what is the last two digits of $2^{999}$
my asnwer is 32
digit again again
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Re: digit again again
no one!.......................
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Re: digit again again
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)..
$24^{2n} \equiv 76$(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.
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Re: digit again again
my ans is 88
I'm not sure because i found it by an ugly way.........
I'm not sure because i found it by an ugly way.........
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Re: digit again again
$a \equiv b$(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
mobile e lekhlam...Vul h0ite parre...Tar jnno sry
$\Rightarrow \frac{a-b}{m}=x \epsilon Z$
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
mobile e lekhlam...Vul h0ite parre...Tar jnno sry