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Number Theory Level 2

$4^{61} + 4^{62} + 4^{63} + 4^{64}$ is divisible by

11 17 13 3

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Prasun Biswas
Apr 9, 2015

$4^{61}+4^{62}+4^{63}+4^{64} \\ = 4^{61+0}+4^{61+1}+4^{61+2}+4^{61+3} \\= 4^0\cdot 4^{61}+4^1\cdot 4^{61}+4^2\cdot 4^{61}+4^3\cdot 4^{61} \\ = 4^{61}(4^0+4^1+4^2+4^3) \\ = 4^{61}(1+4+16+64) = 4^{61}\cdot 85 = 4^{61} \cdot 5 \cdot \boxed{17}$

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Brilliant Mathematics Staff - 6 years, 2 months ago

Aakash Singh
Dec 17, 2014

$={ 4 }^{ 61 }+{ 4 }^{ 62 }+{ 4 }^{ 63 }+{ 4 }^{ 64 }\\ ={ 4 }^{ 61 }(1+4+16+64)\\ ={ 4 }^{ 61 }(85)\\ ={4}^{61}(17)(5)\\$

it can be clearly seen that 17 is a factor of the given expression

nice solution

Abhishek Chopra - 6 years, 5 months ago

Himanshu Tuteja
Dec 20, 2014

see the cyclicity as 4+16+64+256are divisible by 17 so the no is also

Nam Cao Vũ Hoàng
Dec 22, 2014

$A = 4^{61} &plus; 4^{62} &plus; 4^{63} &plus; 4^{64} \Leftrightarrow 4A = 4^{62} &plus; 4^{63} &plus; 4^{64} &plus; 4^{65} \Leftrightarrow 4A - A = 3A = 4^{65} - 4^{61} \Leftrightarrow A = \frac{4^{65} - 4^{61}}{3} =\frac{4^{61}\left ( 4^{4} - 1 \right )}{3} =\frac{4^{61}\left ( 255 \right )}{3} =4^{61}\left ( 85 \right ) =4^{61}\left ( 17 \right )\left ( 5 \right )\vdots 17$

Mukul Rathi
Dec 24, 2014

Note that 4^2=16 which is congruent to -1 (mod 17), so if the power of 4 is odd, it is congruent to -1 mod 17 and if it is even it is congruent to 1 mod 17. Since there are two odd and two even powers of 4, we have -1 + 1 - 1 + 1 =0 (mod 17) so 17 divides the given expression

Moderator note:

Your solution is contradictory. You mentioned that $4^2 = 16 \equiv -1 \pmod {17}$ and you also mention that if the power of $4$ is even, it is congruent to $1 \bmod {17}$ . Which is it? You should clarify that $\begin{cases}{4^{2n} \bmod{17} \equiv 1} && {} \\ {4^{2n+1} \bmod{17} \equiv -1} && {}\end{cases}$ for positive integer $n \geq 2$ .

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