Question

Find the HCF of 2^120-1 and 2^100-1.

Answer 1

[(2^100)-1]

Answer 2

how do u find it?

Answer 3

[(2^20)-1]

Answer 4

as per the rules x^n/ x^m = x^n-m
so 2^120-100= 2^20-1

Answer 5

u r chinese right!

Answer 6

ya

Answer 7

is that -1 at the end??

Answer 8

i dont understand the notation
is 2^120 - 1 = -1 / 2^120 ?

Answer 9

looks like it's not easy as it looks
\( 2^{120} - 1\) and \( 2^{100} - 1\)

Answer 10

1 / 2^120 ?

Answer 11

oh - right - no thats not easy

Answer 12

hmm..i've think thru it and i've a different explaination:


2^(120)-1= (2^(100)-1)(2^(20)+1)
Hence, 2^(120)-1/2^(100)-1= (2^(100)-1)(2^(20)+1)/(2^(100)-1)
= 2^(20)+1 (remainder)

Thus, 2^(100)-1 is the HCF

Answer 13

this approach might work.
a^n - x^n = (a-x)(a^(n-1) + a^(n-2)x + .... + a x^(n-1) + x^n)

Answer 14

Oo ... how foolish of me.

Answer 15

(2^(100)-1)(2^(20)+1) = 2^120 - 2^20 + 2^(100) - 1, not 2^(120)-1

Answer 16

oo ya, just realised...

Answer 17

r u sure its not (2^120)^-1 = 1 / 2^120? - in which case its easy to solve.

Answer 18

in that case the HCF is (2^100)^-1

Answer 19

@Shu_Yun what is your question??
\( ( 2^{100})^{-1} \) or \( (2^{100})-1\)

Answer 20

would'nt 2^100 - 1 be prime?? - not sure

Answer 21

@experimentX (2^100)-1

Answer 22

(2^100)-1 = (2^20)^50-1^50 = (2^20 - 1){some junk terms}
(2^120)-1 = (2^20)^60-1^60 = (2^20 - 1){some junk terms}
until now the highest common factor is 2^20 - 1 <--- it might even get higher.

keep trying

Answer 23

i just checked 2^120 - 1 and 2^100 - 1 on wolframalpha - neither are prime

Answer 24

yeah ... there are some junk terms ...
a^n - x^n = (a-x)(a^(n-1) + a^(n-2)x + ..stupid junk terms.. + a x^(n-1) + x^n)

Answer 25

i've cheated!!!!
wolfram came up with 2^20 - 1

Answer 26

huh??? so my guess was right??

Answer 27

yes

Answer 28

well, thanks for checking :)

Answer 29

wolfram solved arithmetically - writing the values as product of their prime factors

Answer 30

thats ok