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How can one determine if $$28 | 89^{135} $$

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    Can an even number divide an odd number? – 2017-02-24
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    Note also that $89$ is prime. – 2017-02-24

3 Answers 3

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The parity argument is enough for yes/no answer, so to get a more detailed answer:

$89^{135}\equiv 5^{135} \bmod 28$, and since $\lambda(28)=6$, and $5,28$ are coprime, $5^6\equiv 1 \bmod 28$. Thus $89^{135}\equiv 5^{135}\equiv 5^3\equiv 13 \bmod 28$.

$89^{135}$ has a remainder of $13$ when divided by $28$.

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    It would help to explicitly mention that you are (intentionally?) doing much more than needed, since a trivial parity argument proves indivisibility. – 2017-02-24
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Last digit of $89^{135}$ = last digit of $9^{135}$ = last digit of $1^{67} \times 9= 9$

i.e., $89^{135}$ is an odd number as mentioned in @egreg in his comment.

So, it is obvious that $89^{135} \not\equiv 0 \pmod {28}$ because $28$ is even

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Hint:

Last digit of $89^1 = 9$

Last digit of $89^2 = 1$

Last digit of $89^3 = 9$

Last digit of $89^4 = 1$

Do you see a pattern here? The numbers cycle from $9,1,9,1$ and so on. So now consider your case, which is $89^{135}$. Can you take it from here?