Why is Six Afraid of Forty Nine?

Note first that the given expression can be written as $(7 - 1)^{273} + (7 + 1)^{273}.$

Now in the binomial expansion of each of these terms, the exponent of $7$ will exceed $2$ for all but the last two terms in each expansion. Thus calculating modulus $49$ , we are left with

$\binom{273}{1}*7*(-1)^{272} + (-1)^{273} + \binom{273}{1}*7*1^{272} + 1^{273} =$

$273*7*2 = 39*7*7*2 \equiv 0 \pmod{49}.$

Thus the desired remainder is $\boxed{0}.$

$7\mid 6+8;\, 7\nmid 6,8$ , so by Lifting the Exponent Lemma (LTE) :

$\upsilon_7(6^{273}+8^{273})=\upsilon_7(6+8)+\upsilon_7(273)=1+1=2$

So $7^2\mid 6^{273}+8^{273}$ (and in fact $7^3\nmid 6^{273}+8^{273}$ , but we don't need it for this problem).