Question

Prove or disprove for all integers n greater than 0:
\(133^n\) has the same last three digits as \(375+(-1)^n 250 + 8^n\)

Answer 1

Basically, we want to find
\[133^n \mod 1000 = \left\{
\begin{array}{lr}
125 + (8 \mod 1000) && : n \text{ is odd}\\
625 + (8 \mod 1000) && : n \text{ is even}
\end{array}
\right.\]

Answer 2

Not "Basically." Definitely not basic. But that is the equation form for this problem.

Answer 3

from arraging a little

(5^3+8)^n=133^n

375+(-1)^n*250+8^n = 3*5^3+(-1)^n*2*5^3+8^n
so when n is even then
5^4+8^n
when
n is odd
5^3+8^n

Answer 4

(5^3+2^3)^n=133^n

Answer 5

(5^3+2^3)^n mod 10^3

(a^3+b^3)^n mod (ab)^3

Answer 6

\[375+(-1)^n 250 + 8^n -133^n \equiv 0+(-1)^n0+8^n-8^n\equiv 0 \pmod{5^3} \]
\[375+(-1)^n 250 + 8^n -133^n \equiv -1+(-1)^n2+0-5^n\equiv 0 \pmod{2^3} \]
so \(2^35^3\mid 375+(-1)^n 250 + 8^n -133^n\)

Answer 7

hmm
@rational how u managed 2^3 ?

Answer 8

ya i already diiddd it up theere :)

Answer 9

soo on
even its 5^4 + 8^n and
odd 5^3+8^n

Answer 10

5^4=375+250
5^3=375-250

Answer 11

(a^3+b^3)^n mod (ab)^3 = a^3n + b^3n mod (ab)^3

Answer 12

was typing then suddenly random music appears so i wanna type ans share\
https://www.youtube.com/watch?v=28sdV_DXSrU