“congruence modulo $7$” is an equivalence relation on $\mathbb Z.$ Find three elements in the equivalence class $[3].$ so $3$ is congruent to $mod\ 7$..
My attempt:
a = bq + r = 7(1) + 3 = 10 , .. 7(0) + 3 = 3, .. 7(2) + 3 = 17
Thus, three elements in $[3]$ are $\{ 10, 3, 17 \}$
Is this correct?
b) Again consider the equivalence class $[3]$ for the equivalence relation “congruence modulo $7$” on $\mathbb Z.$ Suppose that $S = \{1, 2,\dots,N\}$, where $N$ is a positive integer. Find all possible
values of $N$ so that $[3] \cap S$ contains exactly $10$ elements.
I know it should go from $1 – (mod \ -1)$ so there are $9$ values? I'm so confused
Best Answer
Yes for (a) U are right
Since $$10 \equiv 3 \pmod 7$$ and $$17 \equiv 3 \pmod 7$$ and $$3 \equiv 3 \pmod 7$$
Well for (b)
You first have to list the elements in $$[3] = \{3,10,17,24,31,..... \}$$
Notice that this list keeps going on and they are all have the form $3 + 7k$ where $k$ is an integer. For your question, We are only concerned with the positive values of $k$
Now the first positive value in $[3]$ is $3$ This is when $k=0$
Now the $10^{th}$ value in $[3]$ is when $k = 9$ which is $3 + 7(9) =66$
And so if $N = 66$ then $S \cap [3] = 10$ and this will hold true for all values of $N$ up to $66 + 7 -1 = 72$
And so $ 66\leq N \leq 72$
Notice that if $N = 73$ then $S \cap [3] = 11$ because $73 \in [3]$ and so that's why it is strictly less than 73
Take some time to understand this question ! It's pretty neat and will save you a lot of time in the future because it truly tests your understanding of congruences