answer with explanation 101^100-1 this expression will be divisible by?

Question

find it out

Answer 1

101^100 -1 =
270 481382 942152 609326 719471 080753 083367 793838 278100 277689 020104 911710 151430 673927 943945 601434 674459 097335 651375 483564 268312 519281 766832 427980 496322 329650 055217 977882 315938 008175 933291 885667 484249 510000 =

(2 ^ 4) x 3 x (5 ^ 4) x 11 x 17 x 31 x 151 x 491 x 601 x 701 x 1301 x 1381 x 5101 x 39301 x 3146501 x 9367291 x 27595151 x 405048401 x 10827505640180401 x 10313185122435134101 x
244038008012812744006 (new line,same number)
1590332648845701001 x
153982620997066284506 (new line...)
369382388518803703267 (new line...)
7650843152489274354301

Number of divisors (prime and composite ones):
26214400

Answer 2

It is divisible by
a) 101 by the Fermat (not last) Theorem which says:
"if p is prime and p is not a divisor of A, then A^(p-1) -1 is divisible by p"
[Edited: not true because p=101 is naturally a divisors of A=101, so for p=101 one cannot apply the Fermat theorem. Sorry:)]

Being a difference of two squares it is also divisible by
b) 100
c) 101^50 +1
d) 101^25 -1 and then by
(101^24+101^23+ ... +101^2+101+1)*
e) 101^25 +1 which dissociate in
(101+1)*(101^24 -101^23+ ... +101^2 -101+1)
(and then it is also divisible by 102, so
f) 3
g) 17

Answer 3

Any expression of the form x^n-1 can be factored as (x-1)(1+x+x^2+...+x^(n-1)), so 101^100-1 is divisible by 100.

Answer 4

= (101^100 - 1^100)
= (101^10)^2 - (1^10)^2
= (101^10 + 1^10)(101^10 - 1^10)
= (101^10 + 1^10)((101^5)^2 - (1^5)^2)
= (101^10 + 1^10)(101^5 + 1^5)(101^5 - 1^5)
= (101^10 + 1^10)(101^5 + 1^5)(104060401 - 1)
= (101^10 + 1^10)(101^5 + 1^5)(104060400)

So,
(101^100 - 1) will be divisible by 104060400, 86717 etc...

Answer 5

I'm guessing you mean (101^100)-1

otherwise 101^99 is only divisible by every power of 101 from 0 to 99 since 101 is prime.

Answer 6

Nayanmange, there is a "power inadvertence" in your calculations, 101^100 is not equal to (101^10)^2 (which is equal to 101^20 )
but
101^100 = 101^10^2 = 101^(10^2)

Answer 7

101^100 - 1 ...

is divisible by 100.


consider x^n - y^n = (x - y) * (x^n-1 + x^n-2*y + ... +y^n-1)

this is called a "cyclotonic polynomial" or something like that.

x=101
n=100
y=1

(x-y) = (101 - 1) = 100

qed

Answer 8

my ti83 cant handle such big numbers and I suppose normal calculators cant.

if its 101^100-1 then it would be 101^99 which would end in some 9. SO Im guessing it would be divisible by 1, 3, and 9.