Ok now I am even more confused. Can you write down with formulas what you need to prove?
That's why I don't do maths anymore.
For divisibility questions, it can be helpful to work with remainders. Can you show that x^3 mod 6 = x mod 6 for all values of x?
Here's an attempted explanation, using nice math buzz words. I'm sure with more time spent I could make it more graceful. And if I made a typo or a mistake, I apologize in advance. However, I give no warranty or assurance of its perfection, or (since it's late at night and I'm tired) even its correctness! It might provide some help -- but use or refer to it at your own risk. Here goes:
If some INTEGER number is divisible (which means "evenly divides" in this context) by six, is the product of that number times other integers also divisible by 6? (And that could even include the product of that number multiplied TIMES ITSELF, by the way, since we've defined "that number" to be an integer at the start!).
Well, let's see: Let's call "that number" A. Since we are told that A is divisible by 6, that means the following simple equation MUST be true, by the definition of "evenly divides."
A = 6 N, where N is some integer. NOTICE THAT. It's true, by the definition of "evenly divides!" If 6 DIDN'T "divide" A "evenly" THEN A would = (6N + some remainder!) EXAMPLE: 32 is NOT divided evenly by 6. Why?
32 = 6*5 + 2. "N" here is "5." But the extra 2 is a "remainder." Since there's a "remainder," 32 was not evenly divided by 6.
Remember, "divisible by 6" means if you divide by 6, you get no remainder. In case somebody forgot, an "integer" is a number like ... -5, -4, -3, -2, -1, 0, 1, 2, 3, 4 .... (etc.). In other words, the number does not have an attached decimal fraction or other fraction.
Ok, so if our number "A" = 6N in this case -- as it HAS to be if it is evenly divisible by 6 -- then A^2 ( which means A squared, or A to the second power) = A * A = 6N * 6N = 36 * N^2.
Can you divide that by 6? Of course. 36/6 = 6. No remainder. And THAT 6, times the rest of the number (N^2) is going to create another integer. How do we know that?
This is important: the integer family of numbers is "closed under multiplication." If you multiply ANY integer times ANY other integer, you just get another integer, always. That phrase, "closed under multiplication" is an important explanatory phrase. Those are math buzz words. It's usually just taken as an axiom -- something you don't need to prove. It's obviously true. You just state it.
But what if we CUBE our number A. Is that result STILL going to be divisible by 6? Let's see:
A^3 = A * A * A = 6N * 6N * 6N (because A must = 6N since it's "divisible by 6," remember?)
= 6*6*6*N*N*N = 216 * N^3 (we can always rearrange the order of the numbers in a multiplication, because of the "associativity property" of multiplication. We used it in the first example, too, but that example was already too long, so I didn't mention it).
Well , can we divide THAT by 6? Of course. 216/6 = 36. No remainder. SO, we CAN divide A^3, (which equals 216N^3) by 6 to get 36 N^3. Is that clear? If not, I suggest re-reading. Slowly. Carefully. Maybe repeatedly.
But wait, how do we know that 36N^3 is itself a simple "integer" with no fraction or remainder (which it HAS to be, if we are to say that 6 "divided" into the original 216N^3?
Well, AGAIN we just use the fact that the INTEGERS ARE CLOSED UNDER MULTIPLICATION. In other words, if you multiply ANY two integers, you can ONLY get an integer!
And 36 is an integer. And SO is N, remember? Scroll up. We KNOW N has to be an integer, because we defined it that way to begin with! And we could properly DO that, because if we didn't, then A would NOT be "divisible" by 6. But we were told it is!
Do you see where this is going? If you multiply 6 x 6 ANY NUMBER OF TIMES, the resulting product WILL be divisible by 6. It will have a factor of 6 and the OTHER factors are just other 6s. Which are integers. Which will multiply together to get another integer, because of the "closure of integers under multiplication" they MUST do that since 6 is an integer!
So we will end up with 6 x "some integer" -- BUT -- (6 x some integer) / 6 = (6/6) x that integer = (1) x that integer = that integer. Which is, uh, an integer! So we can say 6 DOES "divide evenly" into (6 x some integer)! Because the division MUST produce a simple integer (and therefore no extra fraction or decimal or remainder).
IN CONCLUSION -- haven't we just proven that if some number A is divisible by 6, then A*A is divisible by 6, and A*A*A is divisible by 6, and so forth! And by the same sort of reasoning, if A is divisible by 6, then A times ANY OTHER INTEGER is ALSO divisible by 6!
Let's check some examples. Is 6 x 5 x 18 = 540 divisible by 6? Yes! The quotient is 90!
Is 6 x 223 x 5 = 6,690 divisible by 6? Yes! The quotient is 1,115. No remainders!
OKAY, almost done: If the integers A, B, C are all divisible by 6. Then by the reasoning above, A^3 and B^3 and C^3 are all divisible by 6. And they are all integers!
BUT what if we add those three integers together? Is the SUM divisible by 6? Yep. If you add together ANY three integers that are EACH divisible by 6, then their SUM is divisible by 6, too.
Let's call those three integers X, Y and Z.
Well, if each is divisible by 6, then by DEFINITION of "divides evenly" , X = 6 M (with M being "some integer")
and Y = 6 P (with P being "some integer") and Z = 6 R (with R being "some integer). I used different letters (which were arbitrarily chosen, not special) because since X, Y and Z might be different integers (they don't have to be), then M, P and R might have to be different from each other, too! So we'd better use different letters, just in case!
Okay. Lets add X + Y + Z. That is the same as 6M + 6P + 6R. Which, because of the "distributive property" equals
6 (M + P + R).
Buy guess what? M + P + R added together MUST give an integer, because they are all integers (we said that to start!), and the integer family of numbers IS ALSO "CLOSED UNDER" ADDITION! That's another axiom -- something most people just state and take for granted! Math buzz words! But let's give an example, to show I'm not just trying to pull the wool over anybody's eyes. How about 17 + 5 + 99. Does that equal an integer? Yes. It equals 121. which is an integer.
SO 6 (M + P + R) = 6 x some integer.
And (6 x some integer) / 6 = (6 / 6) x that integer = 1 x that integer = AN INTEGER! Because, remember, the integer family of numbers is "closed under" multiplication, too!
Which assures us that when you divide 6(M+P+R) by 6 you get some nice integer -- and no fraction or remainder! So 6 DOES divide 6(M+P+R) which was the SUM of our integers X, Y and Z, remember?
So haven't we just shown that adding any three integers -- EACH of which is known first to be divisible by 6 -- creates a sum that is ALSO divisible by 6?
Okay, tying it all up -- haven't we just shown that IF A, B and C are each divisible by 6, THEN (A^3 + B^3 + C^3) ALSO equals an integer that is divisible by 6?
I think we have! Others may feel differently and are free to disagree and give me a horsey. I know it was long-winded, but there's a lot of detail to be included, as it tuns out.
Incidentally, I'd use the same sort of process to show that (A^2 + B^2 + C^2) and (A + B + C) ^2 or (A+B+C)^3 are likewise divisible by 6 -- so long as A, B and C are each divisible by 6.
I apologize for the flood of words -- it's just lengthy to explain if one wants to anticipate where confusion might arise, and TRY to prevent that. And I'm not sure of the starting level of any who might be interested in reading it. So I erred on the side of wordiness, I'm afraid. I hope it at least provides some glimmers of hope to somebody.
And clearly anybody with no interest in this much math verbiage is free to say "oh, no THANK you, long-winded noflaps" and ignore the whole thing.
If some INTEGER number is divisible (which means "evenly divides" in this context) by six, is the product of that number times other integers also divisible by 6? (And that could even include the product of that number multiplied TIMES ITSELF, by the way, since we've defined "that number" to be an integer at the start!).
Well, let's see: Let's call "that number" A. Since we are told that A is divisible by 6, that means the following simple equation MUST be true, by the definition of "evenly divides."
A = 6 N, where N is some integer. NOTICE THAT. It's true, by the definition of "evenly divides!" If 6 DIDN'T "divide" A "evenly" THEN A would = (6N + some remainder!) EXAMPLE: 32 is NOT divided evenly by 6. Why?
32 = 6*5 + 2. "N" here is "5." But the extra 2 is a "remainder." Since there's a "remainder," 32 was not evenly divided by 6.
Remember, "divisible by 6" means if you divide by 6, you get no remainder. In case somebody forgot, an "integer" is a number like ... -5, -4, -3, -2, -1, 0, 1, 2, 3, 4 .... (etc.). In other words, the number does not have an attached decimal fraction or other fraction.
Ok, so if our number "A" = 6N in this case -- as it HAS to be if it is evenly divisible by 6 -- then A^2 ( which means A squared, or A to the second power) = A * A = 6N * 6N = 36 * N^2.
Can you divide that by 6? Of course. 36/6 = 6. No remainder. And THAT 6, times the rest of the number (N^2) is going to create another integer. How do we know that?
This is important: the integer family of numbers is "closed under multiplication." If you multiply ANY integer times ANY other integer, you just get another integer, always. That phrase, "closed under multiplication" is an important explanatory phrase. Those are math buzz words. It's usually just taken as an axiom -- something you don't need to prove. It's obviously true. You just state it.
But what if we CUBE our number A. Is that result STILL going to be divisible by 6? Let's see:
A^3 = A * A * A = 6N * 6N * 6N (because A must = 6N since it's "divisible by 6," remember?)
= 6*6*6*N*N*N = 216 * N^3 (we can always rearrange the order of the numbers in a multiplication, because of the "associativity property" of multiplication. We used it in the first example, too, but that example was already too long, so I didn't mention it).
Well , can we divide THAT by 6? Of course. 216/6 = 36. No remainder. SO, we CAN divide A^3, (which equals 216N^3) by 6 to get 36 N^3. Is that clear? If not, I suggest re-reading. Slowly. Carefully. Maybe repeatedly.
But wait, how do we know that 36N^3 is itself a simple "integer" with no fraction or remainder (which it HAS to be, if we are to say that 6 "divided" into the original 216N^3?
Well, AGAIN we just use the fact that the INTEGERS ARE CLOSED UNDER MULTIPLICATION. In other words, if you multiply ANY two integers, you can ONLY get an integer!
And 36 is an integer. And SO is N, remember? Scroll up. We KNOW N has to be an integer, because we defined it that way to begin with! And we could properly DO that, because if we didn't, then A would NOT be "divisible" by 6. But we were told it is!
Do you see where this is going? If you multiply 6 x 6 ANY NUMBER OF TIMES, the resulting product WILL be divisible by 6. It will have a factor of 6 and the OTHER factors are just other 6s. Which are integers. Which will multiply together to get another integer, because of the "closure of integers under multiplication" they MUST do that since 6 is an integer!
So we will end up with 6 x "some integer" -- BUT -- (6 x some integer) / 6 = (6/6) x that integer = (1) x that integer = that integer. Which is, uh, an integer! So we can say 6 DOES "divide evenly" into (6 x some integer)! Because the division MUST produce a simple integer (and therefore no extra fraction or decimal or remainder).
IN CONCLUSION -- haven't we just proven that if some number A is divisible by 6, then A*A is divisible by 6, and A*A*A is divisible by 6, and so forth! And by the same sort of reasoning, if A is divisible by 6, then A times ANY OTHER INTEGER is ALSO divisible by 6!
Let's check some examples. Is 6 x 5 x 18 = 540 divisible by 6? Yes! The quotient is 90!
Is 6 x 223 x 5 = 6,690 divisible by 6? Yes! The quotient is 1,115. No remainders!
OKAY, almost done: If the integers A, B, C are all divisible by 6. Then by the reasoning above, A^3 and B^3 and C^3 are all divisible by 6. And they are all integers!
BUT what if we add those three integers together? Is the SUM divisible by 6? Yep. If you add together ANY three integers that are EACH divisible by 6, then their SUM is divisible by 6, too.
Let's call those three integers X, Y and Z.
Well, if each is divisible by 6, then by DEFINITION of "divides evenly" , X = 6 M (with M being "some integer")
and Y = 6 P (with P being "some integer") and Z = 6 R (with R being "some integer). I used different letters (which were arbitrarily chosen, not special) because since X, Y and Z might be different integers (they don't have to be), then M, P and R might have to be different from each other, too! So we'd better use different letters, just in case!
Okay. Lets add X + Y + Z. That is the same as 6M + 6P + 6R. Which, because of the "distributive property" equals
6 (M + P + R).
Buy guess what? M + P + R added together MUST give an integer, because they are all integers (we said that to start!), and the integer family of numbers IS ALSO "CLOSED UNDER" ADDITION! That's another axiom -- something most people just state and take for granted! Math buzz words! But let's give an example, to show I'm not just trying to pull the wool over anybody's eyes. How about 17 + 5 + 99. Does that equal an integer? Yes. It equals 121. which is an integer.
SO 6 (M + P + R) = 6 x some integer.
And (6 x some integer) / 6 = (6 / 6) x that integer = 1 x that integer = AN INTEGER! Because, remember, the integer family of numbers is "closed under" multiplication, too!
Which assures us that when you divide 6(M+P+R) by 6 you get some nice integer -- and no fraction or remainder! So 6 DOES divide 6(M+P+R) which was the SUM of our integers X, Y and Z, remember?
So haven't we just shown that adding any three integers -- EACH of which is known first to be divisible by 6 -- creates a sum that is ALSO divisible by 6?
Okay, tying it all up -- haven't we just shown that IF A, B and C are each divisible by 6, THEN (A^3 + B^3 + C^3) ALSO equals an integer that is divisible by 6?
I think we have! Others may feel differently and are free to disagree and give me a horsey. I know it was long-winded, but there's a lot of detail to be included, as it tuns out.
Incidentally, I'd use the same sort of process to show that (A^2 + B^2 + C^2) and (A + B + C) ^2 or (A+B+C)^3 are likewise divisible by 6 -- so long as A, B and C are each divisible by 6.
I apologize for the flood of words -- it's just lengthy to explain if one wants to anticipate where confusion might arise, and TRY to prevent that. And I'm not sure of the starting level of any who might be interested in reading it. So I erred on the side of wordiness, I'm afraid. I hope it at least provides some glimmers of hope to somebody.
And clearly anybody with no interest in this much math verbiage is free to say "oh, no THANK you, long-winded noflaps" and ignore the whole thing.
ChatGPT is visibly not the only one that overcomplicates things.
@Noflaps said in #24:
>
this is way better than every explaination I seen so far instead of guessing what it means
good job mate
>
this is way better than every explaination I seen so far instead of guessing what it means
good job mate
@xDoubledragon if you wanted an explanation, you could just have asked for one :)
[PS: the one given by @Noflaps does not solve the problem that was asked to you, see my post below]
Here is mine:
Claim 1: if a is an integer, then a^3-a is even.
Proof: if a is even then so is a^3, and hence a^3-a is also even. On the other hand, if a is odd then so is a^3, but the difference of two odd itegers is even, so a^3-a is again even.
Claim 2: if a is an integer, then a^3-a is divisible by 3.
Proof: note that every integer a can be written a=3q+r, where q is an integer, and r is either -1, 0 or 1. Then a^3=27q^3+3(9q^2r)+3(3qr^2)+r^3.
Note that in each possible case for r, we have r^3=r.
So a^3-a=3(9q^3+(9q^2r)+(3qr^2)-q), which is divisible by 3.
Claim 3: if n is an integer, then n is divisible by 6 if and only if n is divisible by 2 and 3.
Proof: if n is divisible by 6, then there exists an integer m such that n=6m=2*3*m, hence m is divisible by 2 and 3. Conversely, if n is divisible by 2 and 3, there is an integer k such that n=3k and 3k is divisible by 2. But since 2 is prime and 2 does not divide 3, then 2 must divide k. So there is an integer h such that k=2h and so n=3*2*h=6h.
Claim 4: if a is any integer then a^3-a is divisible by 6.
Proof: this follows directly from claims 1, 2, and 3 applied to n=a^3-a
Consequence: let S=a+b+c and T=a^3+b^3+c^3. Then by claim 4, S-T is divisible by 6. So there is an integer m such that S-T=6m. But then S=6m+T, hence if T is divisible by 6, then so is S.
[PS: the one given by @Noflaps does not solve the problem that was asked to you, see my post below]
Here is mine:
Claim 1: if a is an integer, then a^3-a is even.
Proof: if a is even then so is a^3, and hence a^3-a is also even. On the other hand, if a is odd then so is a^3, but the difference of two odd itegers is even, so a^3-a is again even.
Claim 2: if a is an integer, then a^3-a is divisible by 3.
Proof: note that every integer a can be written a=3q+r, where q is an integer, and r is either -1, 0 or 1. Then a^3=27q^3+3(9q^2r)+3(3qr^2)+r^3.
Note that in each possible case for r, we have r^3=r.
So a^3-a=3(9q^3+(9q^2r)+(3qr^2)-q), which is divisible by 3.
Claim 3: if n is an integer, then n is divisible by 6 if and only if n is divisible by 2 and 3.
Proof: if n is divisible by 6, then there exists an integer m such that n=6m=2*3*m, hence m is divisible by 2 and 3. Conversely, if n is divisible by 2 and 3, there is an integer k such that n=3k and 3k is divisible by 2. But since 2 is prime and 2 does not divide 3, then 2 must divide k. So there is an integer h such that k=2h and so n=3*2*h=6h.
Claim 4: if a is any integer then a^3-a is divisible by 6.
Proof: this follows directly from claims 1, 2, and 3 applied to n=a^3-a
Consequence: let S=a+b+c and T=a^3+b^3+c^3. Then by claim 4, S-T is divisible by 6. So there is an integer m such that S-T=6m. But then S=6m+T, hence if T is divisible by 6, then so is S.
@Noflaps said in #24:
> numbers in a multiplication, because of the "associativity property" of multiplication.
And more importantly, because of the commutativity of multiplication.
@Noflaps said in #24:
> If the integers A, B, C are all divisible by 6.
Well, the problem DOES NOT ASSUME that A, B, C are divisible by 6.
Example: 1+4+7=12 is divisible by 6, 1^3+4^4+7^3=408=6*68 is divisible by 6, but 1,4,7 are NOT divisible by 6.
> numbers in a multiplication, because of the "associativity property" of multiplication.
And more importantly, because of the commutativity of multiplication.
@Noflaps said in #24:
> If the integers A, B, C are all divisible by 6.
Well, the problem DOES NOT ASSUME that A, B, C are divisible by 6.
Example: 1+4+7=12 is divisible by 6, 1^3+4^4+7^3=408=6*68 is divisible by 6, but 1,4,7 are NOT divisible by 6.
@thence -- what I said about associativity was fine. Of course, if you'd prefer, we can arrive at associativity, "through the back door" by making a sequence of the uses of commutativity and then using transitivity.
For example: A*B*C*D = A*D*B*C is an example of associativity. One and done.
One could, of course, also arrive at the same result by REPEATED uses of commutativity and then transitivity.
For example: A*B*C*D = A*B*D*C by commutativity. Then A*B*D*C = A*D*B*C by a second use of commutativity.
THEN Since A*B*C*D = A*B*D*C AND A*D*B*C = A*B*D*C, we can say, via transitivity, that A*B*C*D must equal A*D*B*C.
QED
And so far as I understood the problem posed (and it is certainly the problem I attempted to solve) it was:
If A, B and C are all divisible by 6, THEN
show that (A^3 + B^3 + C^3) is likewise divisible by 6.
I also attempted to solve some slight variations of that problem, at the end (since they are solvable using essentially the same mechanisms).
Like "show (A^2 + B^2 + C+2) is likewise divisible by 6.
Or like "show (A+B+C)^3 is likewise divisible by 6.
If we do NOT assume (are not "given") that A, B and C are divisible by 6 to begin with then I don't believe we can solve ANY of those problems -- absent being told that ONE of them is divisible by 6 AND given some additional relationships between that one and the others.
AS TO WHETHER I SOLVED THE "CORRECT" problem, I will scroll up again and check the original problem (as I did last night, too).
The problem wasn't initially stated with great clarity -- I think because the dragon was attempting to translate from Polish, which added a layer of linguistic complexity -- so I had to try to figure out what was being asked, based on what sort of problems math teachers GENERALLY ask in this area. (I even stated some possible variants, which could be solved using the same principles, since the original statement wasn't, as I said, crystal clear).
EDIT: okay, I went back again -- here is what the OP originally said:
"Prove if sum of any 3 natural number can be divided by 6 when [typo? did he mean "then" not "when"?] sum of those numbers to the power of 3 can be divided by 6" (NOTE: material in brackets added by me, to highlight a likely typo that added a bit of confusion -- since the sentence STARTED with "if" -- making a later "when" too indecisive and ambiguous).
I suppose that could be interpreted to be: If (A+B+C) is evenly divisible by 6 then (A+B+C)^3 can be evenly divisibly by 6.
THAT is a much easier problem than the one I solved, although it can be solved using the same mechanisms I showed in my solution.
So I assumed, to be on the safe side, knowing the sort of stuff math teachers USUALLY ask, that he meant to state the very problem I solved, or one of the other variants I mentioned that can be solved by the same mechanisms.
But, let me quickly solve the "much simpler" problem stated above -- to show it can be solved using those mechanisms.
Let A+B+C = X be divisible by 6. Then X = 6N, with N being some integer. So X^3 = (6N * 6N * 6N) =
(6*6*6*N*N*N) by associativity (or by repeated use of commutativity and then transitivity, if you prefer)
= 6 x (6*6*N*N*N) = 6 x "some integer" since 6*6*N*N*N is multiplying nothing together but integers, and by closure that must result in an integer.
(6 x "some integer") / 6 = ((6/6) x some integer) = 1 x some integer = that integer, which is a, well, duh, an integer ...
so divisibility is proven.
Now, perhaps STILL A DIFFERENT problem might have been intended. This string is long, and it seems that the very statement of the problem was debated, with some back and forth, with all hoping to arrive at one statement. It was not clear to me that one, unambiguous, certain statement of the problem was ever really agreed upon. Perhaps I missed it in the long string.
If I got wrong what the problem was, I apologize to all concerned.
For example: A*B*C*D = A*D*B*C is an example of associativity. One and done.
One could, of course, also arrive at the same result by REPEATED uses of commutativity and then transitivity.
For example: A*B*C*D = A*B*D*C by commutativity. Then A*B*D*C = A*D*B*C by a second use of commutativity.
THEN Since A*B*C*D = A*B*D*C AND A*D*B*C = A*B*D*C, we can say, via transitivity, that A*B*C*D must equal A*D*B*C.
QED
And so far as I understood the problem posed (and it is certainly the problem I attempted to solve) it was:
If A, B and C are all divisible by 6, THEN
show that (A^3 + B^3 + C^3) is likewise divisible by 6.
I also attempted to solve some slight variations of that problem, at the end (since they are solvable using essentially the same mechanisms).
Like "show (A^2 + B^2 + C+2) is likewise divisible by 6.
Or like "show (A+B+C)^3 is likewise divisible by 6.
If we do NOT assume (are not "given") that A, B and C are divisible by 6 to begin with then I don't believe we can solve ANY of those problems -- absent being told that ONE of them is divisible by 6 AND given some additional relationships between that one and the others.
AS TO WHETHER I SOLVED THE "CORRECT" problem, I will scroll up again and check the original problem (as I did last night, too).
The problem wasn't initially stated with great clarity -- I think because the dragon was attempting to translate from Polish, which added a layer of linguistic complexity -- so I had to try to figure out what was being asked, based on what sort of problems math teachers GENERALLY ask in this area. (I even stated some possible variants, which could be solved using the same principles, since the original statement wasn't, as I said, crystal clear).
EDIT: okay, I went back again -- here is what the OP originally said:
"Prove if sum of any 3 natural number can be divided by 6 when [typo? did he mean "then" not "when"?] sum of those numbers to the power of 3 can be divided by 6" (NOTE: material in brackets added by me, to highlight a likely typo that added a bit of confusion -- since the sentence STARTED with "if" -- making a later "when" too indecisive and ambiguous).
I suppose that could be interpreted to be: If (A+B+C) is evenly divisible by 6 then (A+B+C)^3 can be evenly divisibly by 6.
THAT is a much easier problem than the one I solved, although it can be solved using the same mechanisms I showed in my solution.
So I assumed, to be on the safe side, knowing the sort of stuff math teachers USUALLY ask, that he meant to state the very problem I solved, or one of the other variants I mentioned that can be solved by the same mechanisms.
But, let me quickly solve the "much simpler" problem stated above -- to show it can be solved using those mechanisms.
Let A+B+C = X be divisible by 6. Then X = 6N, with N being some integer. So X^3 = (6N * 6N * 6N) =
(6*6*6*N*N*N) by associativity (or by repeated use of commutativity and then transitivity, if you prefer)
= 6 x (6*6*N*N*N) = 6 x "some integer" since 6*6*N*N*N is multiplying nothing together but integers, and by closure that must result in an integer.
(6 x "some integer") / 6 = ((6/6) x some integer) = 1 x some integer = that integer, which is a, well, duh, an integer ...
so divisibility is proven.
Now, perhaps STILL A DIFFERENT problem might have been intended. This string is long, and it seems that the very statement of the problem was debated, with some back and forth, with all hoping to arrive at one statement. It was not clear to me that one, unambiguous, certain statement of the problem was ever really agreed upon. Perhaps I missed it in the long string.
If I got wrong what the problem was, I apologize to all concerned.
I will try, i have some university experience tho
> Prove if sum of any 3 natural number can be divided by 6 when sum of those numbers to the power of 3 can be divided by 6
So we are asked to prove that if:
((n1 + n2 + n3)^3) / 6 = n5
Then
(n1 + n2 + n3) / 6 = n4
Where n1, n2, ..., n5 are natural numbers - i.e. integers greater than zero. Usually a technique employed in math is proof by contradiction. We can use that here. First you rewrite the equation you are given in a format that you can derive some meaningful relation with which to "test" the second equation.
Furthermore, to establish a contradiction, lets say that the implication would not be a natural number but a fraction, written in improper form, with no common denominators (completely reduced). So then we'd have:
(n1 + n2 + n3) / 6 = f4 for some improper fraction f4.
Multiply first eq. by 6^2 and divide by 6^2 (1/1 = 1).
6^2 * ((n1 + n2 + n3) / 6)^3 = n5
Now substitute the second equation we use to suggest a contradiction:
f4 ^ 3 = n5 / 6^2
or
f4 = n5^(1/3) / 36^(1/3)
f4 being a reduced fraction, we can write it as the quotient of two natural numbers, call them n6 and n7.
n6 / n7 = n5^(1/3) / 36^(1/3)
For this equation to be true, it must be the case that n6 = n5^(1/3), and n7 = 36^(1/3).
However there is no natural number n7 that can satisfy the second condition.
Therefore the assumption that (n1 + n2 + n3) / 6 = f4 is wrong, leaving as the only alternative that it must be n4 instead (if its not a fraction, it is an integer - it is a mutually exhaustive set of all the real numbers > 0). Just to check it works, in that case we have:
n4 = n5^(1/3) / 36^(1/3) -> n4^3 = n5 / 36. So n5 is some multiple of 36 whose cubed root is a natural number; for example choose n5 = 6^3 * 36. Then n4 = 6 (or 7^3 and so on).
> Prove if sum of any 3 natural number can be divided by 6 when sum of those numbers to the power of 3 can be divided by 6
So we are asked to prove that if:
((n1 + n2 + n3)^3) / 6 = n5
Then
(n1 + n2 + n3) / 6 = n4
Where n1, n2, ..., n5 are natural numbers - i.e. integers greater than zero. Usually a technique employed in math is proof by contradiction. We can use that here. First you rewrite the equation you are given in a format that you can derive some meaningful relation with which to "test" the second equation.
Furthermore, to establish a contradiction, lets say that the implication would not be a natural number but a fraction, written in improper form, with no common denominators (completely reduced). So then we'd have:
(n1 + n2 + n3) / 6 = f4 for some improper fraction f4.
Multiply first eq. by 6^2 and divide by 6^2 (1/1 = 1).
6^2 * ((n1 + n2 + n3) / 6)^3 = n5
Now substitute the second equation we use to suggest a contradiction:
f4 ^ 3 = n5 / 6^2
or
f4 = n5^(1/3) / 36^(1/3)
f4 being a reduced fraction, we can write it as the quotient of two natural numbers, call them n6 and n7.
n6 / n7 = n5^(1/3) / 36^(1/3)
For this equation to be true, it must be the case that n6 = n5^(1/3), and n7 = 36^(1/3).
However there is no natural number n7 that can satisfy the second condition.
Therefore the assumption that (n1 + n2 + n3) / 6 = f4 is wrong, leaving as the only alternative that it must be n4 instead (if its not a fraction, it is an integer - it is a mutually exhaustive set of all the real numbers > 0). Just to check it works, in that case we have:
n4 = n5^(1/3) / 36^(1/3) -> n4^3 = n5 / 36. So n5 is some multiple of 36 whose cubed root is a natural number; for example choose n5 = 6^3 * 36. Then n4 = 6 (or 7^3 and so on).
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