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# Official TSR Mathematical Society Watch

1. Yes but you can't say that 33 (mod 10) = 3*11 (mod 10) and therefore 33 mod 10 is divisible by 3 and 11 (but not 9 so it can't be a square, which is what the proof says).
2. (Original post by SsEe)
Yes but you can't say that 33 (mod 10) = 3*11 (mod 10) and therefore 33 mod 10 is divisible by 3 and 11 (but not 9 so it can't be a square, which is what the proof says).
No, 33 (mod 10) is not necessarily divisible by 3 or 11.
43 = 33 (mod 10)...
3. Oh! Yeah, what you're saying makes perfect sense. The mix-up was on my part. Here's another way to look at the question:

For n>4, S always ends in 3 and hence isn't a perfect square.

Proving that if a number ends in 3 it isn't a perfect square basically uses SsEe's method.
4. (Original post by dvs)
I don't get what you're saying. The proof isn't really dividing per se, it's considering the prime decomposition of 33 (mod 10).
just to add, some care is needed when saying prime in mod arithmetic.
for example 3=3.1 and 3=9.7 mod 10.
5. Even better and slicker, mod 5.
6. Can 801,345,230,914 be written as the sum of 3 cubes?
7. Is that a guess or do you have some reasoning to go with that?
8. I have some reasoning. (That I don't want to show in case I'm talking BS again. )
9. Well let's see the reasoning then.
10. Let's consider perfect cubes modulo 7:
a³ = b (mod 7)
0³=0, 1³=1, 2³=1, 3³=-1, 4³=1, 5³=-1, 6³=-1 and 7³=0 (all mod 7)

So perfect cubes leave remainders 0, 1 or -1 when divided by 7. In other words, perfect cubes are of the form:
7k+1, 7k-1 or 7k.

Now consider the sum of 3 cubes of the form 7k-1, it'd be of the form 7p-3=7n+4. So numbers of the form 7n+4 can be written as the sum of 3 perfect cubes. Let's check this for n={1,2,3}:
7*1+4 = 11 = 3³ + (-2)³ + (-2)³
7*2+4 = 18 = 3³ + (-2)³ + (-1)³
7*3+4 = 25 = 3³ + (-1)³ + (-1)³

Finally, let's consider 801,345,230,914 modulo 7:
801,345,230,914 = 4 (mod 7), which means it's of the form 7k+4 and can be written as the sum of 3 cubes.
11. Nice. Sadly you've only proved that a solution exists modulo 7.

I have another question.
a² + b² = 6ab. Find the value of (a+b)/(a-b).
This is from an SMC so shouldn't be too hard.
12. a² + b² = (a+b)² - 2ab = (a-b)² + 2ab = 6ab

So (a+b)/(a-b) = sqrt((a+b)²/(a-b)²) = sqrt(8ab/4ab) = sqrt(2)
13. Well done . I knew the efforts of the TSR mathmos wouldn't fail to solve that really hard question hehe
14. (Original post by SsEe)
Nice. Sadly you've only proved that a solution exists modulo 7.
But isn't that sufficient? I mean, I showed that numbers of the form 7k+4 can be written as the sum of 3 perfect cubes, and then showed that the number in the question is of the form 7k+4.

How'd you do it?
15. You showed that numbers of the form 7k+4 can be written as 3 cubes modulo 7. In fact it can't be written as 3 cubes:
Consider modulo 9:
cubes modulo 9 can only be 0, 1 or 8.
If a solution exists:
a³ + b³ + c³ = 801,345,230,914
+-(0, 1 or 8) +- (0, 1 or 8) +- (0, 1 or 8) = 4 (mod 9)
(+- because 1=-8 and 8=-1)
No combination of these make 4 mod 9 so no solution can exist.
If it was congruent to any of these there would be a solution mod 9 (but not necessarily mod 1) : 0, 1, 2, 3, 6, 7, 8 (I think that's all).

Somebody should check this as I'm having some doubts.
16. (Original post by SsEe)
You showed that numbers of the form 7k+4 can be written as 3 cubes modulo 7. In fact it can't be written as 3 cubes:
Consider modulo 9:
cubes modulo 9 can only be 0, 1 or 8.
If a solution exists:
a³ + b³ + c³ = 801,345,230,914
+-(0, 1 or 8) +- (0, 1 or 8) +- (0, 1 or 8) = 4 (mod 9)
(+- because 1=-8 and 8=-1)
No combination of these make 4 mod 9 so no solution can exist.
If it was congruent to any of these there would be a solution mod 9 (but not necessarily mod 1) : 0, 1, 2, 3, 6, 7, 8 (I think that's all).

Somebody should check this as I'm having some doubts.
Sorry, I cant see how you need to get 4 (mod 9) .
801 = 0 (mod 9), 345 = 3 (mod 9), so you cant say anything about these two... However,
230 and 914 = 5 (mod 9), but since cubes modulo 9 are -1,0,1, no sum of 3 can make 5, so these 2 cannot be written as the sum of 3 cubes.
17. (Original post by dvs)
But isn't that sufficient? I mean, I showed that numbers of the form 7k+4 can be written as the sum of 3 perfect cubes, and then showed that the number in the question is of the form 7k+4.

How'd you do it?
No, it is not a sufficient condition. You have not proved that *every* single number which can be written in the form 7k+4 can be written as the sum of 3 cubes, only that the sum of 3 cubes will always be a number of the form 7k+4.
18. (Original post by JamesF)
No, it is not a sufficient condition. You have not proved that *every* single number which can be written in the form 7k+4 can be written as the sum of 3 cubes, only that the sum of 3 cubes will always be a number of the form 7k+4.
I don't think he's proven that, because it isn't true. e.g. 0^3 + 0^3 + 0^3.

Rather he has shown that mod 7 considerations don't provide reasons why a number of the form 7k+4 couldn't be written as the sum of three cubes.
19. (Original post by RichE)
I don't think he's proven that, because it isn't true. e.g. 0^3 + 0^3 + 0^3.

Rather he has shown that mod 7 considerations don't provide reasons why a number of the form 7k+4 couldn't be written as the sum of three cubes.
Oops yea, I didnt read the actual post lol
20. 345 = 7^3 + 1^3 + 1^3.

9^3 < 801 < 10^3, so a,b,c < 10
One of the cubes must be larger than 801/3 = 267, therefore
10 > a > 6
Then try a = 7,8,9
801 - 9^3 = 72,
801 - 8^3 = 289
801 - 7^3 = 458
Then its the same but showing each of those can or cannot be written as the sum of 2 cubes.
Starting with 72, 5 > b > 3, so b = 4, c = 2
So we have a solution
801 = 9^3 + 4^3 + 2^3

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The content structure of the text passage is a discussion thread from a forum, specifically the "Official TSR Mathematical Society" where users are discussing and solving various mathematical problems.
multiple_questions_with_answers

the passage contains multiple mathematical questions and their corresponding answers, such as:
1. can 801,345,230,914 be written as the sum of 3 cubes?
2. a² + b² = 6ab. find the value of (a+b)/(a-b).

for each question, there are detailed discussions and solutions provided by different users.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
You are Here: Home >< Maths

# Official TSR Mathematical Society Watch

1. Yes but you can't say that 33 (mod 10) = 3*11 (mod 10) and therefore 33 mod 10 is divisible by 3 and 11 (but not 9 so it can't be a square, which is what the proof says).
2. (Original post by SsEe)
Yes but you can't say that 33 (mod 10) = 3*11 (mod 10) and therefore 33 mod 10 is divisible by 3 and 11 (but not 9 so it can't be a square, which is what the proof says).
No, 33 (mod 10) is not necessarily divisible by 3 or 11.
43 = 33 (mod 10)...
3. Oh! Yeah, what you're saying makes perfect sense. The mix-up was on my part. Here's another way to look at the question:

For n>4, S always ends in 3 and hence isn't a perfect square.

Proving that if a number ends in 3 it isn't a perfect square basically uses SsEe's method.
4. (Original post by dvs)
I don't get what you're saying. The proof isn't really dividing per se, it's considering the prime decomposition of 33 (mod 10).
just to add, some care is needed when saying prime in mod arithmetic.
for example 3=3.1 and 3=9.7 mod 10.
5. Even better and slicker, mod 5.
6. Can 801,345,230,914 be written as the sum of 3 cubes?
7. Is that a guess or do you have some reasoning to go with that?
8. I have some reasoning. (That I don't want to show in case I'm talking BS again. )
9. Well let's see the reasoning then.
10. Let's consider perfect cubes modulo 7:
a³ = b (mod 7)
0³=0, 1³=1, 2³=1, 3³=-1, 4³=1, 5³=-1, 6³=-1 and 7³=0 (all mod 7)

So perfect cubes leave remainders 0, 1 or -1 when divided by 7. In other words, perfect cubes are of the form:
7k+1, 7k-1 or 7k.

Now consider the sum of 3 cubes of the form 7k-1, it'd be of the form 7p-3=7n+4. So numbers of the form 7n+4 can be written as the sum of 3 perfect cubes. Let's check this for n={1,2,3}:
7*1+4 = 11 = 3³ + (-2)³ + (-2)³
7*2+4 = 18 = 3³ + (-2)³ + (-1)³
7*3+4 = 25 = 3³ + (-1)³ + (-1)³

Finally, let's consider 801,345,230,914 modulo 7:
801,345,230,914 = 4 (mod 7), which means it's of the form 7k+4 and can be written as the sum of 3 cubes.
11. Nice. Sadly you've only proved that a solution exists modulo 7.

I have another question.
a² + b² = 6ab. Find the value of (a+b)/(a-b).
This is from an SMC so shouldn't be too hard.
12. a² + b² = (a+b)² - 2ab = (a-b)² + 2ab = 6ab

So (a+b)/(a-b) = sqrt((a+b)²/(a-b)²) = sqrt(8ab/4ab) = sqrt(2)
13. Well done . I knew the efforts of the TSR mathmos wouldn't fail to solve that really hard question hehe
14. (Original post by SsEe)
Nice. Sadly you've only proved that a solution exists modulo 7.
But isn't that sufficient? I mean, I showed that numbers of the form 7k+4 can be written as the sum of 3 perfect cubes, and then showed that the number in the question is of the form 7k+4.

How'd you do it?
15. You showed that numbers of the form 7k+4 can be written as 3 cubes modulo 7. In fact it can't be written as 3 cubes:
Consider modulo 9:
cubes modulo 9 can only be 0, 1 or 8.
If a solution exists:
a³ + b³ + c³ = 801,345,230,914
+-(0, 1 or 8) +- (0, 1 or 8) +- (0, 1 or 8) = 4 (mod 9)
(+- because 1=-8 and 8=-1)
No combination of these make 4 mod 9 so no solution can exist.
If it was congruent to any of these there would be a solution mod 9 (but not necessarily mod 1) : 0, 1, 2, 3, 6, 7, 8 (I think that's all).

Somebody should check this as I'm having some doubts.
16. (Original post by SsEe)
You showed that numbers of the form 7k+4 can be written as 3 cubes modulo 7. In fact it can't be written as 3 cubes:
Consider modulo 9:
cubes modulo 9 can only be 0, 1 or 8.
If a solution exists:
a³ + b³ + c³ = 801,345,230,914
+-(0, 1 or 8) +- (0, 1 or 8) +- (0, 1 or 8) = 4 (mod 9)
(+- because 1=-8 and 8=-1)
No combination of these make 4 mod 9 so no solution can exist.
If it was congruent to any of these there would be a solution mod 9 (but not necessarily mod 1) : 0, 1, 2, 3, 6, 7, 8 (I think that's all).

Somebody should check this as I'm having some doubts.
Sorry, I cant see how you need to get 4 (mod 9) .
801 = 0 (mod 9), 345 = 3 (mod 9), so you cant say anything about these two... However,
230 and 914 = 5 (mod 9), but since cubes modulo 9 are -1,0,1, no sum of 3 can make 5, so these 2 cannot be written as the sum of 3 cubes.
17. (Original post by dvs)
But isn't that sufficient? I mean, I showed that numbers of the form 7k+4 can be written as the sum of 3 perfect cubes, and then showed that the number in the question is of the form 7k+4.

How'd you do it?
No, it is not a sufficient condition. You have not proved that *every* single number which can be written in the form 7k+4 can be written as the sum of 3 cubes, only that the sum of 3 cubes will always be a number of the form 7k+4.
18. (Original post by JamesF)
No, it is not a sufficient condition. You have not proved that *every* single number which can be written in the form 7k+4 can be written as the sum of 3 cubes, only that the sum of 3 cubes will always be a number of the form 7k+4.
I don't think he's proven that, because it isn't true. e.g. 0^3 + 0^3 + 0^3.

Rather he has shown that mod 7 considerations don't provide reasons why a number of the form 7k+4 couldn't be written as the sum of three cubes.
19. (Original post by RichE)
I don't think he's proven that, because it isn't true. e.g. 0^3 + 0^3 + 0^3.

Rather he has shown that mod 7 considerations don't provide reasons why a number of the form 7k+4 couldn't be written as the sum of three cubes.
Oops yea, I didnt read the actual post lol
20. 345 = 7^3 + 1^3 + 1^3.

9^3 < 801 < 10^3, so a,b,c < 10
One of the cubes must be larger than 801/3 = 267, therefore
10 > a > 6
Then try a = 7,8,9
801 - 9^3 = 72,
801 - 8^3 = 289
801 - 7^3 = 458
Then its the same but showing each of those can or cannot be written as the sum of 2 cubes.
Starting with 72, 5 > b > 3, so b = 4, c = 2
So we have a solution
801 = 9^3 + 4^3 + 2^3

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Updated: August 3, 2015
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user
The content structure of the text passage is a discussion thread from a forum, specifically the "Official TSR Mathematical Society" where users are discussing and solving various mathematical problems.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

The passage contains multiple mathematical questions and their corresponding answers, such as:
1. Can 801,345,230,914 be written as the sum of 3 cubes?
2. a² + b² = 6ab. Find the value of (a+b)/(a-b).

For each question, there are detailed discussions and solutions provided by different users.