Dear math forum members,

I have a problem

How many integers divisble by 7 can you find in the interval [4500, 7000] ?

Could you please give me just a hint on how to proceed?

Thank you in advance!

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- Mar 16th 2008, 02:43 AMCoachSequences and series
Dear math forum members,

I have a problem

How many integers divisble by 7 can you find in the interval [4500, 7000] ?

Could you please give me just a hint on how to proceed?

Thank you in advance! - Mar 16th 2008, 02:57 AMMoo
Hello,

Firstly, i'll note that 7000 is a multiple of 7.

So we'll start from there.

The previous multiple of 7 will be obtained by substracting 7 to 7000, and so on.

So if you divide the interval into intervals of 7 numbers, ]a;b], these interval will contain one and only one multiple of 7.

Hence you "just" have to see how many intervals of this sort there are in [4500;7000], and this number is given by (7000-4500)/7. - Mar 16th 2008, 03:02 AMCoach
Except, that the number comes out as a decimal...

- Mar 16th 2008, 03:14 AMMoo
I know :-)

But it's as if you could restrict your study to the interval [first multiple of 7 coming after 4500 = N; 7000]

The number of multiples of 7 in [N;7000] will be (7000-N)/7 + 1 (because you count the extremity).

And this number is the same as the number of multiples of 7 in [4500;7000] because there is no multiple of 7 between 4500 and N.

So this shows that if you truncate (7000-4500)/7 by the inferior value (because (7000-4500)/7 = (7000-N)/7 + APositiveNumber) and add 1, you'll have the number you want.

This is the same as taking the superior integer of (7000-4500)/7

I don't know if it's clear enough (Blush)

You can take a smaller example, such as [10;35] - Mar 16th 2008, 06:06 AMSoroban
Hello, Coach!

Quote:

How many integers divisble by 7 can you find in the interval [4500, 7000] ?

Since every seventh number is divisible by 7,

. . there are: .$\displaystyle \frac{7000}{7} \:=\:1000$ of them on the interval [0, 7000]

We must eliminate the multiple that are less than 4500.

How many are there?

. . There are: .$\displaystyle \frac{4500}{7} \:=\:642.857...\:\Rightarrow\:642$ of them.

Therefore, there are: .$\displaystyle 1000 - 642 \:=\:358$ multiples of 7 on [4500, 7000]

- Mar 16th 2008, 08:59 AMPlato
Coach, do you know the

*floor*function (aka: the greatest integer function)?

The floor of x, $\displaystyle \left\lfloor x \right\rfloor $, equals the largest integer which does exceed x.

Some examples: $\displaystyle \left\lfloor \pi \right\rfloor = 3\,,\,\left\lfloor { - e} \right\rfloor = - 3\,\& \,\left\lfloor 3 \right\rfloor = 3$.

Most calculators have such a function built in.

For positive integers $\displaystyle d\,\& \,N,\,d < N$ the number of multiples ofin $\displaystyle [1,N]$ is $\displaystyle \left\lfloor {\frac{N}{d}} \right\rfloor $.**d**

Thus to use a calculator to solve your problem we would calculate:

$\displaystyle \left\lfloor {\frac{{7000}}{7}} \right\rfloor - \left\lfloor {\frac{{4499}}

{7}} \right\rfloor $.