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What is the remainder when 3^12 + 5^12 is divided by 13?
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3^12 + 5^12 = 27^4 + 25^6 now 27 = 1 mod 13 =>27^4 = 1 mod 13 and 25 = -1 mod13 =>25^6 = 1 mod 13 bod leave remainder 1 each,,hence net remainder will be 1+1 or 2..
or one can use fermat little theorem :)
Fermat's little theorem states that if \(p\) is a prime number, then for any integer \(a\)\[a^{p-1} \equiv1 \ \ \ \text{mod} \ p\]
would you just show me that in this question?
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sure p=13 is a prime number so \[3^{13-1}=3^{12} \equiv1 \ \ \ \text{mod} \ 13\]\[5^{13-1}=5^{12} \equiv1 \ \ \ \text{mod} \ 13\]so\[3^{12}+5^{12} \equiv2 \ \ \ \text{mod} \ 13\]
learnt something new @mukushla ..thanks..
np man :)
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