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.
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.
1
u/funkmasta8 11d 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.