5
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$$2^a - 5^b7^c = 1$$ So, $a,b,c>0$.
For solving this I observed that last digit of $ 2^a$ are among any of $2, 4, 8$ and $6$ and last digit of $5^b7^c$ is always $5$ . Therefor if a is only in the form of 4n this equation may be valid. now arbitrarily I put $a=8$ and $b=1$ and $c=2$ and I found $$2^8-5^17^2=11$$ this result gives a feelings thet there is no such solution of this problem. I want to know wheather I am in the appropriate track or I am wrong. Please help.

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3
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Hint:since $$2^a-1=5^b\cdot 7^c$$ if $b>0$ then we have $$5|2^a-1\Longrightarrow 4|a\Longrightarrow 3|2^a-1$$ impossible

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