also doesn't work for the year 2020. And only half for the year 2017. Solutions: for year 2005 : birth year = 1979 , age = 26 for year 2029 : birth year = 2018 , age = 11 for year 2020 : birth year = 2009 , age = 11 for year 2017 : birth year = 1994 , age = 23 (your method will find this solution) OR birth year = 2012 , age = 5 (not found by your method).

[@fryingpan94]

@@yurenchu yeah, there has to be a missing rule and I just got lucky with the original year. Obviously there is some sort of pattern being made, and I'm sure if I followed up more into it, I could solve that missing piece. But I really don't care too much. Like I said though, I think the missing piece is in the 'reset' of the sum for every new decade.

[@MSJ_7]

Great puzzle Presh. Asnwer is 2014 and can be obtained via setting the conditions to equations 11x + 2y = 19 for 2000s and 11x+2y=111 for 1922 to 1999. Solving for the pair in the first equation gave x=1 and y=4 thus giving 2014 as the answer while the for the second case. it came out to be x=9 and y=6 giving 1996. These were the ones I found on top of my head. Maybe there are more pairs if one is to dig deeper.

[@earthbind83]

Nice! I just went through the years manually until the age became too big for the digits. But why did you have to specify the age to be less than 100? I mean, you can't get the year beyond 100 anyway.

[@orionspur]

Min and max possible digit sums are 2 (at 2000) and 28 (at 1999). This is also the age range. Solutions pop out at 7 and 25.

[@evreatic3438]

What's the purpose of "I am less than 100 years old" condition? 4 digits can't possibly add to more than 100 anyway...

[@carultch]

Probably just the way the riddle was structured to focus on a realistic possibility. Even if you consider historical figures from a BCE birth year, you still can't add digits to equal their age. The largest 4-digit number has digits that add up to 36, and adding more digits will just exacerbate the gap between their age and the sum of their digits.

[@Laggron93]

Search for the biggest sum of digits of an integer between 1922 and 2021. 19xy : it's 1+9+9+9 = 28 20xy: it's 2+0+1+9 = 12 Then the person is at most 28 years old, which means he's born at least in 1993. Let's write A the birth year. We know that any integer is congruent to the sum of its digits modulo 9. We also know that the sum of the digits of its birth year A is equal to his age. We have: 2021 - A = A modulo 9 2A = 5 modulo 9 Try and search, we get A = 7 modulo 9. At this point, we can limit A to 1996, 2005 and 2014. All we have to do is calculate the sum of the digits and the age for each birth year. 1996: 1+9+9+6 = 25 and 2021 - 1996 = 25 2005: 2+0+0+5 = 7 and 2021 - 2005 = 16 2014: 2+0+1+4 = 7 and 2021 - 2014 = 7 We eliminate 2005, both 1996 and 2014 work, they're our two solutions.

[@dvilardi]

Got lazy and found 1996 and 2014 with python lol def sum_digits(n): sum = 0 for d in str(n): sum = sum + int(d) return sum for i in range(1921,2021): age = 2021 - i sd = sum_digits(i) if age == sd: print(str(i) + " - " + str(age))

[@stuknox7468]

Here is a case where it is quicker to just add the sum of digits of each year and compare the sum to the age from that year, all can be quite easily done just using mental arithmetic.

[@alextunneyware]

Definitely one of the easiest ones on the channel, both answers can easily be found through basic trial and error worst comes to worst

[@cybersekkin]

I did it in python: year = 2021 max = 100 temp = 0 while temp < max: cur_year = str(year - temp) total = int(cur_year[0]) + int(cur_year[1]) + int(cur_year[2]) + int(cur_year[3]) if total == temp: print(f"the year {cur_year} looks good at {total}") #print(f"tested {cur_year} and got {total}") temp = temp + 1 With the output: the year 2014 looks good at 7 the year 1996 looks good at 25

[@shha22]

There is no need to use 2 unknowns and try to figure out the age. 1) If the birth is in 1900s the highest sum is 1+9+9+9=28 - which means the birth year has to actually be in the last decade. So we have 2021-1990-x=1+9+9+x ; 31-x=19+x ; 2x=12 ;x=6 DOB=1996 2) if the birth is in 2000s, the highest sum would be 2+0+1+9 = 12 which doesnt add up, so actually has to be in 2nd decade. So we have 2021-2010-x=2+1+x; 11-x=3+x; 8=2x; x=4 DOB =2014

[@friday6426]

From 1000 - 1999 AD, which year all the digits sums up to number 29? Is it possible?

[@user-hu9rd7jk8v]

Really cool solution

[@icystrangers5482]

< 100 years old is redundant information if your age is the sum of the digits of your birth year.

[@BlazingLarvesta]

1996 is my guess before watching the video. After watching, I was surprised 2014 would be a valid solution, as well. Didn’t expect this to be true if someone was only 7.

[@neuralwarp]

Those Shorty Awards are looking a bit dated now. Maybe do a puzzle about that year?

[@lstsnwfll]

0:22 I brute forced it in Excel. Got two answers: 1) 1996 Age: 25 Sum of digits: 25 || 2) 2014 Age: 7 Sum of digits: 7 I should learn how to solve this properly.

[@MrWarlls]

I missed the "2000s" case. I'm too old.

[@Wildcard71]

case distinction if younger than 21 even number results in odd age -> sum is odd odd number results in even age -> sum is even odd age would between 3 and 11 (including limits), so 7 would be a possible solution (2014) odd number and even sum together require two odd digits, so it would be something between 4 and 10 same result and intitial guess falsified if older than 21 the maximum sum is 28 and the minimum age 22 so 25 would match (1996) Summary: both 7 (2014) and 25 (1996) are possible solutions.

[@rohangeorge712]

ngl, the answers legit came to my head in less than 10 seconds lol

[@RJSRdg]

Yes - got this one right!

[@The1RandomFool]

My solution before watching the video is as follows: 2021 - x = y = a + b + c + d, x = 10^3*a + 10^2*b + 10*c + d, where x is the birth year, y is the age, and a through d are the digits of x. Additionally, with the use of modular arithmetic, x ≡ 10^3*a + 10^2*b + 10*c + d (mod 9), x ≡ a+b+c+d (mod 9). Well, a+b+c+d is y, which is also 2021 - x, so, x ≡ 2021 - x (mod 9), 2x ≡ 5 (mod 9). The smallest x (between 0 and 8) which satisfies this congruence is 7, so using the properties of modular arithmetic, x is of the form 7 + 9n where n is an integer. Age must be strictly between 0 and 100, so this places bounds on n, which is strictly between 212 and 224. Using a calculator or computer, this range is easy to search, and we get two solutions: 1996, 2014. After looking through the comments, I saw n can be further restricted by noticing that the max possible sum of digits is 1+9+9+9 = 28.

[@Qermaq]

If the year of birth is 1900+10x+y, then per the conditions 11x + 2y = 111, and x and y are integers between 0 and 9 inclusive. If x is 8 or smaller, then 2y would have to be at least 23 which is more than twice any valid value for y, so x = 9 and y = 6. Birth year is 1996.

[@Qermaq]

I reject the 2014 answer as the person is 7 and would not utter such a precocious statement. :P

[@Mythraen]

This one I solved with a bit of brute force and s bit of logic. As an example, once you solve 2014, you know nothing between 2000 and 2013 is going to work. They obviously can't add to a larger number, and they have to do that to qualify. After determining 1996, just had to check 1989, and then the rest of the century was excluded.

[@tommytwotakes]

Finally one I could figure out

[@gordonspond]

Figured it out... yay!!!

[@mahdi7d1rostami]

Here's a one-liner Python program to solve this problem: answers = [born for born in range(1921, 2021) if sum(map(int, list(str(born)))) == 2021 - born]

[@md2perpe]

Python one-liner: [y for y in range(1900, 2022) if sum(int(d) for d in str(y)) == 2021-y]

[@juandalebingle]

Imagine some random 7 year old just straight up asking you to guess his age using this problem 💀

[@xupermike]

Between 1921-2021 the largest sum from all digits comes from 1999 , so the age must be equal or less than 28. From here, try from 1993(28 years old) forward, year of 1995, 2014 both satisfy the conditions.

[@redpug5042]

My answer is 2014. I just counted down from 2020. 2020 adds to 4 with 1 yo then i went to 2015, which adds to 8 which is 2 off, then i went to 2014 which adds to 7 which is also the difference in years.

[@78rera]

Yeah, we know that we be twinning because of it.. Hadeuuuuhhh..

[@mr.bacteria7148]

You could come up with some equations for certain intervals Between 2019 and 2010 the sum of your (age) and the (sum of the birthyear digits) is constant X + Y = 14 And Between 2009 and 2000 X + Y = 23 (you can’t find the age here since the result is odd) And Between 1999 and 1990 X + Y = 50 Checking other intervals (1989-1980) the result is odd (1979-1970) the age is larger than the sum of the digits meaning they’ll never meet (age increases and the sum decreases) (this same problem actually applies to the odd intervals) and to every other interval of interest beyond this one. You can find where X and Y are equal by using a common variable For [2019,2010] 2x = 14 , x = 7, so your age would be 7. 2021 - 7 = 2014 For [1999,1990] 2x = 50, x =25, so your age would be 25. 2021 - 25 = 1996

[@yurenchu]

Where did you get the formula "X + Y = 14" for years between 2019 and 2000? For example, if the birth year was 2001, then {age} + {sum of birth year digits} = 20 + 3 = 23, which doesn't equal 14 . Likewise, with "X + Y = 50" between 1999 and 1900. For example, if the birth year was 1970, then {age} + {sum of birth year digits} = 51 + 17 = 68 , which doesn't equal 50 . Or if the birth year was 1985 , then {age} + {sum of birth year digits} = 36 + 23 = 59 .

[@mr.bacteria7148]

@@yurenchu I wrote this in a bad situation (probably shouldn’t have). I tried to fix the mistakes. Tell me if anything is out of place.

[@yurenchu]

@@mr.bacteria7148 Looks okay to me now. :-)

[@AutisticAnn]

1996/2014. Take your pick

[@gabrielgauchez9435]

You can never be older than 100 yo 4 digits can only sum to 36

[@aroundandround]

“I am less than 100 years old” is redundant.

[@simoncrabb]

The line over 19ab is for what?

[@canniballissimo]

by excel is 7 or 25 years

[@YTSparty]

My solution was a lot simpler. You have ABCD is DOB. AB = 19 or 20. 10C+D = 121-AGE or 21-AGE And AGE = 10+C+D or 2+C+D. You get a simple formula D=(111-11C)/2 or D=(19-11C)/2. Just plug in C=0 to 9. You see C,D = 9,6 for AB=19 and C,D=14 for AB=20. So ABCD = 1996 or 2014. No other solution yields integers for both C and D.

[@moreon340]

Considering that the maximal sum of 4 digits is 36 (limited to numbers below 2021, its only 28), it seems unnecessary to include the 100 years old constraint.

[@toolebukk]

I immediately thought it would be 199x intuitively and tried 1996 and it worked out

[@mahamed3485]

nice puzzle