A nine year old wrote this problem. Can you solve it? Viral Math Problem

This math problem went viral after being posted in 2020.
Celeste Ng tweeted 'At dinner, nine year old made up a math problem and made his father and me solve it. Our attempt at simultaneous equations got really complicated. We figured it out via guess and check, but I still think there has to be a more elegant way.'
Can you solve the nine year old math problem using both guess and check and the elegant way - using algebra?

#mathproblems
#mathriddles
#mathsforfun

[j.fitness]

A = 4
E = 7
F = 1
C = 5


Just started with a lucky guess (A = 4) and found other 3 as integers as probably one of the solutions of this matrix.

assume A = 4, and then in this case F must be equal to 1 according to 2nd equation.

Since we also found F, C have to be 5 according to 3rd equation.

and also substituting C and A in 4th equation, we find that E is equal to 7

and 1st equation exist just for justification(linearly dependent?) as we already found A and E which are 4 and 7.

[obsidiandwarf]

Well, I found this laughably easy to solve using my own, tried, trusted method; I skipped forward to about the 9 minute mark and...voila. Problem solved.

[chegumathschannel]

by above equations we get a cubic equation in variable 'a'
that is a*3_22a+24=0
from this we get thrre answers for 'a' those are 4, 2+√10, 2-√10
when 'a' is 4 then e=7;f=1;c=5

[romangonzalezadrianmaurici]

Ok now I feel bad reading the comments because I didn´t use algebra, but I want to share my train of thought to internet strangers. First I assumed there were integers and positive, because a 9 years old couldn´t think about decimals or negative numbers, then in the second equation "a" must be less than five because if "a" is greater than five "f" would be greater than six making "c" necesarily a negative number for the third equation to be true. So "a" needs to be less than five then I just try a=4 and it was all solved.

[brothapipp]

Not sure if this was the right approach, but I completed the square on a*a-f=15
f= a^2-15
f-1 = (a^2 -16)
f-1 = (a-4)(a+4)
f=(a-4)(a+4) +1
=> a = 4, f = 1
=> c = 5 from c+f = 6
=> e = 7 from a+e
c*a-e = 13
5*4-7= 13

[kjaideep]

a= 4, e= 7, f = 1, c= 5
Honestly I've done and I didn't copied any comment's answer after I got ... I am commenting Nice problem btw 🙂

[cerwe]

I tried (1)+(4) and (2)+(3)
Added the results and solved for a with the Quadratic formula.
Then i solved for f in (3) and plugged a and f in (2). Then i got only c as a variable, but was to lazy to solve for it (tried Squaring, trinomial expansion and that stuff). Nice Video :) the solution was easier then i thought.

[nNicok]

Paused at the start and I came up with a=4 f=1 e=7 c=5. I started with second and third equations then did the first and last. Getting A seemed to be the best course of action because it is the most used. Getting A pretty much solves the whole thing.

[clc]

Add all 4 equations together, you will get (a+1)(a+c)=45=5*9
So a+1=5 or 9
Then a=4 or 8 and c=5 or -3. However, all numbers must be positive then a=4 and c=5
Substitute into other functions and get e=7, f=1

[cresttutor]

Now... Here's how I solved it on first look
C can't be equal to zero because then the 4th equation will be negative and it's not, so I tried, let C equals 1 then F will equals 5 but that won't work for equation 2 because there's nothing you will multiply twice that'll give you 10.
So I tried out a = 4
Which will mean f = 1
And c = 5
Going to equation 4...
5 × 4 - e = 13
Will give e = 7
Finally checking out equation 1
4 + 7 = 11
🕺🕺😇

[Mike-]

Ay, finally got something correct for once, cool
Needing Proof; *im gonna stop you right there*

[rexboggs]

I made it as far as the cubic equation but then got stuck. So I used a graphics calc to solve for the zeroes. Looking forward to seeing your algebraic solution to this.

[exz_rblx]

I was a little confused on the way you do it but then i watched this for a while and reload this a lot just to understand... thanks because when i see a problem like this i was dead inside... thanks again to you ill keep on watching you to understand ^^ btw have you done a vid about circle???

[황예성-gv]

I lost my words.... how can everyone be so

[SeventhSaucer]

And how exactly does one factorize a cubic? I did it by trying long division with the factors (a-1), (a-2), (a-3) until finally (a-4) worked. This is essentially guess and check again, unless you let an app do it for you.

[PlayandTalkwithSean]

I love Math subject! Btw, he alnost reaching 1M subs

[JadeDragon]

Was trying to go the long route to solve for at least one letter, gaining what each letter equalled in equations 1-3 and substituting all that into equation 4 so that only 1 letter "f" would then exist, whereas I got: I simplified that down to f=13f-((17-f)√15)+11 before I felt like I was heading down a bad rabbit hole. I didn't really understand your arbitrary method for creating equations 5 & 6 but it's been a while since I done algebra in a more complicated way. That 9 year old probably leaves math riddles as answers to why they can't clean their room mwahahahaha

[pragnanmaharshi]

a, e, f, c are definitely whole numbers because this is created by a 9 year old, to solve this add (1, 4), (2, 3) equations a^2+c=21, a(c+1)=24, a is less than 5 from equation 1, and 4 satisfies this equation

[skiser]

Don’t we still have to guess one answer for factorising?

[pickachu]

A=4
E=7
F=1
C=5





If I’m wrong don’t mind me I’m only 11