digit again
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
find the last two digits of $6^{2007}+7^{2007}$
Re: digit again
The last two digits of \[6^{2007}\equiv 36(mod 100) and 7^{2007}\equiv43(mod 100) \]
so the ans is (36+43)=79............
so the ans is (36+43)=79............
Re: digit again
yap you are correct. but give the full solution
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Re: digit again
rrre ekhon new year er jonno $6^{2011}+7^{2011}$ er last duita onko dao....ans ta khub e mojar....
[new post e dilam na cz 2tai similar onko]
[new post e dilam na cz 2tai similar onko]
Re: digit again
\[6^{2011}\equiv 56(mod 100)
7^(2011)\equiv 43(mod 100)
so the ans is 6^{2011}+7^{2011}\equiv 99(mod 100)\]
7^(2011)\equiv 43(mod 100)
so the ans is 6^{2011}+7^{2011}\equiv 99(mod 100)\]
Last edited by prodip on Thu Dec 30, 2010 10:46 pm, edited 1 time in total.
Re: digit again
@prodip
$56+43=109$ ??!!
It should be
$56+43=109$ ??!!
It should be
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"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Re: digit again
Tnx labib vaia.9এর জায়গায় 0 বসেছে খেয়াল করি নি।
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Re: digit again
@prodip da ai j 62011 56(mod100) ata kivabe holo .ami korle to arokom hoi: 62011 11(mod100) or 89(mod100) .jodi vul hoy tahole vul ta ki .
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Re: digit again
vai ei post er 3rd comment taar hide ongsho theke hints niite parro.........atiqur_jhe wrote:@prodip da ai j $6^{2011} \equiv 56(mod100)$ ata kivabe holo .ami korle to arokom hoi: $6^{2011} \equiv 11(mod100) or 89(mod100)$ .jodi vul hoy tahole vul ta ki .