problem

[tushar7]

what is the remainder if $3^0+3^1+3^2+3^3..... +3^{2011}$ is divided by $8$?

source:AMC(just changed the last power)

[AntiviruShahriar]

$3^{2n} \equiv 1 (mod 8)$.............[1]
$3^{2n} \equiv 9 (mod 8)$
$3^{2n-1} \equiv 3 (mod 8)$....................[2]
$2011=2 \cdot 1006-1$ and $2010=2 \cdot 1005$
from [1] and [2]
$3^{2n}+3^{2n-1} \equiv 1+4 (mod 8)$
so,$3^{2n} \cdot 1005+3^{2n-1} \cdot 1006 \equiv 1 \cdot 1005+3 \cdot 1006 (mod 8) \equiv 1005+3018(mod 8) \equiv 4023 \equiv 7 (mod 8)$
ans:$7$???

[AntiviruShahriar]

$3^{2n} \equiv 1 (mod 8)$.............[1]
$3^{2n} \equiv 9 (mod 8)$
$3^{2n-1} \equiv 3 (mod 8)$....................[2]
$2011=2 \cdot 1006-1$ and $2010=2 \cdot 1005$
from [1] and [2]
$3^{2n}+3^{2n-1} \equiv 1+4 (mod 8)$
so,$3^{2n} \cdot 1005+3^{2n-1} \cdot 1006 \equiv 1 \cdot 1005+3 \cdot 1006 (mod 8) \equiv 1005+3018(mod 8) \equiv 4023 \equiv 7 (mod 8)$
ans:$7$???

ooopppppssss,

$3^{2n} \cdot 1005+3^{2n-1} \cdot 1006 \equiv 1 \cdot 1005+3 \cdot 1006 (mod 8) \equiv 1005+3018(mod 8) \equiv 4023 \equiv 7 (mod 8)$
$3^1+3^2+3^3+.........+3^{2011} \equiv 7 (mod 8)$
so, $3^0+3^1+3^2+........+3^{2011} \equiv 7+3^0 \equiv 8 \equiv 0 (mod 8)$
Ans:$0$..........

what's the wrong ....my computer is saying that ans will be 4[!!!!!!!!!!]but i can't find any wrong here

[AntiviruShahriar]

Yaaaahuuu.....
At last ei easy prob taar solution krte paarsi.But aager gulay vul ki chilo???Plz keu b0lo.
$3^{4} \equiv 1$(mod 8)
$3^{2012} \equiv 1$(mod 8)
$3^{2012}-1 \equiv 0 \equiv 8$(mod 8)
$ \frac{3^{2012}-1}{2} \equiv 4$(mod 8)
Ans:4.....

[bristy1588]

Eikhane shamanno ektu shomoshha ase:
Shahriar, TUMI jokhon 2 diye divide korso, tokhon tumi (mod 8) ke divide korte bhule geso.

[sm.joty]

বৃষ্টি ১০০% ঠিক। সহমত।

[Labib]

In this case, you just have to notice that,
$8|(3^{4n}+3^{4n+1}+3^{4n+2}+3^{4n+3})$
and so, $8|(3^0+3^1+3^2+3^3..... +3^{2011})$.
Thus the remainder is $0$.

[sm.joty]

আরে এটা তো খেয়াল করিনি। তাহলে শাহরিয়ার ভুলটা করল কোথায় ???