# Sequences and series

• March 16th 2008, 02:43 AM
Coach
Sequences 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?

• March 16th 2008, 02:57 AM
Moo
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.
• March 16th 2008, 03:02 AM
Coach
Except, that the number comes out as a decimal...
• March 16th 2008, 03:14 AM
Moo
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]
• March 16th 2008, 06:06 AM
Soroban
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: . $\frac{7000}{7} \:=\:1000$ of them on the interval [0, 7000]

We must eliminate the multiple that are less than 4500.
How many are there?
. . There are: . $\frac{4500}{7} \:=\:642.857...\:\Rightarrow\:642$ of them.

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

• March 16th 2008, 08:59 AM
Plato
Coach, do you know the floor function (aka: the greatest integer function)?
The floor of x, $\left\lfloor x \right\rfloor$, equals the largest integer which does exceed x.
Some examples: $\left\lfloor \pi \right\rfloor = 3\,,\,\left\lfloor { - e} \right\rfloor = - 3\,\& \,\left\lfloor 3 \right\rfloor = 3$.
Most calculators have such a function built in.

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

Thus to use a calculator to solve your problem we would calculate:
$\left\lfloor {\frac{{7000}}{7}} \right\rfloor - \left\lfloor {\frac{{4499}}
{7}} \right\rfloor$
.