Counting the powers

Number Theory Level 2

Let $a$ be the remainder of $\large {{3}}^{1234567890}$ upon division by 8.
Similarly let $b$ be the remainder of $\large { {5}}^{1234567890}$ upon division by 6.

Then, what is $a+b$ ?

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The answer is 2.

3 solutions

Abhay Tiwari
Jun 13, 2016

$\large \dfrac{{\color{#3D99F6} {3}}^{1234567890}}{8} + \dfrac{{\color{#D61F06} {5}}^{1234567890}}{6}$

$\large \dfrac{{\color{#3D99F6} {9}}^{617283945}}{8} + \dfrac{{\color{#D61F06} {25}}^{617283945}}{6}$

$\large \dfrac{{\color{#3D99F6} {(8+1)}}^{617283945}}{8} + \dfrac{{\color{#D61F06} {(24+1)}}^{617283945}}{6}$

$\large \dfrac{{\color{#3D99F6} {1}}^{617283945}}{8} + \dfrac{{\color{#D61F06} {1}}^{617283945}}{6}$

So remainder in both the fractions is $1$

So, $1+1=2$

Nicely done:) +1!

Rishabh Tiwari - 5 years ago

Thanks a lot, Tiwari ji ;).

Abhay Tiwari - 5 years ago

Lol, two of my tiwari friends here.

Ashish Menon - 5 years ago

@Ashish Menon – Ya right lol!,;)

Rishabh Tiwari - 5 years ago

Haaha ;) Thnx u2 tiwariji ! :-):-)

Rishabh Tiwari - 5 years ago

Perfect solution!!! (+1)

Ashish Menon - 5 years ago

Thanks a lot :)

Abhay Tiwari - 5 years ago

Terry Smith
Jul 2, 2016

3 taken to an even power is always 1 mod 8

5 taken to an even power is always 1 mod 6

1 plus 1 is two. I'm delighted to write this to a bunch of postgrad (or WILL be) math types....

Rushikesh Jogdand
Jul 1, 2016

Relevant wiki: Modular Arithmetic - Multiplication

Observe that $9=3^2\equiv 1 \mod{8}$ $\therefore 3^{1234567890}=9^{617283945}\equiv 1^{617283945}\equiv 1 \mod{8}\Rightarrow \boxed{a=1}$ And $5\equiv -1\mod{6}$ $\therefore 5^{1234567890}\equiv (-1)^{1234567890}\equiv 1 \mod{6}\Rightarrow \boxed{b=1}$

Counting the powers

Want some interesting questions like this?. Enter the world where Calculators will not help

The answer is 2.

3 solutions

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