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I feel a bit stupid

  1. Aug 25, 2009 #1
    So my mother mentioned to a co-worker, who is a "math guy," that I was interested in math. The co-worker subsequently gave her a piece of paper with the following written on it to give me to to "solve" or "prove" in my mother's words, yet I am unsure what to do. Here is what was on the paper:

    [tex]3^{2010}+15^{2010} \div 13[/tex]

    I figured the person wanted me to reduce this in some way as it is not an equation. The only thing I think I could prove is the reduction if I could figure out how to do it.


    My first thought was to try and reduce [tex]15^{2010}/13[/tex] in some way be solving the equation [tex]15^{2010/ x}/13=y[/tex]but I could not find a whole number that worked for x that gave a whole number for y so I am a bit stuck...

    Any help? I feel a bit stupid because it seems to be simple, yet I am having trouble figuring it out. Also, I don't really know if reducing this is what the person wanted me to do.

    Thanks
     
  2. jcsd
  3. Aug 25, 2009 #2
    Wait, do you need to prove if
    [tex]
    3^{2010}+15^{2010}
    [/tex] is divisible by 13 i.e

    [tex]13|3^{2010}+15^{2010}[/tex]

    Or you need to reduce
    [tex]
    3^{2010}+\frac{15^{2010}}{13}
    [/tex]

    ?

    Regards
     
  4. Aug 25, 2009 #3
    I believe your mother's friend wants you to "explore" math. The exponents lead me to believe he wants to introduce you to the properties of logarithms.

    Edit: Eh, its not as simple as I was thinking. Forgive me.
     
    Last edited by a moderator: Aug 25, 2009
  5. Aug 25, 2009 #4

    sylas

    User Avatar
    Science Advisor

    If this is meant to be an integers problem, then possibly you are being asked if 13 divides the sum. It does. Can you prove it? In other words, consider
    [tex]\frac{3^{2010} + 15^{2010}}{13}[/tex]​

    Can you prove that this is an integer?

    Here is a hint. Consider successive powers of 5, modulo 13.

    [tex]\begin{align*}
    5^0 & \equiv 1 (\mod 13) \\
    5^1 & \equiv 5 (\mod 13) \\
    5^2 & \equiv 12 (\mod 13) \\
    5^3 & \equiv 8 (\mod 13) \\
    5^4 & \equiv 1 (\mod 13) \\
    5^5 & \equiv 5 (\mod 13) \\
    5^6 & \equiv 12 (\mod 13) \\
    5^7 & \equiv 8 (\mod 13)
    \end{align*}[/tex]​

    And so on. (You can prove this if you need to).

    Cheers -- sylas
     
  6. Aug 25, 2009 #5
    But we don't know if that is the task.

    Anyways

    [tex]
    13|3^{2010}+15^{2010}=3^{2010}+5^{2010}\cdot 3^{2010}=3^{2010}(1+5^{2010})
    [/tex]
    [tex]
    3^3 & \equiv 1 (\mod 13) \\
    [/tex]
    hence
    [tex]3^{2010} \equiv 1^{670} (\mod 13)[/tex]

    [tex]5^{2} \equiv -1 (\mod 13)[/tex]

    [tex]5^{2010} & \equiv (-1)^{1005} (\mod 13) \\[/tex]

    [tex]5^{2010} + 1 & \equiv (-1)+1 (\mod 13) \[/tex]

    hence

    [tex]3^{2010}(5^{2010} + 1) & \equiv 0 (\mod 13) \[/tex]

    Hence we proved it is divisible by 13.

    Regards.
     
  7. Aug 25, 2009 #6
    So here is exactly what was written:

    [tex]3^{2010} + 15^{2010} \vdots 13[/tex]

    The three dots looked like a division symbol on the paper

    Also, maybe I should learn what mod means?

    Any suggestions on texts that could help me with that?
     
  8. Aug 25, 2009 #7
    Mod just mean's Modulo. Basically when you say x mod y it means that as you increase x, once it gets to y it reverts to zero. Think of a clock, which is expressed in mod 12. The only thing of note is that a clock has no zero but starts at one so let's just subtract 1 from everything and pretend time goes 0 o' clock, 1 o'clock (which is actually 2 o'clock) and so on until 11 o' clock (which is actually 12 o' clock). What is one hour after 11 o' clock in our scheme? Well we're back to 0 o' clock. Mathematically we say 11 mod 12 + 1 = 0 mod 12. So what's 7 mod 12 + 10? (10 hours after 7 o'clock)? It's 5 o' clock. You can think of mod as "clock arithmetic"
     
  9. Aug 25, 2009 #8

    DaveC426913

    User Avatar
    Gold Member

    Never thought of it that way. I like that.
     
  10. Aug 26, 2009 #9

    sylas

    User Avatar
    Science Advisor

    I've never seen that notation. A standard notation is
    [tex]a \;\; | \;\; b[/tex]​
    which means "a divides b", or alternatively, a is a factor of b. So my guess at what your problem means could have been written
    [tex]13 \;\; | \;\; 3^{2010} + 15^{2010}[/tex]​
    The order is important, and your notation is in the reverse order to this.

    The other common notation that I used in my first reply is
    [tex]a \equiv b (\mod c)[/tex]​
    which has the same meaning as
    [tex]c \;\; | \;\; a-b[/tex]​
    or, a and b have the same remainder when divided by c.

    My reading of your problem could be written in this notation as
    [tex]3^{2010} + 15^{2010} \equiv 0 ( \mod 13)[/tex]​

    Sometimes the brackets are omitted. This uses three dashes, but requires the additional 0 in there.

    A page introducing modular arithmetic is here: Modular Arithmetic, or see the wikipedia article.

    But I'm still only guessing that this was the intent of the original question.

    Cheers -- sylas
     
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