digit again again

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tushar7
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digit again again

Unread post by tushar7 » Tue Jan 18, 2011 8:59 pm

what is the last two digits of $2^{999}$
my asnwer is 32

tushar7
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Re: digit again again

Unread post by tushar7 » Thu Jan 20, 2011 11:38 pm

no one!.......................

AntiviruShahriar
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Re: digit again again

Unread post by AntiviruShahriar » Fri Jan 21, 2011 1:26 am

Solving this i have observed some nice things....
$24^{2n} \equiv 76$(mod 100) $24^{2n+1} \equiv 24$ (mod 100) where$ n \epsilon N$
$11^i \equiv 10 \cdot i +1$(mod100)......put, i=10n+1
$11^{10n+i} \equiv 10 \cdot i+1$(mod 100).
another one is:
we know in modular arithmetic, if G.C.D(n,d)=k in
$a \equiv b $(mod n) then, $ \frac{a}{d} \equiv \frac{b}{d} (mod \frac{n}{k})$.....(1)
let's try it without changing n......
from $(1)\Rightarrow \frac{a}{d}- \frac{b}{d} \cdot (k/n)$ $=x \epsilon Z$
so, $( \frac{ak}{d}- \frac{bk}{d}) \cdot \frac {1}{n}$
$ \frac{ak}{d} \equiv \frac{bk}{d}$(mod n).. ;)
Last edited by AntiviruShahriar on Fri Jan 21, 2011 2:25 pm, edited 2 times in total.

AntiviruShahriar
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Re: digit again again

Unread post by AntiviruShahriar » Fri Jan 21, 2011 1:55 am

my ans is 88 :?
I'm not sure because i found it by an ugly way.........
$2^{10} \equiv 24\equiv 3 \cdot 2^3$ (mod 100)
$\Rightarrow 2^{100} \equiv 3^{10} \cdot 2^{30}$(mod 100)
$\Rightarrow 2^{100} \equiv 3^4 \cdot 3^4 \cdot 3^2 \cdot 2^{10 \cdot 3}$(mod 100)
$\Rightarrow 2^{100} \equiv 49 \cdot 24^3 \equiv 49 \cdot 24$(mod 100)[উপরের পোস্ট এ দেখও]
$\Rightarrow 2^{1000} \equiv 76^{10} \equiv 2^{10 \cdot 2} \cdot 19^{10}$(mod 100)
$\Rightarrow 2^{1000} \equiv 24^2 \cdot 61^4 \cdot 61$(mod 100)
$\Rightarrow 2^{1000} \equiv 76 \cdot 21 \cdot 21 \cdot 61 \equiv 76 \equiv 176$(mod 100)
applying (1)$\Rightarrow \frac{2^{1000} \cdot 4}{8} \equiv \frac{176 \cdot 4}{8}$(mod 100) [G.C.D(8,100)=4]
$ \Rightarrow 2^{999} \equiv 88$(mod 100).....
ভাইয়ারা আমি ঘুমাইতে ঘুমাইতে লেখলাম। কোনো ভুল থাকলে নিজে থেইক্কা সশোধন কইরা নিও...।
তুষার ভাই তোমার \[ \sum_{i=0}^{2011} 3^i \equiv x\](mod 8) এর সমস্যা টা দেখো[ফোরামে দিসিলা মনে হয়]

AntiviruShahriar
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Re: digit again again

Unread post by AntiviruShahriar » Sun Jan 23, 2011 10:51 am

$a \equiv b$(mod m)
$\Rightarrow \frac{a-b}{m}=x \epsilon Z$
now,$ \frac{ak-bk}{mk}=x$
so,$ak \equiv bk$(mod mk)..........(1)
$2^{ \phi(25)} \equiv 1$(mod 25)
$2^{20} \equiv 2^{980} \equiv 1$(mod 25)
$2^{982} \equiv 4(mod 25 \cdot 4)$[using 1]
$2^{982} \cdot 2^{10} \cdot 2^7 \equiv 4 \cdot 24 \cdot 28$(mod100)
$2^{999} \equiv 88$(mod 100)
Ans:88
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protik
Posts:35
Joined:Wed Dec 08, 2010 7:21 am

Re: digit again again

Unread post by protik » Sun Jan 23, 2011 2:49 pm

nice solve Antivirus..

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