-2

This is what I've done to solve it:

Since $k$ is congruent to $2 \mod 4$, any number in the set $\{...,-10,-6,-2,2,4,18,22,...\}$ will solve the original congruence. By simple algebra, if $5k + 13$ is divided by $4$, then if $k \leq -6$ the remainder is $0.25$, and if $k$ is greater than $-6$ the remainder is $0.75$.

This is all I have for my solution. Does it seem like the correct way to answer the question? Is it complete?

I ALSO have a part $2$ to this question, which asks "Given that $k$ is congruent to $1 \mod 4$, determine the remainder when $7k^{333}+11$ is divided by $4$":

Again as above, I determine that any number in the set $\{...,-9, -5, 1, 5, 9 , ...\}$ will solve the original congruence. But I am not sure where to go from here, as I cannot calculate that large of a number using my previous example.

Any help?

Bill Dubuque
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Brittaney
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  • One usually writes $k=2\mod 4$ or $k\equiv 2 \mod 4$, the last notation being due to Gauss. – Pedro Nov 26 '13 at 02:25
  • thanks, will write it as such from now on – Brittaney Nov 26 '13 at 02:41
  • $!\bmod 4!:\ \color{#0a0}{5\equiv 1},,\color{#c00}{k\equiv2},\color{#0af}{13\equiv 1}\Rightarrow \color{#0a0}5\color{#c00}k+\color{#0af}{13}\equiv \color{#0a0}1\cdot\color{#c00}2+\color{#0ad}1\equiv 3$ by the linked congruence Sum & Product Rules. Similarly for the 2nd problem using $,k\equiv 1\Rightarrow k^n\equiv 1^n,$ by the Congruence Power Rule. $\ \ $ – Bill Dubuque Oct 11 '24 at 18:11

2 Answers2

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HINT (for the first question): $$ 5k \equiv k \, (\bmod \, 4) \quad \text{and} \quad 13 \equiv 1 \, (\bmod \, 4) $$ Combining this with the given information $k \equiv 2 \, (\bmod \, 4)$, what can you say about $5k + 13 \, (\bmod \, 4)$?

HINT (for the second question): $$ k \equiv 1 \, (\bmod \, 4) \implies k^n \equiv 1 \, (\bmod \, 4) $$ for any nonnegative integer $n$.

tylerc0816
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  • Taking your suggestion I continue as follows: 13≡1(mod 4) will give me 12R0. Working with 5k≡k(mod 4) I get: 5k-k≡0(mod4)... k(5-1)≡0(mod4)... k(4)≡0(mod4)... so k=0 (keeping in mind that k-2 must be divisible by 4) Combining all of this information: plug k=0 into the equation 5k+13, and when divided by 4 I am left with 3.25 (a remainder of 25). What do you think? – Brittaney Nov 26 '13 at 02:40
  • $5k \equiv k (\bmod 4)$ does not imply $k \equiv 0 (\bmod 4)$, rather, it implies $4k \equiv 0 (\bmod 4)$. This would be a contradiction to the assumptions, anyway, as $k \equiv 2 (\bmod 4)$. Instead, try to use the information in the hint to see why $5k + 13 \equiv k + 1 \equiv 2 + 1 \equiv 3 (\bmod 4)$. – tylerc0816 Nov 26 '13 at 02:43
  • i see where you're going. I think it's stuck in my head to treat these as equivalence relations, which they aren't exactly. Will work it out, thanks for the hints. math isn't my strong suit but you've helped :) – Brittaney Nov 26 '13 at 02:49
  • Actually, reduction mod $n$ is an equivalence relation (which is a good exercise!). Also, it may help to try plugging in actual numbers for $k$ to get a better feel of how this operation works. For instance, see what happens when you put in $k = 2$ or $k = -2$. – tylerc0816 Nov 26 '13 at 02:56
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The trick is, you can rewrite $k\equiv2~\text{mod}~4$ as $k=4m+2$, for some integer $m$. It follows:

$$5k+13=5(4m+2)+13=20m+23$$

Taking $\text{mod}~4$, we have:

$$20m+23\equiv(0+3)~\text{mod}~4\equiv\boxed{3}~\text{mod}~4$$