Show that \(19^{93}\) - \(13^{99}\) is a positive integer divisible by 162.

NOTE:Actually I could not understand how to proceed. Can anyone show me the path ?

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## Comments

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TopNewestWhat I had done is that I wrote it as \((18+1)^{93}-(12+1)^{99}\) and now it is easy to see that a very few terms are left, after the binomial expansion, that are not divisible by \(162\),i.e., \(3^4*2\). And then it is very easy to show that the remaining terms add up to be a multiple of \(162\) as well.

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latex not loading proprly pls type it

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Another method is to note that \(13^2 \equiv 7 \pmod {162}\).

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We will have to prove that its divisible by \(81\) and \(2\). But its obviously divisible by \(2\) and if we consider both numbers mod 9, it turns out that its divisible by \(9\).

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Got that it is divisible by 2 and 9. But we need to prove that it is divisible by 162. Anyways, I could not think of approaching it this way.

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it goes for 81 also

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hey vikram bhaiya please mujhe yei problem latex chor kar likh do mere computer par load nehi ho raha latex

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show that 19^93 - 13^99 is a positive integer divisible by 162

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right but first prove 19^93>13^99

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not needed to show the greater than sign.

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A way to reduce the bulk of this problem is to use Euler's totient function. Also, it is inevitably evident that the given number is divisible by \(2\), so we have to worry about \(81\) only.

After applying Euler's totient function, we only need to show that: \(19^{39} \equiv c \pmod{81}\) and \(13^{45} \equiv c \pmod{81}\) i.e \(19^{39}\) and \(13^{45}\) leave the same remainder namely \("c"\) on division by \(81\).

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