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Prove that for any positive integer $t$, there exists a positive integer $n$ such that the base $10$ representation of $\phi(n)$ ends with $t$ zeros.

The base $10$ representation is confusing me a lot. Is that just saying that the value of $\phi(n)$ must end in $t$ zeros's? If so, then I've found that any number ending in $t+1$ zeros will yield a value for $\phi(n)$ ending in $t$ zeros. I am not sure how to prove that in general, I just saw that pattern while messing around with sage.

But then that might not make any sense if I am just not understanding what the base 10 representation is.

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  • $\begingroup$ It is a bit circular to refer to base 10. Whatever base is used, its radix representation is "10". Better to write base ten for clarity. $\endgroup$
    – hardmath
    Jun 1, 2016 at 4:01

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$\phi(10^{t+1}) = \phi(2^{t+1}5^{t+1}) = (2-1)2^t(5-1)5^t=4\cdot 10^t$

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