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Number Theory Problem

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 ?

Note by Nishant Sharma
4 years, 2 months ago

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What 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. Shourya Pandey · 4 years, 2 months ago

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Another method is to note that \(13^2 \equiv 7 \pmod {162}\). Shourya Pandey · 4 years, 2 months ago

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latex not loading proprly pls type it Superman Son · 4 years, 2 months ago

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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 \(81\). Aditya Parson · 4 years, 2 months ago

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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 \(9\). Vikram Waradpande · 4 years, 2 months ago

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@Vikram Waradpande 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. Nishant Sharma · 4 years, 2 months ago

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@Nishant Sharma it goes for 81 also Superman Son · 4 years, 2 months ago

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@Vikram Waradpande hey vikram bhaiya please mujhe yei problem latex chor kar likh do mere computer par load nehi ho raha latex Superman Son · 4 years, 2 months ago

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@Superman Son show that 19^93 - 13^99 is a positive integer divisible by 162 Anoopam Mishra · 4 years, 2 months ago

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@Anoopam Mishra thanks Superman Son · 4 years, 2 months ago

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@Vikram Waradpande right but first prove 19^93>13^99 Superman Son · 4 years, 2 months ago

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@Superman Son not needed to show the greater than sign. Shourya Pandey · 4 years, 2 months ago

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@Shourya Pandey yes if we do bionomial then not but otherwise Show that 19^93 - 13^99 is a positive integer Superman Son · 4 years, 2 months ago

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@Superman Son inequality korchhis bole sob jaegae inequality? positivity or negativity is not at all needed as the problem is asking to prove divisibility. Soham Chanda · 4 years, 2 months ago

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@Soham Chanda hmm thikie bolecho go sohomda Superman Son · 4 years, 2 months ago

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@Superman Son But why is it required? Shourya Pandey · 4 years, 2 months ago

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@Shourya Pandey yeah i got my mistake Superman Son · 4 years, 2 months ago

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