## digit again

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tushar7
Posts: 101
Joined: Tue Dec 07, 2010 3:23 pm

### digit again

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

prodip
Posts: 27
Joined: Sun Dec 19, 2010 1:24 pm

### 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............

tushar7
Posts: 101
Joined: Tue Dec 07, 2010 3:23 pm

### Re: digit again

yap you are correct. but give the full solution

AntiviruShahriar
Posts: 125
Joined: Mon Dec 13, 2010 12:05 pm
Location: চট্রগ্রাম,Chittagong
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### Re: digit again

AntiviruShahriar
Posts: 125
Joined: Mon Dec 13, 2010 12:05 pm
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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]

prodip
Posts: 27
Joined: Sun Dec 19, 2010 1:24 pm

### Re: digit again

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

Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm

### Re: digit again

@prodip

$56+43=109$ ??!!

It should be
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prodip
Posts: 27
Joined: Sun Dec 19, 2010 1:24 pm

### Re: digit again

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

atiqur_jhe
Posts: 13
Joined: Tue Dec 14, 2010 11:49 am
Location: Jhenidah

### 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 .

AntiviruShahriar
Posts: 125
Joined: Mon Dec 13, 2010 12:05 pm
Location: চট্রগ্রাম,Chittagong
Contact:

### Re: digit again

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.........