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Math Help - Divide by 9 Question

  1. #1
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    Divide by 9 Question

    Hi All,

    I am struggling with this question.

    How many integers from 1 to 100 are divisible by 9? How many integers from 1 to 2000 are divisible by 9? If n, a are positive integers, how many integers from 1 to n are divisible by a?

    Is there a rule that can help me with this??

    Thanks
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Joel View Post
    If n, a are positive integers, how many integers from 1 to n are divisible by [I]a?
    Hi Joel.

    The answer is \left\lfloor\frac na\right\rfloor where \lfloor x\rfloor denotes the greatest integer not greater than x.

    Proof:

    Use the division algorithm and write n=qa+r where 0\le r<a. Now any integer divisible by a is of the form ka for some integer k. We have

    1\le ka\le n

    \implies\ ka\le qa+r<qa+a=(q+1)a

    \implies\ k<q+1

    \implies\ k\le q

    Conversely k\le q\ \implies\ 1\le ka\le qa. Hence there are exactly q integers from 1 to n which are divisible by a.

    And \left\lfloor\frac na\right\rfloor=\left\lfloor q+\frac ra\right\rfloor=q as q is an integer and 0\le\frac ra<1. This completes the proof.
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  3. #3
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    Quote Originally Posted by Joel View Post
    How many integers from 1 to 100 are divisible by 9? How many integers from 1 to 2000 are divisible by 9? If n, a are positive integers, how many integers from 1 to n are divisible by a?
    How many multiples of 9 are 100 or less? Divide 100 by 9 to get 11.111.... Then 9(11) = 99 is the largest multiple of 9 less than 100, and there are 11 multiples of 9 less than 100.

    How many multiples of 9 are 2000 or less? Divide 2000 by 9 to get 222.222... Then 9(222) = 1998 is the largest multiple of 9 less than 2000, and there are 222 multiples of 9 less than 2000.

    How many multiples of 9 are n or less? This is where you have to use that funky function thing. Same as for "a" and n. But you can see the general principal above.

    Note: If you're familiar with calculators, you might also refer to the "integer" function, which gives the whole-number part of the above divisions.

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    Thanks heaps. Your post not only gave me the answer, but when I am asked why this works in the future...thanks to you I have to tools to prove it. Cheers
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