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Find the remainder when $3^{29}$ is divided by $12$.

Since $12=3•2^2$, this can be simplified to $3^{28}/4$. And the units digit of powers of 3 follows the pattern of $3,9,7,1$, so we know that $3^{28}$'s units digit is going to be $1$. However, that doesn't help much as $3^{28}$ divided by 4 can have a remainder of 1 or 3. How can I solve this without a calculator (I am not allowed to use one)? I feel like I could use a modulo, but since I'm not that familiar with it, I'm not sure.

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  • $\begingroup$ Good job figuring out that you only need to deal with the remainder modulo $4$ (as long as you "lift" the answer to a remainder divisible by three modulo $12$). You can use the techniques described here. Personally I would vote to close this as a duplicate of either that or another suitable thread, but I have promised not to be the first voter. $\endgroup$ – Jyrki Lahtonen Sep 1 '17 at 15:26
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    $\begingroup$ Said @Jyrki "lifting" can be done formulaically using the mod Distributive Law (see my answer). This powerful method is not in said canned responses thread (nor can every possible insightful solution be, thanks to the richness of number theory and the ingenuity of the human mind). $\endgroup$ – Bill Dubuque Sep 1 '17 at 15:58
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we have $$3^5\equiv 3 \mod 12$$ thus by squaring $$3^{10}\equiv 9\mod 12$$ again by squaring $$3^{20}\equiv 81\equiv 9\mod 12$$ and since $$3^9\equiv 3 \mod 12$$ we get $$3^{29}\equiv 27\equiv 3 \mod 12$$

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    $\begingroup$ Brute force proofs are not enlightening for problems like this. $\endgroup$ – Bill Dubuque Sep 1 '17 at 15:25
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Note $\ 3^{\large 1+2n}\!\bmod\, 12\, =\, 3(3^{\large 2n}\!\bmod 4)\, =\, 3((-1)^{\large 2n}\!\bmod 4)\, =\, 3$
using $\phantom{I^{I^I}}\, ca\bmod cn\, =\, c\ (\,a\,\bmod\, n)\, =\,$ mod Distributive Law.

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$3^1\equiv3\mod12$, $3^2\equiv9\mod12$, and $3^3\equiv27\mod12\equiv3\mod12$.

If we try $3^4$, we already know that it will be $9\mod12$ because we've already tried multiplying $3\mod12$ by $3$ again, and building on top of that, we already know that $3^5\equiv3\mod12$. We can see that this pattern of alternating between $3$ and $9\mod12$ will continue forever.

We can come up with a general pattern. $3^k$ is $3\mod12$ for odd $k$ and $9\mod12$ for even $k$. Applying this to $3^{29}$, we know that it is $3\mod12$ becaue $29$ is odd. $\square$

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Since $12=3\times 4$ consider first $$3^{29}\equiv 0 \bmod 3$$ and $$3^{29}\equiv (-1)^{29}\equiv -1 \equiv 3\bmod 4$$

The Chinese Remainder Theorem guarantees a single common solution to these congriuences modulo $12$, and this is clearly $3$, though can be computed.


We find numbers $X$ and $Y$ with $X\equiv 1\bmod 3; 0\bmod 4$ and $Y\equiv 0\bmod 3; 1 \bmod 4$.

We have $X=3a+1=4b$ so we look for $4b-3a=1$. Also $Y=3c=4d+1$ so $3c-4d=1$. These are possible to solve by Bezout - we have in fact $4-3=1$ so $a=-c=1; b=-d=1$ will do. $X=4, Y=-3\equiv 9\bmod 12$.

If we then have $n\equiv r\bmod 3; s\bmod 4$ then $n\equiv Xr+Ys\equiv 4a+9b\bmod 12$ will work

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  • $\begingroup$ We can eliminate CRT and simplify using mod distributivity, e.g. see my answer. $\endgroup$ – Bill Dubuque Sep 1 '17 at 15:27
  • $\begingroup$ @BillDubuque Nicely done. I almost didn't mention CRT, and almost didn't post it, because it felt like overkill. But CRT gets mentioned a lot with very few clues as to how it works in practice. $\endgroup$ – Mark Bennet Sep 1 '17 at 15:35
  • $\begingroup$ I agree that how CRT works deserves better emphasis than is given in many textbooks. In particular CRT optimizations are rarely mentioned, e.g the mod distributivity version I used (which is often applicable and can lead to great simplifications for more complicated problems). $\endgroup$ – Bill Dubuque Sep 1 '17 at 15:49
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Alternatively: $$3^3\equiv 3 (mod \ 12)$$ $$(3^3)^9\equiv (3^3)^3\equiv 3 (mod \ 12)$$ $$(3^3)^9\cdot 3^2\equiv 3^3\equiv 3 (mod \ 12).$$

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