Mathematics Thread

[ilikecubing]

• We may not know the speed at which Peter travels to pick up his child, but we do know that when the child started walking home an hour earlier than usual, i.e. at 4PM, it resulted in Peter arriving at home a half-hour earlier than usual.
• Assuming Peter was traveling from home and not some other location, and further assuming level ground, symmetrical traffic conditions, and no significant prevailing wind, this means that Peter's trip toward his child was 15 minutes shorter than usual, and their trip back home together was also 15 minutes shorter than usual.
• If Peter's trip toward his child was 15 minutes shorter than usual, and it usually ends at 5PM, then today Peter met up with his child at 4:45PM.
• If the child began walking at 4:00PM and was picked up by Peter at 4:45PM, the child was obviously walking for 45 minutes.

Correct, 45 minutes is the right answer, thanks for the simple explanation. I initially tried making equations and got stuck with the variables

Nowhere is it mentioned how long it takes them to walk home. It does not usually end at 5pm.

Peter took 30 mins LONGER than usual. Since they left 1 hour early but only returned home 30 mins early.

This question is quite unclear in many aspects.

It does end at 5PM usually, on that day, it ended at 4 PM

30 minutes earlier than what?

Than the usual time at which they return home.

 

[ilikecubing]

1. In how many ways can 5 rings be worn on the four particular fingers of the right hand?
(A) 4^5
(B) 5^4
(C) 8C3
(D) 8P3
(E) 4!*5!

2. In how many ways can 5 different rings be worn on the four particular fingers of the right hand?
(A) 4^5*5C4
(B) 5^4*5C4
(C) 8C3
(D) 8P3
(E) 4!*5!

3. In how many ways can 5 different rings be worn on the any four fingers of the right hand?
(A) 4^5*5C4
(B) 5^4*5C4
(C) 8C3*5C4
(D) 8P3*5C4
(E) 4!*5!*5C4

 

[Tempus]

1. In how many ways can 5 rings be worn on the four particular fingers of the right hand?
(A) 4^5
(B) 5^4
(C) 8C3
(D) 8P3
(E) 4!*5!

Questions like these are why so many people hate word problems. They are all too often poorly written, using wording that is either ambiguous in meaning or flat-out wrong. In this particular case, I am assuming that the question means to ask "How many different ways can you wear 5 identical rings when limited to using only 4 specific fingers?" If that is in fact the question, then the answer is 56, which can be expressed as either 8C5 or 8C3. 8C3 is among the options, so the answer would appear to be option C.

So far, so good, but here's where things take a turn for the unfortunate...

2. In how many ways can 5 different rings be worn on the four particular fingers of the right hand?
(A) 4^5*5C4
(B) 5^4*5C4
(C) 8C3
(D) 8P3
(E) 4!*5!

The answer to this one is 6,720, which can be expressed as 8P5, but that is not among the options, so all five options are incorrect. It's possible that the question's original author was confused about how the permutations function was supposed to be implemented, and assumed it was X!/Y! instead of X!/(X-Y)!, causing option D to be 8P3 instead of the correct 8P5.

3. In how many ways can 5 different rings be worn on the any four fingers of the right hand?
(A) 4^5*5C4
(B) 5^4*5C4
(C) 8C3*5C4
(D) 8P3*5C
(E) 4!*5!*5C4

This question is the worst one of the lot. I can only conclude that it means to ask "How many different ways can 5 different rings be worn on 5 specific fingers such that at least one finger is bare." (The fact that questions this garbled appear on standardized tests is unfortunate.) If this is in fact what they are asking, then the answer is 15,119, or (9P5)-1, which is once again not present in the list of options.

Where did you find these questions, if you don't mind my asking?

 

[10461394944000]

ukpeople: anyone else doing step/something similar this year? i'm doing all 3 step papers next month and some of the questions are pretty hard like this silly thing about exponentiating matrices

Spoiler

and this silly integral thing

Spoiler

and this silly thing with zeta(2) inside of it

Spoiler

and this other silly integral thing

Spoiler

 

[KongShou]

Post the ones you can't do. Ill see if I can help out. I can do pretty much all pure maths ones. Dont really look at mech or stats much tho.

I should probably also mention I dont do much STEP 3.

 

[10461394944000]

Post the ones you can't do. Ill see if I can help out. I can do pretty much all pure maths ones. Dont really look at mech or stats much tho.

I should probably also mention I dont do much STEP 3.

lol i dont need help with any of them but some of the questions are weird

 

[KongShou]

lol i dont need help with any of them but some of the questions are weird

Jar it is pretty different to a level. I think its more uni styled. I like them tho! Its like uni maths but you dont have to understand any of the higher stuff.

 

[10461394944000]

Jar it is pretty different to a level. I think its more uni styled. I like them tho! Its like uni maths but you dont have to understand any of the higher stuff.

kinda but "type it into mathematica" isnt an acceptable solution in step

also its no fun if you dont understand the hard stuff

did step1 from 2000 yesterday/today and it was really easy so hopefully this years papers are easy too qi

 

[10461394944000]

okso me and soup are looking for primes that are of some silly form because primes are cool

here are some silly primes:

\( 513!-\lfloor e^{513} \rfloor \)

\( 51^{51}-2 \)

\( \lfloor \sinh 582 \rfloor=\lfloor \frac{e^{582}}{2}\rfloor \)

\( 496^{496}+496!+1 \)

\( \lfloor 615e^{615}\rfloor \)

\( \lfloor 951\pi^{951}\rfloor \)

\( \lfloor 952(\pi e)^{952}\rfloor \)

31415926535897932384626433832795028841 (first 38 digits of pi)

141421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147010955997160597027453459 (225 digits of √2)

325! - !325 + 1

1395! + !1395 - 1

\( 46!^{46}+71 \)

edit: \( 95!^{95}+113 \), 14062 digits!
edit: 2# + 3# + 5# + 7# + 11# + ... + 397# - 1
edit: ocool, 136!+149 and 136!+151 are twin primes

edit: pairs {x, y} where \( 2\leq x<y\leq500 \) and \( y x^x+x y^y+1 \) is prime:

Spoiler

{2, 3}, {2, 5}, {2, 6}, {2, 21}, {2, 130}, {3, 4}, {3, 5}, {3, 12}, {3, 49}, {3, 313}, {4, 5}, {4, 11}, {4, 14}, {4, 72}, {4, 95}, {4, 114}, {5, 7}, {5, 8}, {5, 9}, {5, 14}, {5, 51}, {5, 316}, {6, 12}, {6, 30}, {6, 276}, {7, 12}, {7, 17}, {7, 26}, {7, 38}, {7, 39}, {7, 311}, {7, 345}, {8, 10}, {8, 15}, {8, 20}, {8, 48}, {8, 57}, {9, 10}, {9, 12}, {11, 18}, {11, 122}, {12, 21}, {12, 61}, {12, 315}, {13, 44}, {14, 25}, {15, 30}, {15, 119}, {16, 53}, {18, 23}, {19, 366}, {20, 55}, {20, 71}, {20, 384}, {21, 80}, {21, 298}, {22, 50}, {22, 156}, {23, 67}, {24, 54}, {24, 70}, {24, 72}, {24, 74}, {24, 189}, {24, 350}, {24, 466}, {26, 91}, {26, 226}, {27, 36}, {27, 45}, {27, 50}, {28, 80}, {28, 189}, {29, 190}, {29, 290}, {30, 31}, {30, 160}, {30, 188}, {30, 345}, {31, 35}, {31, 131}, {31, 210}, {31, 356}, {32, 41}, {32, 59}, {34, 69}, {34, 155}, {35, 46}, {35, 86}, {35, 228}, {35, 426}, {37, 53}, {37, 122}, {40, 42}, {40, 86}, {40, 426}, {41, 181}, {42, 117}, {42, 330}, {43, 51}, {43, 150}, {43, 183}, {43, 420}, {44, 63}, {44, 66}, {44, 89}, {44, 335}, {44, 391}, {46, 134}, {47, 80}, {49, 86}, {50, 68}, {50, 252}, {51, 219}, {52, 441}, {53, 61}, {54, 111}, {54, 121}, {54, 320}, {54, 396}, {55, 78}, {55, 108}, {55, 195}, {56, 115}, {56, 142}, {56, 473}, {58, 69}, {58, 201}, {59, 79}, {60, 173}, {62, 106}, {63, 167}, {64, 384}, {65, 70}, {65, 88}, {65, 90}, {66, 80}, {66, 138}, {67, 98}, {67, 116}, {67, 117}, {69, 139}, {70, 74}, {71, 105}, {71, 438}, {72, 94}, {72, 158}, {73, 395}, {73, 443}, {74, 82}, {74, 255}, {76, 185}, {77, 112}, {77, 142}, {78, 312}, {80, 105}, {80, 164}, {81, 112}, {81, 119}, {82, 294}, {84, 212}, {84, 280}, {85, 269}, {87, 252}, {87, 271}, {88, 257}, {89, 196}, {91, 111}, {98, 221}, {99, 192}, {100, 210}, {101, 159}, {102, 195}, {104, 120}, {104, 128}, {104, 169}, {104, 234}, {105, 259}, {105, 332}, {105, 459}, {109, 189}, {109, 476}, {110, 295}, {113, 204}, {115, 234}, {116, 205}, {120, 457}, {121, 378}, {122, 371}, {124, 456}, {126, 258}, {127, 378}, {128, 215}, {128, 460}, {131, 156}, {131, 210}, {132, 240}, {133, 143}, {136, 194}, {138, 289}, {138, 395}, {139, 246}, {141, 164}, {141, 319}, {143, 456}, {147, 213}, {155, 290}, {156, 340}, {158, 166}, {158, 185}, {158, 189}, {159, 450}, {160, 242}, {162, 166}, {162, 330}, {165, 174}, {167, 321}, {174, 175}, {174, 209}, {175, 245}, {176, 463}, {177, 183}, {177, 260}, {182, 330}, {183, 362}, {189, 469}, {192, 395}, {193, 387}, {193, 474}, {195, 351}, {206, 276}, {206, 280}, {210, 500}, {213, 478}, {214, 339}, {215, 318}, {216, 290}, {216, 347}, {225, 226}, {225, 423}, {226, 278}, {227, 423}, {229, 392}, {230, 500}, {234, 293}, {235, 311}, {239, 265}, {244, 249}, {246, 271}, {249, 478}, {252, 309}, {252, 386}, {256, 422}, {257, 319}, {258, 270}, {258, 316}, {260, 329}, {266, 367}, {268, 365}, {269, 298}, {275, 391}, {276, 325}, {276, 425}, {278, 421}, {282, 365}, {285, 472}, {285, 483}, {289, 294}, {292, 305}, {294, 401}, {300, 315}, {302, 441}, {306, 471}, {314, 366}, {318, 345}, {324, 439}, {329, 339}, {332, 454}, {337, 470}, {344, 417}, {345, 439}, {354, 430}, {356, 452}, {372, 385}, {381, 446}, {383, 386}, {392, 470}, {398, 449}, {399, 414}, {435, 462}, {482, 493}, {482, 500}

edit: seems that !x + !y where y>x is only prime if y=x+1 (apart from {0, 0}, {0, 2}, {0, 3}, {1, 3}, {2, 2} csch smallexceptions) but idk how to prove csch dont know anything about subfactorials

 

[ilikecubing]

Questions like these are why so many people hate word problems. They are all too often poorly written, using wording that is either ambiguous in meaning or flat-out wrong. In this particular case, I am assuming that the question means to ask "How many different ways can you wear 5 identical rings when limited to using only 4 specific fingers?" If that is in fact the question, then the answer is 56, which can be expressed as either 8C5 or 8C3. 8C3 is among the options, so the answer would appear to be option C.

So far, so good, but here's where things take a turn for the unfortunate...


The answer to this one is 6,720, which can be expressed as 8P5, but that is not among the options, so all five options are incorrect. It's possible that the question's original author was confused about how the permutations function was supposed to be implemented, and assumed it was X!/Y! instead of X!/(X-Y)!, causing option D to be 8P3 instead of the correct 8P5.


This question is the worst one of the lot. I can only conclude that it means to ask "How many different ways can 5 different rings be worn on 5 specific fingers such that at least one finger is bare." (The fact that questions this garbled appear on standardized tests is unfortunate.) If this is in fact what they are asking, then the answer is 15,119, or (9P5)-1, which is once again not present in the list of options.

Where did you find these questions, if you don't mind my asking?

Yeah, these questions are indeed very ambiguous. They gave me a hard time and I am average in permutations and combinations so I thought I'd get some insight on these questions here. But it is unfortunate that the questions are not framed properly I found these in a Facebook group. Sorry about the ambiguity.

Question to all -

Is there a way I could somehow use the Rubiks Cube to help myself with the concepts of Permutations and Combinations, so that I don't face much problem in solving different kinds of Permutations and Combinations questions.

Another Question -

Find the sum of the roots of the equation x^2013 + (0.5 - x)^2013 = 0

 

[5BLD]

x^2013+(0.5-x)^2013=0
x^2012/2 - (2013C2) x^2011/4 +...=0
(1/2) x^2012 - (1012539/2) x^2011 +...=0

Summing roots for any polynomial is -b/a.

1012539

 

[10461394944000]

x^2013+(0.5-x)^2013=0
x^2012/2 - (2013C2) x^2011/4 +...=0
(1/2) x^2012 - (1012539/2) x^2011 +...=0

Summing roots for any polynomial is -b/a.

1012539

should be 503 csch the coefficient of x^2012 is 2013/2 not 1/2

 

[5BLD]

Oops of course i knew that
But yeah this sum of roots thing is interesting. I wonder whether if it says something to do with vectors (of complex roots)...

 

[KongShou]

Um... Does anyone know a good book/video of a lecture on youtube/resource/etc that explains metric space well? I have long decided that wikipedia is useless when youre trying to learn something.

 

[tozies24]

Um... Does anyone know a good book/video of a lecture on youtube/resource/etc that explains metric space well? I have long decided that wikipedia is useless when youre trying to learn something.

A Metric Space is just a set with a metric (a distance equation relating two points) defined on it that has certain properties. (Wikipedia actually isn't horrible for learning definitions of things, but I agree with your comment. But in this case it isn't horrible. Some of the examples are actually insightful.)

Anyway, the reason why we use Metric Spaces is so that we have a concrete set on which we can work with distances. (This is similar to a topological space however, in topology, we generally don't care about distance unless specified. Also a metric space is a specific case of a topological space) It is just the setting which we are doing the problem. Basically it just means we have a distance function and other properties in our toolbox.

In my real analysis course I took last year, we used Rudin's Principles of Mathematical Analysis (often referred to as Baby Rudin). This book is widely used in beginning Real Analysis courses and most of the problem's solutions can be found online.

 

[Cubeologist]

Multi-variable calculus and calculus of variation. It can take a surprising level of mathematics to simply do classical mechanics.

 

[10461394944000]

what is that area

find a symbolic expression that approximates it

 

[qqwref]

what is that area

find a symbolic expression that approximates it

Spoiler

obviously it is symmetrical
bottom intersection is at x = pi/4
left intersection is at arccos(pi/4) ~= 0.6675
the area is 2*integral from arccos(pi/4) to pi/4 of cos(cos(x)) - sin(cos(x))
wolframalpha cannot into this integral but the value is about .0127631

 

[10461394944000]

Spoiler

obviously it is symmetrical
bottom intersection is at x = pi/4
left intersection is at arccos(pi/4) ~= 0.6675
the area is 2*integral from arccos(pi/4) to pi/4 of cos(cos(x)) - sin(cos(x))
wolframalpha cannot into this integral but the value is about .0127631

yes

Spoiler

it is also almost a rhombus so it is approximately (cos(1/sqrt2)-sin(1/sqrt2))(pi/4-arccos(pi/4)) wich is about 0.01304517

 

[10461394944000]

silly thing:

take 922 digits of pi, including the 3 at the start
remove the decimal point
replace all the 0s with 69
congralutations you have a prime number

Spoiler

3141592653589793238462643383279569288419716939937516958269974944592369781646962862698998628693482534211769679821486986513282369664769938446699556958223172535946981284811174569284169276919385211695559644622948954936938196442881699756659334461284756482337867831652712691969914564856692346693486169454326648213393669726692491412737245876969666963155881748815269926996282925469917153643678925969366969113369536954882694665213841469519415116699433695727693657595919536992186117381932611793169511854869744623799627495673518857527248912279381836911949129833673362446965664369866921394946395224737196976921798669943769277695392171762931767523846748184676694695132696969568127145263566982778577134275778966991736371787214684469969122495343691465495853716956979227968925892354269199561121296921966986469344181598136297747713699966951876972113499999983729786949951695973173281669963185956924459455346969836926425223698253344685693526193118817169169696931378387528865875332698381426961717766914736935982534969428755468731

also i have step2 tomorrow which will be fun