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If the integer n is greater than or equal to 2, can 4/n be written as the sum of three positive fractions?
In other words, is there a solution to 4/n=1/x+1/y+1/z?
Given that you did not qualify this example (i.e n>=2 for all natural numbers etc)
Let n=4 and x=y=z=3
4/n=1/x+1/y+1/z
4/4=1/3+1/3+1/3
1=1
This is a **famous unsolved problem in number theory**.
### Short answer
* **It is conjectured that YES**, for **every integer ( n \ge 2 )** there exist **positive integers** ( x,y,z ) such that
[
\frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}.
]
* This statement is called the **Erdős–Straus Conjecture** (1948).
* **It has not been proven in full generality**, but it has been **verified for all ( n ) up to extremely large bounds** (well beyond (10^{17})) and proven for many infinite classes of ( n ).—
## What *is* known for sure
### ✅ Verified computationally
The equation has been checked and **works for every ( n \ge 2 )** up to enormous limits by computer search.
### ✅ Proven for many infinite families
For example, solutions are known when:
* ( n \equiv 0,1,2 \pmod{4} )
* ( n ) has certain prime factorizations
* ( n ) is even
* ( n ) is divisible by many primesOnly certain rare forms of odd numbers require deeper checking—but all tested cases work.
—
## Example solutions
### Example 1: ( n = 2 )
[
\frac{4}{2} = 2 = \frac{1}{1} + \frac{1}{1} + \frac{1}{\infty}
]
(Trivial case; usually excluded.)### Example 2: ( n = 3 )
[
\frac{4}{3} = \frac{1}{1} + \frac{1}{4} + \frac{1}{12}
]### Example 3: ( n = 5 )
[
\frac{4}{5} = \frac{1}{2} + \frac{1}{4} + \frac{1}{20}
]### Example 4: General construction (one of many)
If ( n \equiv 2 \pmod{4} ), then:
[
\frac{4}{n}
= \frac{1}{\frac{n}{2}}* \frac{1}{n}
* \frac{1}{n}
]—
## Why this is hard
* Writing a rational number as a sum of **unit fractions** (Egyptian fractions) is highly non-unique.
* The difficulty is proving **existence for all integers**, not finding examples.
* No counterexample is known.—
## Final conclusion
* ✔️ **Yes, it is believed that such a decomposition always exists**
* ❌ **But it has not been proven for all ( n \ge 2 )**
* 🧠 This is a major open conjecture in mathematicsYes there are a few solutions. For n=2, x=1, y=2, z=2; For n=3, x=1, y=6, z=6; For n=4, x=2, y=3, z=6, For n=6, x=3, y=6, z=6. And so on …
Do they have to be different fractions?
In other words can x=y?
n = 2, x = 1, y = 2, z = 2
2 = 2
maybe you missed something in the riddle?This proposal (Erdos-Straus Conjecture) has been tested and confirmed by supercomputers up to 10^17, but it has never been PROVEN in the absolute mathematical sense.
Why can’t n=2 x=1 y=1 and z=.5 Am I not getting the riddle?
The original poster demonstrate what happens when you “know” the problem but can not quote it in full.
In fact, it is proven that it is enough to make one false statement, say, 1=2, then you can prove any false statement or something like that!
Sheker, sheker tirchok.
yeah, almost certainly yes, but no one has actually proved it in full.
the question is whether for every integer n greater than or equal to 2, four over n can be written as a sum of three positive unit fractions, meaning fractions of the form one over an integer.
this exact question is a famous open problem in number theory called the erdos straus conjecture (1948). its been around forever and somehow still open.
current status: no counterexample has ever been found. the statement is known to work for tons of cases, including all even n and many infinite families of odd n. on top of that, computers have brute-forced it for absolutely massive ranges of n (like up to 10^18), and it *always* works.
but despite all that evidence, there is still no general proof that it works for every single n. so mathematically speaking: everyone is basically convinced the answer is yes, but strictly speaking its still unproven. classic math moment tbh.
The King’s Poisoned Wine
The Strategy:
Number the Bottles: Label the bottles from 0 to 999. Convert each decimal number to its 10-digit binary equivalent (e.g., 0 = 0000000000, 3 = 0000000011, 999 = 1111100111).
Assign Prisoners: Assign each of the 10 prisoners a single binary digit position (Prisoner 1 for the first digit, Prisoner 2 for the second, and so on, up to Prisoner 10 for the tenth).
Create Mixtures: For each bottle, create a mixture: if a bottle’s binary number has a ‘1’ in a specific position, add a drop of that bottle’s wine to the cup for the corresponding prisoner.
Example: Bottle #3 (0000000011) would contribute a drop to Prisoner 9 and Prisoner 10’s cups. Bottle #1 (0000000001) only contributes to Prisoner 10’s cup.
Administer Simultaneously: Give each prisoner their unique mixture to drink at the same time.Observe & Decode: After 24 hours, the prisoners who die reveal the poisoned bottle. The pattern of dead (1) and alive (0) prisoners forms a binary number. Convert this binary result back to decimal to find the poisoned bottle.
Why It Works: There are \(2^{10}=1024\) possible unique outcomes (combinations of dead/alive prisoners) for 10 prisoners, which is enough to uniquely identify one of the 1000 bottles (since 1024 > 1000). Each bottle’s unique 10-digit binary code directly maps to a distinct set of prisoners who will drink from it, creating a unique “fingerprint” for that bottle.
Question:
How is this a Mathematical Limerick?
( (12 + 144 + 20 + 3 √4) / 7 ) + 5 x 11 = 9^2 + 0
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Answer:
A dozen, a gross, and a score,
plus three times the square root of four,
divided by seven,
plus five times eleven,
is nine squared and not a bit more.Joseph-
You’ve known me here for a looooong time but I no longer go for the same bait that I used to. That ship has long sailed.
You seemed to have figured it out (or was that a coincidence?).
Anyway- I chomped on your bait and visited to point out that your comment was already posted (back in October of 2010).
Lol anything my brain does right these days is sadly purely coincidental 😅😭
for a change, a Torah question: which arithmetic operation is meduaraita?
It looks like all the geniuses are snowed in and having electrical outages (and their batteries burnt out). So we’ll throw out an “easy” freebie:
Find the sum of all distinct real values of x that satisfy the following equation:
.
(x^2 - 5x + 5)^(x^2 - 11x + 30) = 1.
(x² – 5x + 5)⁽ˣ²⁻¹¹ˣ⁺³⁰⁾ = 1
.
(Sorry; I posted the same equation above three times to insure at least one of them posts correctly to this forum’s software.)
21, and gemini solved that too!
ˣ²⁻¹¹ˣ⁺³⁰ = 0
or
x² – 5x + 5 = 1
or
x² – 5x + 5 = -1 and x^2 – 11x + 30 is evento answer my own question: subtraction.
Avraham asks Hashem what if there will be 5 lacking to 50, and Hashem tells him that the number will be 45. Avraham gets the idea and starts counting correctly after that.
spy goes from the center keeping 180 degrees from the sniper as long as possible and when their angular speeds equal, then goes straight to the shore.
it works out even if he uses value of pi computed in the gemora.of course, this problem is as fake as most of your posts – a sniper who can’t shoot from a distance at a spy that hangs out openly in the middle of the lake.
And according to Zenon – the spy will never reach the shore, of course.
great problem that can be solved with N-dimensional Pythagorean theorem, but the message goes beyond junior high-school: high-dimensional spaces are very sparse. That is why brute-force training of AI systems in high-dimensional spaces requires huge amount of data to fill in all those empty spaces
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