Can anyone give a hint of how to go about solving this? Please don't give answer thanks
Find the integer $a$ such that $a \equiv 99 \pmod{41}$ and $100 \le a \le 140$
We did not go over this in class and can really use some start up ways. I know 99 mod 41 is 17 but what do you do after that?
Since you want a hint, $$a\equiv b\bmod m$$ Means that $$km=a-b,\text{ for some integer $k$.}$$ Thus, pick values of $k$ so that the resulting $a$ satisfies $100\leq a\leq140$.
Hint $\,\ a\equiv 99\pmod{41}\iff a = 99\!+\!41n,\ $ so $\ 100\,\le\, a = 99\!+\!41n\, \le\, 140\,\Rightarrow\, n=\,\ldots$
Recall the definition: $a \equiv b \mod n$ iff $n|(a-b)$, which means that there exists an integer $c$ such that $nc = a-b$.
The easy way to think about the modulo is that it just "wraps around": for example, for mod 35, you can count $0,1,2,3,4,\ldots,34,0,1,2,3\ldots,34,0,\ldots$. It's also good to note that for $x \equiv y \mod z$, $x$ is just the remainder of integer division of $y$ by $z$. For example, $35/14$ is equal to $2$ remainder $7$, and $7 \equiv 35 \mod 14$.
Anyway, that was just trying to provide a background. Back to your example:
(I see from your response to the other answer that you have figured it out. So I will now post a full solution.)
If you want to find the $a$ such that $100 \leq a \leq 140$ and $a \equiv 99 \mod 41$, you must have $41|(99-a)$, implying $41c = 99 - a$. Your least possible value for $a$ is $100$, and the greatest possible value for $a$ is $140$. So the right side of the equation is going to be between $99-100 = -1$ and $99-140 = -41$. So you have $-1 \geq 41c \geq -41$. Then you look for the possible integer values of $c$. For this range, there's only one: $c= -1$. Now that we know this, we have $41(-1) = 99 -a$, so we have $-41 = 99 - a$, so $-140 = -a$, so $140=a$.
The way I solved it is a little circuitous. Finding a value of $c$ only helps to narrow it down for more difficult problems; in this problem you could've tried to find $a$ straight-away.
