Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ≡ 99 (mod 41) and 100 ≤ a ≤ 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?

share|cite|improve this question
Well now you know $a-17$ is a multiple of $41$ – David H Mar 29 '14 at 22:35

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 $100\leq a\leq140$.

share|cite|improve this answer
Thanks that makes sense now. 140 is the answer :) – user2166592 Mar 29 '14 at 22:31

Hint $\,\ a\equiv 99\pmod{41}\iff a = 99\!+\!41n,\ $ so $\ 100\,\le\, a = 99\!+\!41n\, \le\, 140\,\Rightarrow\, n=\,\ldots$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.