# Finding the remainder

• Jul 15th 2010, 11:31 AM
Mariolee
Finding the remainder
When the positive integer n is divided by 7, the remainder is 2. What is the remainder when 5n is divided by 7?

How would you set this up?
• Jul 15th 2010, 11:50 AM
wonderboy1953
The first step is to set up 5n/7 = 5/7 + n/7 so to strictly answer your question, except where n is a multiple of 7, then the remainder is n/7. Is this what you're asking about?
• Jul 15th 2010, 12:02 PM
Mariolee
No. These are the choices:

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

The answer is 3. I have no freaking clue how they got that. :D
• Jul 15th 2010, 12:19 PM
wonderboy1953
I think I understand the problem now. When n is 9, then division by 7 leaves a remainder of 2. Now 5 x 9 = 45. Try dividing 45 by 7 and see if you get a remainder of 3.
• Jul 15th 2010, 01:40 PM
eumyang
Suppose you divide a number N by P, and you get a quotient Q with a remainder R. This can be rewritten in fraction form:
$\displaystyle \frac{N}{P} = Q + \frac{R}{P}$
(remember that 0 <= R < P).

In our case, we are dividing n by 7. You get a quotient q with a remainder of 2, or:
$\displaystyle \frac{n}{7} = q + \frac{2}{7}$
Now I want to know what 5n is. Multiply both sides of the equation above by 35 and you'll get
$\displaystyle 5n = 35q + 10$

Divide 5n by 7:
\displaystyle \begin{aligned} \frac{5n}{7} &= \frac{35q + 10}{7} \\ &= \frac{35q}{7} + \frac{10}{7} \\ &= 5q + \frac{10}{7} \\ &= (5q + 1) + \frac{3}{7} \\ \end{aligned}
The quotient would be 5q + 1, and the remainder would be 3.
• Jul 15th 2010, 03:05 PM
Mariolee
Thank you so much!
• Jul 15th 2010, 04:36 PM
Bacterius
Another, easier way to do it is to let $\displaystyle n \equiv 2 \pmod{7}$, it then follows that $\displaystyle 5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $\displaystyle 3$.
• Jul 15th 2010, 05:21 PM
Mariolee
Quote:

Originally Posted by Bacterius
Another, easier way to do it is to let $\displaystyle n \equiv 2 \pmod{7}$, it then follows that $\displaystyle 5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $\displaystyle 3$.

I cannot find the MOD command on my calculator though.
• Jul 15th 2010, 05:44 PM
Bacterius
It really is the remainder function, usually calculators don't have it. You can emulate $\displaystyle a \mod b$ by dividing $\displaystyle a$ by $\displaystyle b$, taking the fractional part only and multiplying it by $\displaystyle b$. For instance, to get $\displaystyle 10 \mod 7$ :

$\displaystyle \frac{10}{7} = 1.428571429 ...$
Take the fractional part which is $\displaystyle 0.428571429$, and times it by $\displaystyle 7$, you get :
$\displaystyle 0.428571429 \times 7 = 3$
• Jul 15th 2010, 08:23 PM
Mariolee
Quote:

Originally Posted by Bacterius
Another, easier way to do it is to let $\displaystyle n \equiv 2 \pmod{7}$, it then follows that $\displaystyle 5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $\displaystyle 3$.

Quote:

Originally Posted by Bacterius
It really is the remainder function, usually calculators don't have it. You can emulate $\displaystyle a \mod b$ by dividing $\displaystyle a$ by $\displaystyle b$, taking the fractional part only and multiplying it by $\displaystyle b$. For instance, to get $\displaystyle 10 \mod 7$ :

$\displaystyle \frac{10}{7} = 1.428571429 ...$
Take the fractional part which is $\displaystyle 0.428571429$, and times it by $\displaystyle 7$, you get :
$\displaystyle 0.428571429 \times 7 = 3$

Oh, ok, but where did you get the 10 from? Especially the 10 in the first post here.
Also, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?
• Jul 15th 2010, 08:45 PM
Bacterius
If $\displaystyle n \equiv 2 \pmod{7}$ then $\displaystyle \mathbf{5} \times n \equiv \mathbf{5} \times 2 \equiv 10 \equiv 3 \pmod{7}$
This is where the $\displaystyle 10$ comes from (Nod)

Quote:

Also, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?
That would only work for numbers up to $\displaystyle 13$ ... for instance if you substract $\displaystyle 7$ from $\displaystyle 23$, you get $\displaystyle 16$, which isn't exactly the remainder of $\displaystyle 23$ divided by $\displaystyle 7$, while $\displaystyle 23 \mod 7 = 2$ as expected. Then you could argue that repeatedly substracting $\displaystyle 7$ is a way to do it, true, but dividing is then more efficient.
• Jul 15th 2010, 08:53 PM
Mariolee
I totally understand now. Thank you!
• Jul 16th 2010, 03:34 AM
Soroban
Hello, Mariolee!

A slightly different approach . . .

Quote:

When the positive integer $\displaystyle n$ is divided by 7, the remainder is 2.
What is the remainder when $\displaystyle 5n$ is divided by 7?

"When $\displaystyle n$ is divided by 7, the remainder is 2."

. . Hence: .$\displaystyle n \:=\:7a + 2\:\text{ for some integer }a.$

Then: .$\displaystyle 5n \:=\:35a + 10$

. . and: .$\displaystyle \dfrac{5n}{7} \:=\:\dfrac{35a+10}{7} \;=\;5a + 1 + \frac{3}{7}$

Therefore, the remainder is 3.