Edit: Nvm u changed it.

oh ok so i was on the right track. I had done a little something using square roots on the side.

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Edit: Nvm u changed it.

oh ok so i was on the right track. I had done a little something using square roots on the side.

oh ok so i was on the right track. I had done a little something using square roots on the side.

101.644 days ago

Feb 28, 2023 - 5:07 PM

If you wanted to factor the x^{2}+i you would rewrite it as x^{2}- (-i), which would factor to (x+sqrt(-1))(x-sqrt(-i)) and those are the 2nd and 3rd solutions

101.642 days ago

Feb 28, 2023 - 5:09 PM

rewrite the plus as subtracting a negative, so you can do the difference of squares factoring

101.641 days ago

Feb 28, 2023 - 5:11 PM

W_Licky said:

x

x

x

(x

(x - √i)(x + √i)(x

(x - √i)(x + √i)(x - i√i)(x + i√i) = 0

Check:

(√i)

(-√i)

(i√i)

(-i√i)

101.576 days ago

Feb 28, 2023 - 6:44 PM

Hm. Why didn't the square root symbols for those two copy over?

Edit: fixed. The root symbols for the two i's just didn't show up for some reason.

Edit: fixed. The root symbols for the two i's just didn't show up for some reason.

101.571 days ago

Feb 28, 2023 - 6:52 PM

I’m also not 100% sure where i * sqrt(i) came from

101.57 days ago

Feb 28, 2023 - 6:53 PM

Oh also, it's y=x^{4}+1

edit: demented i think its bc √-i needs to be simplified to i√i. Like how √-9 is i√9.

edit: demented i think its bc √-i needs to be simplified to i√i. Like how √-9 is i√9.

101.57 days ago

Feb 28, 2023 - 6:53 PM

dementedkermit said:

I’m also not 100% sure where i * sqrt(i) came from

Difference of squares with (x

101.569 days ago

Feb 28, 2023 - 6:55 PM

Ohh that makes sense I didn’t see it that way

101.567 days ago

Feb 28, 2023 - 6:58 PM

It's kinda annoying, because it's a multiplicity-one unique root problem.

101.563 days ago

Feb 28, 2023 - 7:04 PM

Here are the roots plotted on the complex plane

101.554 days ago

Feb 28, 2023 - 7:16 PM

Ok I am trying to solve this problem:

How many rational numbers on the open interval (0,1) have a numerator and denominator that add to 1000 when in simplest form?

My first thought would be however many primes there are between 500 and 999 since that’s how many denominators there would be that would have a numerator less than them that would not simplify, but there are most likely nonprimes between 500 and 999 that are coprime with their compliment.

I honestly have no clue how to approach this now

How many rational numbers on the open interval (0,1) have a numerator and denominator that add to 1000 when in simplest form?

My first thought would be however many primes there are between 500 and 999 since that’s how many denominators there would be that would have a numerator less than them that would not simplify, but there are most likely nonprimes between 500 and 999 that are coprime with their compliment.

I honestly have no clue how to approach this now

99.314 days ago

Mar 3, 2023 - 1:02 AM

For now I'll just leave a hint, but if I've been too cryptic please do let me know and I can write up a full explanation:

For which integers x with 0<x<10 are x and 10-x coprime?

For which integers x with 0<x<12 are x and 12-x coprime?

For which integers x with 0<x<7 are x and 7-x coprime?

For which integers x with 0<x<n are x and n-x coprime?

For which integers x with 0<x<10 are x and 10-x coprime?

For which integers x with 0<x<12 are x and 12-x coprime?

For which integers x with 0<x<7 are x and 7-x coprime?

For which integers x with 0<x<n are x and n-x coprime?

99.21 days ago

Mar 3, 2023 - 3:32 AM

10: 1,3,7,9

12: 1,5,7,11

7: 1,2,3,4,5,6

n: all numbers that are not multiples of factors of n

the prime factorization of 1000 is 2^3 * 5^3, so all factors of 1000 are either even or multiples of 5

500 even numbers + 200 multiples of 5 - 100 overlap = 600 non-viable x values

and since we are looking for pairs of numbers, we need to divide the 400/2, giving an answer of 200

Is that correct?

12: 1,5,7,11

7: 1,2,3,4,5,6

n: all numbers that are not multiples of factors of n

the prime factorization of 1000 is 2^3 * 5^3, so all factors of 1000 are either even or multiples of 5

500 even numbers + 200 multiples of 5 - 100 overlap = 600 non-viable x values

and since we are looking for pairs of numbers, we need to divide the 400/2, giving an answer of 200

Is that correct?

93.2 days ago

Mar 9, 2023 - 3:46 AM

Problems like this, while interesting to think about, are ultimately why I prefer my degree in Statistics over pure mathematics. Multi-Dimensional Probability spaces are much more comforting that guessing and hoping for pattern-recognition.

93.14 days ago

Mar 9, 2023 - 5:12 AM

I'm curious what kind of Multidimensional Probability Space questions you could pose that I wouldn't approach by leaning on similar principles of pattern recognition.

93.132 days ago

Mar 9, 2023 - 5:24 AM

There are numbers 0-9 on a keypad that can be made into 6-digit codes. How many different possible codes are there?

23.63 days ago

May 17, 2023 - 5:26 PM

10^6 = 1,000,000. Am I missing something?

Actually, I guess you did say "numbers," not "integers" or something of the like, so the other answer I can come up with is infinity. Although that's an obviously wrong answer, even if it is technically correct in a stupid sort of way.

Or you also don't specify that you're looking for 6-digit codes when you actually ask the question, which might again imply that the answer is infinity, because codes of any length are acceptable. This feels slightly less wrong than my previous infinity, but is still obviously dumb.

Actually, I guess you did say "numbers," not "integers" or something of the like, so the other answer I can come up with is infinity. Although that's an obviously wrong answer, even if it is technically correct in a stupid sort of way.

Or you also don't specify that you're looking for 6-digit codes when you actually ask the question, which might again imply that the answer is infinity, because codes of any length are acceptable. This feels slightly less wrong than my previous infinity, but is still obviously dumb.

23.61 days ago

May 17, 2023 - 5:56 PM

Nope, it's just 1,000,000. Remember that he's asking a high school math problem, so it's going to be relatively clean by default.

23.558 days ago

May 17, 2023 - 7:10 PM

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Forum > Technical Corner > Math Problem Thread

{
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{
"id": "1271948",
"time": "1677604066",
"html": "Edit: Nvm u changed it.<br /><br />oh ok so i was on the right track. I had done a little something using square roots on the side.",
"user": "wlicky"
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"id": "1271949",
"time": "1677604181",
"html": "If you wanted to factor the x<sup>2</sup>+i you would rewrite it as x<sup>2</sup>- (-i), which would factor to (x+sqrt(-1))(x-sqrt(-i)) and those are the 2nd and 3rd solutions",
"user": "dementedkermit"
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"time": "1677604219",
"html": "What?",
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"time": "1677604298",
"html": "rewrite the plus as subtracting a negative, so you can do the difference of squares factoring",
"user": "dementedkermit"
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{
"id": "1271961",
"time": "1677605355",
"html": "<a href=\"https://ibb.co/5LMvSb3\">This might explain that better</a>",
"user": "dementedkermit"
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"id": "1271984",
"time": "1677609884",
"html": "<div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/wlicky\">W_Licky</a> said:</div><div style=\"border:1px solid #888; padding:20px;\"><br />x<sup>4</sup>+1 = 0</div></div><br />x<sup>4</sup>+1 = 0<br />x<sup>4</sup>-(-1) = 0<br />(x<sup>2</sup> - i)(x<sup>2</sup> + i) = 0<br />(x - \u221ai)(x + \u221ai)(x<sup>2</sup> - (-i)) = 0<br /><span style=\"font-weight:bold;\">(x - \u221ai)(x + \u221ai)(x - i\u221ai)(x + i\u221ai)</span> = 0<br /><br /><span style=\"font-weight:bold;\">Check</span>:<br />(\u221ai)<sup>4</sup> = i<sup>2</sup> = -1 \u221a<br />(-\u221ai)<sup>4</sup> = i<sup>2</sup> = -1 \u221a<br />(i\u221ai)<sup>4</sup> = (-i)<sup>2</sup> = -1 \u221a<br />(-i\u221ai)<sup>4</sup> = (-i)<sup>2</sup> = -1 \u221a",
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{
"id": "1271987",
"time": "1677610097",
"html": "<div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/kylijoy\">KylIjoy</a> said:</div><div style=\"border:1px solid #888; padding:20px;\">i<sup>4</sup> = -1</div></div>what???<br />i^4 = 1",
"user": "dementedkermit"
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"id": "1271990",
"time": "1677610238",
"html": "Ummm i<sup>4</sup> is 1, not -1.",
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"id": "1271991",
"time": "1677610349",
"html": "Hm. Why didn't the square root symbols for those two copy over?<br /><br />Edit: fixed. The root symbols for the two i's just didn't show up for some reason.",
"user": "kylijoy"
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{
"id": "1271992",
"time": "1677610414",
"html": "I\u2019m also not 100% sure where i * sqrt(i) came from",
"user": "dementedkermit"
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{
"id": "1271993",
"time": "1677610431",
"html": "Oh also, it's y=x<sup>4</sup>+1<br /><br />edit: demented i think its bc \u221a-i needs to be simplified to i\u221ai. Like how \u221a-9 is i\u221a9.",
"user": "wlicky"
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{
"id": "1271994",
"time": "1677610515",
"html": "<div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/dementedkermit\">dementedkermit</a> said:</div><div style=\"border:1px solid #888; padding:20px;\">I\u2019m also not 100% sure where i * sqrt(i) came from</div></div><br /><br />Difference of squares with (x<sup>2</sup> - (- i)) = (x<sup>2</sup> - (i<sup>2</sup>)(\u221ai<sup>2</sup>))",
"user": "kylijoy"
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{
"id": "1271998",
"time": "1677610687",
"html": "Ohh that makes sense I didn\u2019t see it that way",
"user": "dementedkermit"
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{
"id": "1271999",
"time": "1677611075",
"html": "It's kinda annoying, because it's a multiplicity-one unique root problem.",
"user": "kylijoy"
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"id": "1272007",
"time": "1677611786",
"html": "Here are the roots plotted on the complex plane<br /><br /><img src=\"https://i.ibb.co/r5Mwf6K/download.gif\" alt=\"\" />",
"user": "dementedkermit"
},
{
"id": "1272631",
"time": "1677805337",
"html": "Ok I am trying to solve this problem:<br /><br />How many rational numbers on the open interval (0,1) have a numerator and denominator that add to 1000 when in simplest form?<br /><br />My first thought would be however many primes there are between 500 and 999 since that\u2019s how many denominators there would be that would have a numerator less than them that would not simplify, but there are most likely nonprimes between 500 and 999 that are coprime with their compliment. <br /><br />I honestly have no clue how to approach this now",
"user": "dementedkermit"
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{
"id": "1272695",
"time": "1677814360",
"html": "For now I'll just leave a hint, but if I've been too cryptic please do let me know and I can write up a full explanation:<br /><br />For which integers x with 0<x<10 are x and 10-x coprime?<br />For which integers x with 0<x<12 are x and 12-x coprime?<br />For which integers x with 0<x<7 are x and 7-x coprime?<br />For which integers x with 0<x<n are x and n-x coprime?",
"user": "justabitjaded"
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"id": "1275512",
"time": "1678333570",
"html": "10: 1,3,7,9<br />12: 1,5,7,11<br />7: 1,2,3,4,5,6<br />n: all numbers that are not multiples of factors of n<br /><br />the prime factorization of 1000 is 2^3 * 5^3, so all factors of 1000 are either even or multiples of 5<br /><br />500 even numbers + 200 multiples of 5 - 100 overlap = 600 non-viable x values<br /><br />and since we are looking for pairs of numbers, we need to divide the 400/2, giving an answer of <span style=\"font-weight:bold;\">200</span><br /><br />Is that correct?",
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"time": "1678336902",
"html": "200 is the answer I came to as well.",
"user": "justabitjaded"
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"html": "Problems like this, while interesting to think about, are ultimately why I prefer my degree in Statistics over pure mathematics. Multi-Dimensional Probability spaces are much more comforting that guessing and hoping for pattern-recognition.",
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"html": "Extremely arbitrary ones, lol",
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"id": "1291819",
"time": "1684344408",
"html": "There are numbers 0-9 on a keypad that can be made into 6-digit codes. How many different possible codes are there?",
"user": "wlicky"
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"html": "10^6 = 1,000,000. Am I missing something?<br /><br />Actually, I guess you did say "numbers," not "integers" or something of the like, so the other answer I can come up with is infinity. Although that's an obviously wrong answer, even if it is technically correct in a stupid sort of way.<br /><br />Or you also don't specify that you're looking for 6-digit codes when you actually ask the question, which might again imply that the answer is infinity, because codes of any length are acceptable. This feels slightly less wrong than my previous infinity, but is still obviously dumb.",
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