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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.

  2. jcsd
  3. Aug 25, 2009 #2
    Wait, do you need to prove if
    [/tex] is divisible by 13 i.e


    Or you need to reduce


  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


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    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.

    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)

    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.


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


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

    Hence we proved it is divisible by 13.

  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


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


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    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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