divisibility factors of x^n-/+y^n problem (1 Viewer)

[ctpengage]

prove that 55^62-13^62-13^62+41^62 is divisibile by 13

 

[PC]

Is this question supposed to read:

55⁶² – 13⁶² – 13⁶² + 41⁶²

???

 

[lyounamu]

Is this question supposed to read:

55⁶² – 13⁶² – 13⁶² + 41⁶²

???

Yeah, I think so.

 

[Timothy.Siu]

i dont think its 2unit unless its really really simple and i'm just missing it
cud u take out the 13's so its 55⁶² + 41⁶² still cant do anything though

 

[ctpengage]

i dont think its 2unit unless its really really simple and i'm just missing it
cud u take out the 13's so its 55⁶² + 41⁶² still cant do anything though

its part of the cubic identites portion of 2 unit

 

[Timothy.Siu]

(55^31-13^31)(55^31+13^31) + (41^31-13^31)(41^31+13^31)
= 55^31(55^31+13^31) - 13^31(55^31+13^31) + 41^31(41^31+13^31) - 13^31(41^31+13^31)
= 55^31(55^31+13^31) + 13^31(-13^31-55^31 + 41^31 + 13^31) + 41^31(41^31 + 13^31)
= 55^31(55^31+13^31) + 13^31(41^31-55^31) + 41^31(41^31+13^31)
= 2. 55^31 + 55^31 . 13^31 + 13^31 . 41^31 - 13^31 . 55^31 + 2 . 41^31 + 41^31 . 13^31
= 2 . 55^31 + 2. 41^31 + 2. 13^31 . 41^31
= 2 (55^31 + 41^31 + 13^31 . 41^31)
= 2 (55^31 + 41^31(1+13^31))
= ???

Let me try different method, cubic method for example.

LOL that seemedlike a lot of work
but from memory in 2unit we haven't had to prove divisibility

 

[lyounamu]

LOL that seemedlike a lot of work
but from memory in 2unit we haven't had to prove divisibility

That didn't really work in the end. I am thinking of a different way...

 

[lyounamu]

prove that 55^62-13^62-13^62+41^62 is divisibile by 13

There we go:

55^62-13^62-13^62+41^62 = n(55^62-13^62-13^62 +41^62)/n

Let n=k

k(55^62-13^62-13^62+41^62)/k = 13m

Now prove n=k+1
(k+1)(55^62-13^62-13^62 +41^62)/(k+1) = 13t
L.H.S. = k(k+1)(55^62-13^62-13^62+41^62)/(k+1)k = k(55^62-13^62-13^62+41^62)/k = 13m

I am goin' around a circle, and I used 3 Unit-method. This is the only way I could think of to get this somewhere.

 

[lolokay]

just considering the 55⁶² + 41⁶² part, since the 13⁶² are obviously divisible by 13;

(52+3)⁶² + (39+2)⁶² (52 and 39 are multiples of 13)
= (52⁶² + 62*52⁶¹*3 +(other terms that are multiples of 52) ...+ 3⁶²] + (39⁶² + (other terms that are multiples of 39) + 2⁶²)

so now we need to only worry about the 3⁶² + 2⁶² since everything else is divisible by 13
3⁶² + 2⁶²
= 9³¹ + 4³¹
now, using the sum of odd powers rule;
= (9+4)(9³⁰ - 9²⁹*4 + 9²⁸*4² - ... + 4³⁰)
= 13(9³⁰ - 9²⁹*4 + 9²⁸*4² - ... + 4³⁰)
so 55⁶² - 13⁶² - 13⁶² + 41⁶² must be divisible by 13 since all the parts we broke it into are

 

[lyounamu]

just considering the 55⁶² + 41⁶² part, since the 13⁶² are obviously divisible by 13;

(52+3)⁶² + (39+2)⁶² (52 and 39 are multiples of 13)
= (52⁶² + 62*52⁶¹*3 +(other terms that are multiples of 52) ...+ 3⁶²] + (39⁶² + (other terms that are multiples of 39) + 2⁶²)

so now we need to only worry about the 3⁶² + 2⁶² since everything else is divisible by 13
3⁶² + 2⁶²
= 9³¹ + 4³¹
now, using the sum of odd powers rule;
= (9+4)(9³⁰ - 9²⁹*4 + 9²⁸*4² - ... + 4³⁰)
= 13(9³⁰ - 9²⁹*4 + 9²⁸*4² - ... + 4³⁰)
so 55⁶² - 13⁶² - 13⁶² + 41⁶² must be divisible by 13 since all the parts we broke it into are

What the hell is the sum of the odd power's rule?

 

[lolokay]

do you know the factorising difference of 2 powers rule?
xⁿ - yⁿ = (x-y)(x^n-1 + x^n-2y + x^n-3y² +...+yⁿ) ?
well it's like that (I don't know if either of these are actually taught)
it's
xⁿ + yⁿ = (x+y)(x^n-1 - x^n-2y + x^n-3y² -...+yⁿ)
eg sum of 2 cubes which you would know
x³ + y³ = (x+y)(x² - xy + y²)

 

[lyounamu]

do you know the factorising difference of 2 powers rule?
xⁿ - yⁿ = (x-y)(x^n-1 + x^n-2y + x^n-3y² +...+yⁿ) ?
well it's like that (I don't know if either of these are actually taught)
it's
xⁿ + yⁿ = (x+y)(x^n-1 - x^n-2y + x^n-3y² -...+yⁿ)
eg sum of 2 cubes which you would know
x³ + y³ = (x+y)(x² - xy + y²)

ok, that's really interesting to know. thanks