mickyeddi:



Nice try dude but this another take:

suppose, p_n(x)=x^6+n=x^6±y^a.
the expressions "x^6+y^a" & "x^6-y^a" can be factorized only for the following values of "a"
x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)
x^6-y^6=(x+y)(x-y)(x^2+xy+y^2)(x^2-x*y+y^2)
x^6-y^4=(x^3+y^2)*(x^3-y^2)
x^6+y^3=(x^2+y)(x^4-x^2y+y^2)
x^6-y^3=(x^2-y)(x^4+x^2y+y^2)
x^6-y^2=(x^3+y)*(x^3-y)
because | n | < 500 and "y" (±y^a=n) must be an integer, for different values of "a" y^6=0,1,64 y^4=0,1,16,81,256 y^3=0,1,8,27,64,125,216,343 y^2=0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324, 361,400,441,484
possible values of "n" =y^6=0,1,64 =-y^6=0,-1,-64 =y^4=0,1,16,81,256 =y^3=0,1,8,27,64,125,216,343 =-y^3=0,-1,-8,-27,-64,-125,-216,-343 =-y^2=0,-1,-4,-9,-16,-25,-36,-49,-64,-81,-100,-121,-144,-169,-196,-225,-256,-289,-324,-361,-400,-441,-484
We can get all possible values of "n" by making an union of these 6 set of numbers (I am using the word "union" because some values have occurred more than one time). That gives 35 distinct values of "n".

mickyeddi:



Nice try dude but this another take:

suppose, p_n(x)=x^6+n=x^6±y^a.
the expressions "x^6+y^a" & "x^6-y^a" can be factorized only for the following values of "a"
x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)
x^6-y^6=(x+y)(x-y)(x^2+xy+y^2)(x^2-x*y+y^2)
x^6-y^4=(x^3+y^2)*(x^3-y^2)
x^6+y^3=(x^2+y)(x^4-x^2y+y^2)
x^6-y^3=(x^2-y)(x^4+x^2y+y^2)
x^6-y^2=(x^3+y)*(x^3-y)
because | n | < 500 and "y" (±y^a=n) must be an integer, for different values of "a" y^6=0,1,64 y^4=0,1,16,81,256 y^3=0,1,8,27,64,125,216,343 y^2=0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324, 361,400,441,484
possible values of "n" =y^6=0,1,64 =-y^6=0,-1,-64 =y^4=0,1,16,81,256 =y^3=0,1,8,27,64,125,216,343 =-y^3=0,-1,-8,-27,-64,-125,-216,-343 =-y^2=0,-1,-4,-9,-16,-25,-36,-49,-64,-81,-100,-121,-144,-169,-196,-225,-256,-289,-324,-361,-400,-441,-484
We can get all possible values of "n" by making an union of these 6 set of numbers (I am using the word "union" because some values have occurred more than one time). That gives 35 distinct values of "n".

mickyeddi:
They are ranked by a site called Brilliant.org, which asks users progressively harder math and science problems.

Think you might be as smart as one of these tweens or teens?

Prove it by solving some of the math problems they've solved.






1.


2.


3.


4.

Dollyak:
Number 4 is quite easy too as well.
If you think about this. The total amount spent in the air by the skier is the same as the time taken to slope down the whole distance of 50meters.
thus distance gives =1/2 at t^2
Therefore 50=1/2 × gt^2 = 1/2 (9.8 )t^2 =50 solve for t
T= 3.2 seconds

Very easy because time is only depends on gravity in this case.

Obinoscopy:
Nobody has even attempted the questions. Na wa. I'm busy but let me give the mathematicians in the house my suggested hint on how to solve qxn 1:



x³ + a³ can be factorised into (x + a)(x² - ax + a²)
Therefore we can express x⁶ + a⁶ as (x² + a²)(x⁴ - a²x² + a⁴)
Thus if we can find the various numbers of n that can be expressed in the form a⁶ (note that both n and a should be integers)
This means we need to find the various values of a such that a⁶ is less than 500.

Now that means the values of n are 0 (0⁶), 1 (1⁶) and 64 (2⁶). 3 is not included because 3⁶ is 729 which is greater than 500.

Lordave:
Jesus is the Answer.

mickyeddi:



Nice try dude but this another take:

suppose, p_n(x)=x^6+n=x^6±y^a.
the expressions "x^6+y^a" & "x^6-y^a" can be factorized only for the following values of "a"
x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)
x^6-y^6=(x+y)(x-y)(x^2+xy+y^2)(x^2-x*y+y^2)
x^6-y^4=(x^3+y^2)*(x^3-y^2)
x^6+y^3=(x^2+y)(x^4-x^2y+y^2)
x^6-y^3=(x^2-y)(x^4+x^2y+y^2)
x^6-y^2=(x^3+y)*(x^3-y)
because | n | < 500 and "y" (±y^a=n) must be an integer, for different values of "a" y^6=0,1,64 y^4=0,1,16,81,256 y^3=0,1,8,27,64,125,216,343 y^2=0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324, 361,400,441,484
possible values of "n" =y^6=0,1,64 =-y^6=0,-1,-64 =y^4=0,1,16,81,256 =y^3=0,1,8,27,64,125,216,343 =-y^3=0,-1,-8,-27,-64,-125,-216,-343 =-y^2=0,-1,-4,-9,-16,-25,-36,-49,-64,-81,-100,-121,-144,-169,-196,-225,-256,-289,-324,-361,-400,-441,-484
We can get all possible values of "n" by making an union of these 6 set of numbers (I am using the word "union" because some values have occurred more than one time). That gives 35 distinct values of "n".

Slickest:
We taking cash u toking maths...I only calculate currencies...

Dollyak:
Number 4 is quite easy too as well.
If you think about this. The total amount spent in the air by the skier is the same as the time taken to slope down(starting from the horizontal position) the whole distance of 50meters.
thus distance gives =1/2 at t^2
Therefore 50=1/2 × gt^2 = 1/2 (9.8 )t^2 =50 solve for t
T= 3.2 seconds

Very easy because time is only depends on gravity in this case.

Adultnextdoor:
......Interesting but still not there yet.......i'll give you 6/10 for laying out a logical and semi-correct calculation....

Obinoscopy:
He's the OP of this thread. Are you sure he's semi-correct?

Obinoscopy:


To solve 4:

I can use the conservation of energy formula: KE + Frictional Energy = PE
Therefore 0.5mv² = 0.95mgh. From there we can calculate v.

To get t: We can use the formula s = ut + 0.5gt²
Where u = v and g = -9.8
This gives us a parabola.
The parabola made intersections with the straight line (which made an angle 30 with the horizontal)
Equating the parabola with the straight line should give us the value of t

Adultnextdoor:
......it doesn't matter whether he is the op or not......his calculation is incomplete but its obvious he has an idea!

Donmichaelz:
the people I know that would give it a shoot has less Dan 100k in their account....All am saying is mathematical problems are not real life problems ,I am more of the latter,dangote can't do these ,neither can I ,,,,

Obinoscopy:
I know you don't know the answer so no need...