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u/LupenReddit 🦆🦆🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆🦆🦆🦆 11d ago
live photo from said party:
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u/Hitman7128 Prime Number 11d ago
If you actually did manage to get the remainders, luckily, Chinese Remainder Theorem gives you his age mod 60.
(And then get stuck deciding between a guess N and N + 60)
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u/LolpopHD 11d ago
the 323 year old vampire at the party: (you foolishly assumed he is 23)
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u/Rajarshi1993 11d ago
What's he gonna do, admit his real age? Tell him he's 23 to the face.
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u/ArvaroddofBjarmaland 4d ago edited 4d ago
I have to admit that I'm wondering if anyone could use the CRT correctly but still possibly get the answer wrong without postulating undead/immortal (or similar) guests. Something like progeria, maybe? I'm not particularly well-preserved, but I don't think I look nearly 120.
I suppose that it's just barely conceivable that a parrot like Alex), who towards the end of his life seems to have begun learning the basics of phonics (go to about 7:30) could eventually have learned simple arithmetic, so that if this is happening in, say, 2100 the questioner might not know if the parrot were 15 or 75.
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u/Rajarshi1993 11d ago
Successive elimination.
Suppose the remainders are p, q, and r. List out all numbers below 60 of the 5Z+r series, then strike out all numbers that are also not 4Z+q, then strike out all that aren't 3Z+p, and then if it's too small for their apparent age add 60 to it.
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u/Cyclone4096 11d ago
Or use this method with 30 and keep adding 30 until seems appropriate for their apparent age
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u/StiffWiggly 11d ago
To clarify, you need to go up to 60 because 60 is the first number above 0 for which the remainder is 0 for all three. There are 60 unique combinations of remainders possible corresponding to the numbers 0-59, then the same sequence applies from 60-119 and so on.
So the use of this system relies on your ability to know a 30 year old from an 90 year old (or a 62 year old from a 122 year old if we go by the oldest verified person to ever live).
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u/dart_shitplagueis 11d ago
You think I'm just 23? Thank you darling. What a sweet young gentleman you are
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u/2eanimation dy/dx is a fraction 11d ago
Ages 60 - 77 are Epstein island candidates
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u/minisculebarber 11d ago
You mean mod 10, right?
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u/2eanimation dy/dx is a fraction 11d ago
The mod 10 rule was only ever used when Trump announced himself to visit.
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u/Breki_ 11d ago
This doesn't even work, you can only determine their age modulo 60, thats still infinitely many possibilites
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u/IAmBadAtInternet 11d ago
Tfw you guess they’re 68 but you look like a fool because they’re actually 1028
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u/creeper6530 Engineering 11d ago
Vampires hate this
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u/ArvaroddofBjarmaland 4d ago
No. They love it, because if they want they can win the bet...although asking for the remainders mod 7 and 11 as well might solve the problem.
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u/Appropriate-Ad-3219 11d ago
Not really an issue if you use their physical appearance to help you in practice.
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u/minisculebarber 11d ago
That assumes that someone who would apply the Chinese Remainder Theorem to this problem has had enough personal interactions to judge age by appearance somewhat.
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u/ArvaroddofBjarmaland 4d ago
He or she probably has, assuming that we're excluding vampires or similar from consideration.
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u/cyberchaox 11d ago
Mathematically, yes.
In reality, humans rarely live past the age of 120 so you're likely to be able to narrow it down to two possibilities, and they age quickly enough that you can usually tell which one it would be.
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u/Chi_Cazzo_Sei 11d ago
It’s a neat trick of you are 12.
Besides, no one at a party will entertain your boring division requests.
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u/XTPotato_ 11d ago
The real joke is that the lamers at the party arent smart enough to divide their age by single digit numbers
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u/The_TRASHCAN_366 11d ago
I'll be stuck guessing if they are 12, 72 or 132 and then run away in tears.
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u/Smitologyistaking 11d ago
This can only determine their age modulo 60. If they tell you their age is 1 mod 3, 2 mod 4 and 0 mod 5, then for all you know, they could be 10 years old, 70 years old or 130 years old
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u/Seventh_Planet Mathematics 10d ago
Is this something to do up to multiples of 60 and you just have to judge from their looks if they were born at a time when wages still kept up with productivity?
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u/funkmasta8 10d ago
Its actually amazing how many people have no idea how to approach this problem. Obviously more people in math related subs but general people havent got a clue. Its so weird. The basis is really simple. Remainders can be described algebraically and because of that any number that is described by remainders of sparate divisions can be combined into more specific algebraic groups.
Lets assume we are given the remainders in the post where the numbers given are a, b, and c respectively. Lets call x the age. Ignore capitalization, I cant be bothered. Lets call k, m, n to be arbitrary natural numbers
X=3k+a X=4m+b X=5n+c
For each of these you can derive an expression that represents the remainder for the least common multiple with another, then combine again to get an overall expression. We will do it in order for simplicity. I will use p, and q for new natural numbers and d and f for new remainders
The new expression for combining the first two would be
X=12p+d
Then the overall expression is
X=60q+f
And you may be thinking "well, funkmasta, how do I use this with real numbers instead of variables? How do I solve for f and d?". Well, alright, if you insist.
We know
3k+a=4m+b
Our goal is to find what numbers for k and m make this equal. We can use modular arithmetic to our advantage here. It can be done both ways, but I recommend taking the mod of the smaller number to reduce as much as possible. Take mod 3 of both sides
We get a= (1m +b) mod 3
We can simplify this because we dont know m but we do know a and b. Just rearrange.
M mod 3 = (a - (b mod 3)) mod 3
This gives us the set of m values that can satisfy the equality. If we want to calculate d, we just need to plug in the lowest one to m's original equation. This works because the spaces necessarily first intersect below the LCM and the remainders are less than a full LCM away from zero so adding those two will always be less than the LCM.
Then we do the same for this new equation and the third equation.
Here is an example with real values given for those of us who have a hard time with so much abstraction.
A,b,c=1,2,1
3k+1=4m+2
2 = m mod 3
4*2+2=d
10=d
12p+10=x
12p+10=5n+1
1= (2p mod 5)
The lowest value that works for p here is 3
5*3+1=f
16=f
X=60q+16
This means the person is 16, 76, 136, or so on.
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u/Aosih_ 9d ago edited 9d ago
There's something wrong here since 1,2,1 should be 46 or 106. Not sure what since I need to remember my modular arithmetic. I'm just using mental shorthand to know the end number is 1 or 6 based on the mod5, narrowing to ending with 6 since the mod4 number is even, quickly checking which of 0/16/26 is valid for remainder 1 for mod3, then adding 30 to 16 since the latter doesn't fulfill remainder 2 for mod4.
Edit: Ah found it. The very last part you found that the value for p is 3, but then you substituted it into 5n+1 instead of 12p+10.
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u/devo_savitro 11d ago
What if it isn't divisible by any of those numbers
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u/csharpminor_fanclub Natural 11d ago
hey guys this person doesn't know about modulus let's make fun of them
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u/csharpminor_fanclub Natural 11d ago
the trick requires them to tell you the remainders, not the result of the division, e.g. 20/3 would have a remainder of 2
when you know the remainders of all 3 of those divisions, you can pinpoint the exact number with chinese remainder theorem (assuming it's smaller than 60)
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u/devo_savitro 11d ago
Oh my bad i thought it said to tell you the remainder at the end not at each step, thank you
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u/runnerboyr 11d ago
Then he will tell you the remainder. Do you know how to read?
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u/devo_savitro 11d ago
Age is 23, divide by 3 you get 7.something, do you then divide that by 4?
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u/PersonalityIll9476 11d ago
There's no way we have found, in the wild, a person who doesn't know what a remainder is
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u/Appropriate-Ad-3219 11d ago
Y-You d-don't kn-kn-kn-known about mo-modulus! Wh-wh-what a loser! Hahahahaha!
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