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 ?

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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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. – 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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– Nishant Sharma · 4 years, 2 months ago

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.Log in to reply

– Superman Son · 4 years, 2 months ago

it goes for 81 alsoLog in to reply

– Superman Son · 4 years, 2 months ago

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

– Anoopam Mishra · 4 years, 2 months ago

show that 19^93 - 13^99 is a positive integer divisible by 162Log in to reply

– Superman Son · 4 years, 2 months ago

thanksLog in to reply

– Superman Son · 4 years, 2 months ago

right but first prove 19^93>13^99Log in to reply

– Shourya Pandey · 4 years, 2 months ago

not needed to show the greater than sign.Log in to reply

– Superman Son · 4 years, 2 months ago

yes if we do bionomial then not but otherwise Show that 19^93 - 13^99 is a positive integerLog in to reply

– Soham Chanda · 4 years, 2 months ago

inequality korchhis bole sob jaegae inequality? positivity or negativity is not at all needed as the problem is asking to prove divisibility.Log in to reply

– Superman Son · 4 years, 2 months ago

hmm thikie bolecho go sohomdaLog in to reply

– Shourya Pandey · 4 years, 2 months ago

But why is it required?Log in to reply

– Superman Son · 4 years, 2 months ago

yeah i got my mistakeLog in to reply