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ATOT Maths Challenge

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Alternately,

x = distance Mary walked
Y = distance Nora walked

While Mary walked, the bike carried Nora 65-Y and also doubled back for 65-X-Y:
EQ1:
x/5 = (65-Y)/50 + (65-Y-x)/50
10x = 130-2Y -x
11x=130-2Y



Meanwhile, the distanced walked/5 and distanced biked/50 for Mary and Nora should be the same:
EQ2
x/5 + (65-x)/50 = Y/5 +(65-y)/50
(65+9x)=65+9Y
X=Y

Going back to EQ1:
13x=130
x=10

10/5 = 2 hours
(65-10)/50 = 1.1 hours

3.1 hours total.
 
Lisa being a girl forgot that doubling back would use more petrol, and so ran out. Thus they never made it to point B. Lisa and Nora ended up walking round for days, getting hungrier and thirstier... until there was a terrible accident in which Nora was killed. Then desperate, hungry, thirsty and on the brink of death Lisa bumped into a snake... but that's a different story.
 
Originally posted by: PlasmaBomb
Lisa being a girl forgot that doubling back would use more petrol, and so ran out. Thus they never made it to point B. Lisa and Nora ended up walking round for days, getting hungrier and thirstier... until there was a terrible accident in which Nora was killed. Then desperate, hungry, thirsty and on the brink of death Lisa bumped into a snake... but that's a different story.
Click to expand...

Lisa makes it at the end. Better nate than lever.
 
Originally posted by: 91TTZ
Originally posted by: Barack Obama
Back in the 10th grade I undertook a a highly prestigous mathematics challenge in which I <SNIP>
Click to expand...


Stop stroking your ego on a public forum.
Click to expand...

Some may call me an elitist.

Question 2 has now been posted. This one should be a piece of cake (or pie?).
 
The quotient when you divide 32N + 97 by 4N - 1 is 8 + 105/(4N-1) (simple polynomial division)
So the question is, for what positive integers N is 105 divisible by (4N-1)

The factors of 105 are: 1, 3, 5, 7, 15, 21, 35, and 105
So, solve 4N-1= 1, 4N-1=3, 4N-1=5, and 4N-1=7, 4N-1=15, 4N-1=21, 4N-1=35, and 4N-1=105
Not all of these give integer solutions.
The integers are 1,2,4,and 9

edit: heh, fixed the typo part way through, and the erroneous answer it produced. I should know better than to try to type math before writing it on paper. edit edit: whoa - two typos. Fixed the 106 that should have been a 105

Also, check: this results in 129 divided by 3 = 43, 161 divided by 7 = 23, 225 divide by 15 = 15, and 385 divided by 35 = 11.
 
Originally posted by: DrPizza
The quotient when you divide 32N + 97 by 4N - 1 is 8 + 106/(4N-1) (simple polynomial division)
So the question is, for what positive integers N is 106 divisible by (4N-1)

The factors of 106 are: 1, 3, 5, 7, 15, 21, 35, and 105
So, solve 4N-1= 1, 4N-1=3, 4N-1=5, and 4N-1=7, 4N-1=15, 4N-1=21, 4N-1=35, and 4N-1=105
Not all of these give integer solutions.
The integers are 1,2,4,and 9

edit: heh, fixed the typo part way through, and the erroneous answer it produced. I should know better than to try to type math before writing it on paper.

Also, check: this results in 129 divided by 3 = 43, 161 divided by 7 = 23, 225 divide by 15 = 15, and 385 divided by 35 = 11.
Click to expand...

hah, pwned. go dr. pizza 😉

i'm so used to diff eq. i forgot all the basics. and i'm forgetting diff eq too 😱
 
Originally posted by: DrPizza
The quotient when you divide 32N + 97 by 4N - 1 is 8 + 106/(4N-1) (simple polynomial division)
So the question is, for what positive integers N is 106 divisible by (4N-1)

The factors of 106 are: 1, 3, 5, 7, 15, 21, 35, and 105
So, solve 4N-1= 1, 4N-1=3, 4N-1=5, and 4N-1=7, 4N-1=15, 4N-1=21, 4N-1=35, and 4N-1=105
Not all of these give integer solutions.
The integers are 1,2,4,and 9

edit: heh, fixed the typo part way through, and the erroneous answer it produced. I should know better than to try to type math before writing it on paper.

Also, check: this results in 129 divided by 3 = 43, 161 divided by 7 = 23, 225 divide by 15 = 15, and 385 divided by 35 = 11.
Click to expand...


Damn, beat me to it.
 
Originally posted by: 91TTZ
Originally posted by: Barack Obama
Back in the 10th grade I undertook a a highly prestigous mathematics challenge in which I <SNIP>
Click to expand...


Stop stroking your ego on a public forum.
Click to expand...

1. He didn't spell prestigious correctly; you weren't capable of pointing out the irony, since you wanted to attack him.
2. Noting that it was a "highly prestigious" 10th grade math exam makes it more attractive to those who enjoy solving math problems.
3. I haven't seen you answer one of these questions yet. Jealous of everyone else's mathematics ability?
 
Originally posted by: DrPizza
The quotient when you divide 32N + 97 by 4N - 1 is 8 + 106/(4N-1) (simple polynomial division)
So the question is, for what positive integers N is 106 divisible by (4N-1)

The factors of 106 are: 1, 3, 5, 7, 15, 21, 35, and 105
So, solve 4N-1= 1, 4N-1=3, 4N-1=5, and 4N-1=7, 4N-1=15, 4N-1=21, 4N-1=35, and 4N-1=105
Not all of these give integer solutions.
The integers are 1,2,4,and 9

edit: heh, fixed the typo part way through, and the erroneous answer it produced. I should know better than to try to type math before writing it on paper.

Also, check: this results in 129 divided by 3 = 43, 161 divided by 7 = 23, 225 divide by 15 = 15, and 385 divided by 35 = 11.
Click to expand...

see, thats what a smart person would do, but i'm sure 90% of us would just use Excel or Matlab or some other math program to solve that in 30 seconds. Thats legal right?
 
I forgot most of this stuffs since I left high school, and more than 14 years later, I can tell you with certainty that none of this will make me any money.

So when is your homework due?
 
Originally posted by: BrownTown
Originally posted by: DrPizza
The quotient when you divide 32N + 97 by 4N - 1 is 8 + 106/(4N-1) (simple polynomial division)
So the question is, for what positive integers N is 106 divisible by (4N-1)

The factors of 106 are: 1, 3, 5, 7, 15, 21, 35, and 105
So, solve 4N-1= 1, 4N-1=3, 4N-1=5, and 4N-1=7, 4N-1=15, 4N-1=21, 4N-1=35, and 4N-1=105
Not all of these give integer solutions.
The integers are 1,2,4,and 9

edit: heh, fixed the typo part way through, and the erroneous answer it produced. I should know better than to try to type math before writing it on paper.

Also, check: this results in 129 divided by 3 = 43, 161 divided by 7 = 23, 225 divide by 15 = 15, and 385 divided by 35 = 11.
Click to expand...

see, thats what a smart person would do, but i'm sure 90% of us would just use Excel or Matlab or some other math program to solve that in 30 seconds. Thats legal right?
Click to expand...

Show your work. Partial credit will be given for answers only.
 
Originally posted by: DrPizza

1. He didn't spell prestigious correctly; you weren't capable of pointing out the irony, since you wanted to attack him.
2. Noting that it was a "highly prestigious" 10th grade math exam makes it more attractive to those who enjoy solving math problems.
3. I haven't seen you answer one of these questions yet. Jealous of everyone else's mathematics ability?
Click to expand...



False.
 
Originally posted by: ivan2
Dr pizza by 106 do u mean 105
Click to expand...

Yep; wow, I fixed one typo; didn't notice I did that. The 105 appears elsewhere in that response.

Originally posted by: BrownTown
Originally posted by: DrPizza
The quotient when you divide 32N + 97 by 4N - 1 is 8 + 105/(4N-1) (simple polynomial division)
So the question is, for what positive integers N is 105 divisible by (4N-1)

The factors of 106 are: 1, 3, 5, 7, 15, 21, 35, and 105
So, solve 4N-1= 1, 4N-1=3, 4N-1=5, and 4N-1=7, 4N-1=15, 4N-1=21, 4N-1=35, and 4N-1=105
Not all of these give integer solutions.
The integers are 1,2,4,and 9

edit: heh, fixed the typo part way through, and the erroneous answer it produced. I should know better than to try to type math before writing it on paper.

Also, check: this results in 129 divided by 3 = 43, 161 divided by 7 = 23, 225 divide by 15 = 15, and 385 divided by 35 = 11.
Click to expand...

see, thats what a smart person would do, but i'm sure 90% of us would just use Excel or Matlab or some other math program to solve that in 30 seconds. Thats legal right?
Click to expand...

Give me a similar problem and I'm pretty sure I could solve it in less than 30 seconds on a piece of paper. I usually don't write everything. (Plus, I wouldn't make a typo, then follow the typo through) :embarrassed;
 
Just dropping in this thread because it's a math thread. I'd work on this, but summer is over and I have my own Calc III to do 🙂
 
32N + 97 = 8*(4N - 1) + 105

Therefore (32N + 97)/(4N - 1) = 8 + 105/(4N - 1)

So 4N - 1 has to be a factor of 105.

4N - 1 = 1
4N - 1 = 3, N = 1
4N - 1 = 5
4N - 1 = 7, N = 2
4N - 1 = 15, N = 4
4N - 1 = 21
4N - 1 = 35, N = 9
4N - 1 = 105
 
Originally posted by: chuckywang
32N + 97 = 8*(4N - 1) + 105

Therefore (32N + 97)/(4N - 1) = 8 + 105/(4N - 1)

So 4N - 1 has to be a factor of 105.

4N - 1 = 1
4N - 1 = 3, N = 1
4N - 1 = 5
4N - 1 = 7, N = 2
4N - 1 = 15, N = 4
4N - 1 = 21
4N - 1 = 35, N = 9
4N - 1 = 105
Click to expand...


Why are you solving his homework?
 
GOod stuff, Q2 was solved. Although it was more of a no-brainer. Question 3 is posted up now and it should be a piece of cake. There's some tough questions coming up afterwards.
 
Another easy one. 🙂
I'll write it as a product of two fractions, rather than division of two fractions, and factor as I go

(x+3y)(x+3y) [x²(x-3y) - 4(x-3y)]
----------------x-----------------------
(x+3y) (x-3y) [x(x+3y) - 2(x+3y)]


(x+3y)(x+3y) [(x²-4)(x-3y)]
----------------x----------------
(x+3y) (x-3y) [(x-2)(x+3y)]


(x+3y)(x+3y) [(x+2)(x-2)(x-3y)]
----------------x---------------------
(x+3y) (x-3y) [(x-2)(x+3y)]

Everything cancels except the x+2


So, the answer is:
x+2, with a bunch of restrictions about the x and y values which would result in division by zero in the original fraction.

*note - factoring on a TI-89 is for math sissies who cannot factor in their heads. I even showed the steps.
 
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