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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?

Bill Dubuque
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    How does $6^3 \equiv 20 \pmod{49} \implies 6^{273}\equiv 20 \pmod{49}$? – VIVID May 08 '21 at 09:05
  • $$6^{273}=(6^3)^{91}$$ My teacher said we can do such manipulations and the mod would be same. I did same with $8$ also, is that wrong? –  May 08 '21 at 09:07
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    $6^{273} = (6^3)^{91} \equiv 20^{91}$ – VIVID May 08 '21 at 09:08
  • Then why is $61 \equiv1\pmod{31}$ and even $61^{100} \equiv1\pmod{31}$ ? –  May 08 '21 at 09:12
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    Python: print pow(6,273,49) + pow(8,273,49) gives $49$ so the result is disivisible by $49$ and the reminder is $0$. – Henno Brandsma May 08 '21 at 09:16
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    The standard version of that trick is to calculate $a^{xy}=(a^x)^y$ when $a^x\equiv 1$. Because that makes things really easy. When $a^x\not\equiv \pm 1$, then things aren't that easy. – Arthur May 08 '21 at 09:16
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    It suffices to reduce $273$ to its remainder modulo $\phi(49)$ throughout. – Henno Brandsma May 08 '21 at 09:20
  • @Arthur oh ok I got it now. Thanks, I think I might have misinterpreted that trick. –  May 08 '21 at 09:22
  • The 1st approach is correct then? Is there a shorter method because polynomial formulae don't usually click. –  May 08 '21 at 09:23
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    The trick you mention that didn't work could be replaced by another, not-always-useful trick. $6^{273} = 6^{256}6^{16}6^1$. Exponents that are powers of 2 are easy-ish to find mod m by repeated modulo squaring. But the shorter method you're looking for is in the answer below. – Eric Snyder May 08 '21 at 09:56

1 Answers1

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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 $0$.

TheSimpliFire
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Christian Blatter
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