What is the remainder when $6^{273} + 8^{273}$ is divided by $49$?

What is the remainder when $6^{273} + 8^{273}$ is divided by $49$?

I tried this question through two methods and both are giving different answers so I wanted to know which is the correct one, and why the other is incorrect.

Approach $1$:

Here, I have tried to express everything in terms of $\pmod7$

For odd numbers$$a^n+b^n = (a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-....+b^{n-1})$$So$$6^{273}+8^{273} = (6+8)(6^{272}-6^{271}\cdot8+6^{270}\cdot8^2-....+8^{272})$$

$6^{272}\equiv(-1)\pmod7$

$8^{272}\equiv1\pmod7$

So, the second bracket reduces to:

$$((-1)^{272} - (-1)^{271}\cdot1 +(-1)^270\cdot1....+1^{271})\pmod7$$

Which is $273\pmod7$, basically it is divisible by $7$ and even the $(6+8)$ part is divisible by $7$.

$$6^{273}+8^{273} = 7^2k$$ so $$6^{273}+8^{273}\equiv 0 \pmod{49}$$

Approach $2$:

$6^3 \equiv 20 \pmod{49}$

So by cyclicity: $6^{273}\equiv 20 \pmod{49}$

Similarly,

$8^3 \equiv 22\pmod{49}$

So, $8^{273}\equiv 22 \pmod{49}$

Therefore:

$$6^{273}+8^{273}\equiv 42 \pmod{49}$$

Help, which is correct?

1 Answer

The binomial formula gives$$(7\pm1)^{273}={273\choose1}7^1(\pm1)^{272}+{273\choose0}7^0(\pm1)^{273}=273\cdot 7\pm1=\pm1\qquad({\rm mod}\ 49)\ ,$$since all other terms are divisible by $7^2$. It follows that the answer to your question is $0$.