# Thread: Divide by 9 Question

1. ## 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

2. Originally Posted by Joel
If n, a are positive integers, how many integers from 1 to n are divisible by [I]a?
Hi Joel.

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

Proof:

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

$\displaystyle 1\le ka\le n$

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

$\displaystyle \implies\ k<q+1$

$\displaystyle \implies\ k\le q$

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

And $\displaystyle \left\lfloor\frac na\right\rfloor=\left\lfloor q+\frac ra\right\rfloor=q$ as $\displaystyle q$ is an integer and $\displaystyle 0\le\frac ra<1.$ This completes the proof.

3. Originally Posted by Joel
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.

4. 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