5
$\begingroup$

The problem is to find the number of zeros at the end of the sum $4^{5^6}+6^{5^4}$.

I tried $2^{2 \cdot 5^6}+3^{5^4} \cdot 2^{5^4}= 2^{5^4} \cdot ( 2^{2 \cdot 5^6 -5^4}+ 3^{5^4} )$.

$\endgroup$
3
$\begingroup$

Use the following lemma, which can be proven by induction 'on $b$':

For any odd prime $p$ and $a,b \in \mathbb{N}_{>0}$ such that $p^a \mid\mid x-y$ and $p^b \mid\mid k$:

  $p^{a+b} \mid\mid x^k-y^k$.

For this question we don't need the "strictly divides" in the conclusion, but in general this lemma is very useful.

$4^{5^6}+6^{5^4} = (4^{5^2})^{5^4}+6^{5^4} = (4^{5^2})^{5^4}-(-6)^{5^4}$ and $5 \mid 4^{5^2}-(-6)$ by Fermat's little theorem.

Therefore by the lemma $5^5 \mid 4^{5^6}+6^{5^4}$.

Combining with what you already have gives the number of zeros.

$\endgroup$
  • $\begingroup$ What's the difference between $||$ and $|$? And how do you read it? $\endgroup$ – GFauxPas Jan 18 '15 at 14:57
  • $\begingroup$ $p^k \mid\mid x \Leftrightarrow p^k \mid x \wedge p^{k+1} \nmid x$. $\endgroup$ – user21820 Jan 18 '15 at 14:57
  • $\begingroup$ @GFauxPas: Note also that the lemma is slightly different if $p$ is the special even prime $2$. In particular the conclusion would be $p^{a+b+1} \mid\mid x^k-y^k$ if I remember correctly. $\endgroup$ – user21820 Jan 18 '15 at 14:59
  • $\begingroup$ @GFauxPas: Read "$\mid$" as "divides" and "$\mid\mid$" as "strictly divides", and note that "strictly divides" only makes sense for (prime) powers. It's technically a shorthand abuse of notation but makes it easy to remember. $\endgroup$ – user21820 Jan 18 '15 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.