Don't Multiply It Out!

Number Theory Level 2

Find the remainder when $3^{2015} + 7^{2015}$ is divided by 50.

This problem is part of the set AMSP .

The answer is 0.

5 solutions

Chew-Seong Cheong
Apr 27, 2015

$\begin{aligned} 3^{2015} + 7^{2015} & \equiv (3^5)^{403} \pmod {50}+ 7(7^2)^{1007} \pmod {50} \\ & \equiv \left[ (243)^{403} + 7(49)^{1007} \right] \pmod {50} \\ & \equiv \left[ (-7)^{403} + 7(50-1)^{1007} \right] \pmod {50} \\ & \equiv \left[ -7(49)^{201} + 7(-1) \right] \pmod {50} \\ & \equiv \left[ -7(-1) + 7(-1) \right] \pmod {50} \\ & \equiv \boxed{0} \pmod{50} \end{aligned}$

Moderator note:

I was thinking of Chinese Remainder Theorem. But this gets the job done fairly quick as well. Nice!

mod $25\!:\ \$ $3^{2015\!\pmod{\!20}}+7^{2015\!\pmod{\!20}}$ by Euler's theorem.

$\equiv 3^{-5}+7^{-5}\equiv \frac{1}{\underbrace{81}_{6}\cdot 3}+\frac{1}{\underbrace{49^2}_{(-1)^2}\cdot 7}\equiv \frac{1}{18}+\frac{1}{7}\equiv \frac{1}{-7}+\frac{1}{7}\equiv 0$ and clearly is even, so divisible by $50$ too.

mathh mathh - 6 years ago

Ramesh Goenka
Apr 25, 2015

Austin Seiberlich
Apr 30, 2015

$3^{2015} + 7^{2015} \, \, \equiv [(3^{5})^{403} + (7^{5})^{403}] \pmod{50}$ $\qquad \qquad \qquad \equiv [243^{403} + 16807^{403} \pmod{50}$ $\qquad \qquad \qquad \equiv [(-7)^{403} + (7)^{403}] \pmod{50}$ $\qquad \qquad \qquad \equiv \boxed{0} \pmod{50}$

Shubham Poddar
Apr 27, 2015

The last digit of 3^2015 will be 7.

The last digit of 7^2015 will be 3.

Hence the last digit of the sum will be 0.

Hence it will be completely divisible by 50.

Moderator note:

As Aalap Shah has pointed out, you did not prove that it can further be divisible by 5.

No, this only ensures divisibility by 10, not 50.

Aalap Shah - 6 years, 1 month ago

William Isoroku
Apr 27, 2015

Perfect website: https://www.mtholyoke.edu/courses/quenell/s2003/ma139/js/powermod.html

Don't Multiply It Out!

This problem is part of the set AMSP .

The answer is 0.

5 solutions

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