Results 1 to 13 of 13

Math Help - Finding the remainder

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    45

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    769
    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?
    Last edited by mr fantastic; July 15th 2010 at 06:20 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    45
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    769
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    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:
    \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:
    \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
    5n = 35q + 10

    Divide 5n by 7:
    \begin{aligned}<br />
\frac{5n}{7} &= \frac{35q + 10}{7} \\<br />
&= \frac{35q}{7} + \frac{10}{7} \\<br />
&= 5q + \frac{10}{7} \\<br />
&= (5q + 1) + \frac{3}{7} \\<br />
\end{aligned}
    The quotient would be 5q + 1, and the remainder would be 3.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2009
    Posts
    45
    Thank you so much!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Another, easier way to do it is to let n \equiv 2 \pmod{7}, it then follows that 5n \equiv 10 \equiv 3 \pmod{7}, thus the remainder is 3.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Aug 2009
    Posts
    45
    Quote Originally Posted by Bacterius View Post
    Another, easier way to do it is to let n \equiv 2 \pmod{7}, it then follows that 5n \equiv 10 \equiv 3 \pmod{7}, thus the remainder is 3.
    I cannot find the MOD command on my calculator though.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    It really is the remainder function, usually calculators don't have it. You can emulate a \mod b by dividing a by b, taking the fractional part only and multiplying it by b. For instance, to get 10 \mod 7 :

    \frac{10}{7} = 1.428571429 ...
    Take the fractional part which is 0.428571429, and times it by 7, you get :
    0.428571429 \times 7 = 3
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Aug 2009
    Posts
    45
    Quote Originally Posted by Bacterius View Post
    Another, easier way to do it is to let n \equiv 2 \pmod{7}, it then follows that 5n \equiv 10 \equiv 3 \pmod{7}, thus the remainder is 3.
    Quote Originally Posted by Bacterius View Post
    It really is the remainder function, usually calculators don't have it. You can emulate a \mod b by dividing a by b, taking the fractional part only and multiplying it by b. For instance, to get 10 \mod 7 :

    \frac{10}{7} = 1.428571429 ...
    Take the fractional part which is 0.428571429, and times it by 7, you get :
    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?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    If n \equiv 2 \pmod{7} then \mathbf{5} \times n \equiv \mathbf{5} \times 2 \equiv 10 \equiv 3 \pmod{7}
    This is where the 10 comes from

    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 13 ... for instance if you substract 7 from 23, you get 16, which isn't exactly the remainder of 23 divided by 7, while 23 \mod 7 = 2 as expected. Then you could argue that repeatedly substracting 7 is a way to do it, true, but dividing is then more efficient.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Aug 2009
    Posts
    45
    I totally understand now. Thank you!
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,657
    Thanks
    598
    Hello, Mariolee!

    A slightly different approach . . .


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

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

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


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

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


    Therefore, the remainder is 3.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the remainder
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: April 20th 2011, 04:24 AM
  2. finding remainder
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 7th 2009, 10:57 AM
  3. finding the remainder
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 1st 2009, 03:16 AM
  4. finding P(x) and the remainder
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 14th 2009, 02:02 AM
  5. finding the remainder by mod
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 2nd 2008, 07:31 AM

Search Tags


/mathhelpforum @mathhelpforum