digit again

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tushar7
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digit again

Unread post by tushar7 » Thu Dec 30, 2010 12:53 am

find the last two digits of $6^{2007}+7^{2007}$

prodip
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Re: digit again

Unread post by prodip » Thu Dec 30, 2010 2:14 am

The last two digits of \[6^{2007}\equiv 36(mod 100) and 7^{2007}\equiv43(mod 100) \]
so the ans is (36+43)=79............ :geek:

tushar7
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Re: digit again

Unread post by tushar7 » Thu Dec 30, 2010 12:59 pm

yap you are correct. but give the full solution

AntiviruShahriar
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Re: digit again

Unread post by AntiviruShahriar » Thu Dec 30, 2010 2:17 pm

$6^{40} \equiv 1 (mod 100)$
$6^{40 \cdot 50} \equiv 1^{50} (mod 100)$
$6^{2000} \equiv 1 (mod 100)$
$6^{2000} \cdot 6^3 \cdot 6^3 \cdot 6 \equiv 1 \cdot 16 \cdot 16 \cdot 6 (mod 100)$
$6^{2007} \equiv 36 (mod 100)$
and, $7^{40} \equiv 1 (mod 100)$
$7^{40 \cdot 50} \equiv 1^{50} (mod 100)$
$7^{2000} \cdot 7^3 \cdot 7^3 \cdot 7 \equiv 1 \cdot 43 \cdot 43 \cdot 7 (mod 100)$
$7^{2007} \equiv 43 (mod 100)$
so, the ans is $7^{2007}+6^{2007} \equiv [36+43] \equiv 79 (mod 100)$
ans:$79$
প্রদীপ দা'র বদলে দিলাম...।

AntiviruShahriar
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Re: digit again

Unread post by AntiviruShahriar » Thu Dec 30, 2010 2:38 pm

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]

prodip
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Re: digit again

Unread post by prodip » Thu Dec 30, 2010 6:23 pm

\[6^{2011}\equiv 56(mod 100)
7^(2011)\equiv 43(mod 100)
so the ans is 6^{2011}+7^{2011}\equiv 99(mod 100)\]
Last edited by prodip on Thu Dec 30, 2010 10:46 pm, edited 1 time in total.

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Labib
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Re: digit again

Unread post by Labib » Thu Dec 30, 2010 10:33 pm

@prodip

$56+43=109$ ??!! :?

It should be
$99$
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prodip
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Re: digit again

Unread post by prodip » Thu Dec 30, 2010 10:49 pm

Tnx labib vaia.9এর জায়গায় 0 বসেছে খেয়াল করি নি।

atiqur_jhe
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Re: digit again

Unread post by atiqur_jhe » Fri Dec 31, 2010 8:38 pm

@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 .

AntiviruShahriar
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Re: digit again

Unread post by AntiviruShahriar » Sat Jan 01, 2011 12:10 pm

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 .
vai ei post er 3rd comment taar hide ongsho theke hints niite parro......... :)

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