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number of ordered pair,s $(a,b,c)$ in $2^a+2^b=c!\;,$ Where $a,b,c$ are non negative integers

$\bf{My\; Try::}$ If $a=0,b=0\;,$ Then $c=2$

Similarly for other values of $a$ and $b,$ But i did not understand how can i calculate upper bond of $c$

Help required, Thanks

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See $$2^x\equiv 1,2,4\pmod 7$$ So if we take equation $\pmod 7$ and if $c\geq 7$ then we get that $c!$ is divisble by 7 because it contains at least one 7 but the sum of $2^a+2^b$ never si divisible by 7 beacuse of $2^x\equiv 1,2,4\pmod 7$ ($2^a+2^b$ can give only ($1,2,3,4,5,6\pmod 7$) so $c\leq 6$. Then perhaps you can solve it just by case checking like taking c=6, then c=5.

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