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how to find remainder when $13^{99}-19^{93}$ is divided by 162. I tried writing 19 as 18+ 1 and 13 as 18-5 but still not able to solve.

maveric
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4 Answers4

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$162=3^4\cdot2$

$$13^{99}=(1+12)^{99}\equiv1+\binom{99}112\pmod{162}$$

$$19^{93}=(1+18)^{93}\equiv1+\binom{93}118\pmod{162}$$

lab bhattacharjee
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  • +1. The choices $12=\color{red}2^2\cdot \color{red}3$ and $18=\color{red}2\cdot \color{red}3^2$ along with $1$ were efficient. – farruhota Sep 07 '19 at 09:06
  • @farruhota Even more efficient is to cube a few times, i.e. (Hensel) lift the exponent - see my answer. – Bill Dubuque Sep 07 '19 at 22:05
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Use the binomial theorem, along with the powers. But we have to be clever : to ensure in a binomial expansion that many terms are multiples of $162$ , we must ensure that the summands(i.e. the $x,y$ in $(x+y)^n$) either are divisors by $162$ or share factors in common, so that raising them to high powers end up capturing the required power of a prime factor they share with $162$, compensating for that prime factor. Note that $162 = 2 \times 3^4$ : it is enough to show that it is a multiple of $2$ and of $3^4 = 81$ then.

However, note that $13^{99}-19^{93}$ is even, so we only need to show that it is a multiple of $81$ to conclude. In this regard, all we need to do is note that $13=9+4$ and $19=27-8$ provide good binomial break ups since $9$ and $27$ are powers of $3$ whose squares have $81$ as a divisor.

So write $19^{93} = (27-8)^{93}$. By the binomial expansion : $$ 19^{93} = \sum_{i=0}^{93} \binom{93}{i}27^i8^{93-i} = \color{green}{\sum_{i=2}^{93} \binom{93}{i} 27^i8^{93-i}} + \mbox{ the rest} $$

where the green part is a multiple of $81$. The rest consists of $8^{93} + 27 \times 8^{92} \times 93$, the latter of which is a multiple of $81$. So we only have to compute $8^{93} \mod 81$. Use Euler's theorem to conclude that $8^{72} \equiv 1 \mod 81$, since $72 = \phi(81)$.

Now, you have to check $8^{21} \mod 81$, this you can do.

Your task : do the same for $13^{99}$ using the suggested expansion, and complete the question.

Sarvesh Ravichandran Iyer
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  • wont green summation be till 93? – maveric Sep 07 '19 at 06:29
  • even in rest part finally i m left with $8^{93}+ 27^{93}$?? – maveric Sep 07 '19 at 06:39
  • Actually, I am changing my answer a little bit to aid you : the other answer is not wrong, but there is a simplification. The green summation won't be till $93$, but that part is also getting changed now. Yes, I have done the edit. – Sarvesh Ravichandran Iyer Sep 07 '19 at 08:53
  • I would say "the other answer uses the same method, but simpler". However, +1 for the detailed explanation of the method. – farruhota Sep 07 '19 at 09:02
  • Oh, I was going to delete the answer! Certainly, the other method is far better in terms of capturing : I went for powers of $3$ hoping for quick capture, but then you realize that $99$ and $93$ also contribute powers of $3$ from their binomial expansions, so that does it. – Sarvesh Ravichandran Iyer Sep 07 '19 at 09:03
  • @астон Actually iteratiing powers of $3$ works nicely here - see my answer. – Bill Dubuque Sep 08 '19 at 02:27
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$\begin{align}{\bf Hint}\ \, &\bmod 3\!:\ \ \ \ 13^{\large 4}\ \equiv\ (-2)^{\large 10}\ \ {\rm by}\ \ \ 13\equiv 1\equiv -2\\ \overset{\rm\large cube}\Longrightarrow\ &\bmod 9\!:\ \ \ \ 13^{\large 12}\equiv (-2)^{\large 30} \\ \overset{\large -2\,\times}\Longrightarrow\ & \bmod 9\!:\ \ \ \ 13^{\large 11}\equiv (-2)^{\large 31}\ \ {\rm by}\,\ {-}2 \times {\rm prior,\,\ {-2}(13)\equiv 1} \\ \overset{\rm\large cube}\Longrightarrow\ &\bmod 27\!:\,\ 13^{\large 33}\equiv (-8)^{\large 31}\equiv 19^{\large 31}\\[.3em] \overset{\rm\large cube}\Longrightarrow\ &\bbox[6px,border:1px solid #c00]{\!\!\bmod 81\!:\ 13^{\large 99}\equiv\ 19^{\large 93}} \end{align}$

Bill Dubuque
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13^99=(18-5)^99 expanding binomially every term will have 18^2 except last term hence all terms %162 will result to sum of last term %162 which gives -5^99%162 now 5^99=125^33=(126-1)^33 again expanding this will have 18^2 in every terms except last two which is (33*126-1)%162 affectively we will be left with -107 eqivalent to 55 as remainder 13^99%162=55 similarly second part is relatively easier to handle 19^93=(18+1)^93 expanding again will have 18^2 for every term except last two terms resulting in 55 as remainder 19^93%162=55 hence answer has to be 0 so 13^99-19^93=0 MOD 162

Rohit Kumar
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