# Homework Help: Find the number of multiples of 7 between 30 and 300

1. May 21, 2004

### mustang

Problem 3. Find the number of multiples of 7 between 30 and 300.

Problem 4. How many 4-digits numbers are not divisible by 11?

Problem 5. True or false? If the sequence a, b, c is arithmetic, then the sequence 2^a, 2^b,2^c is geometric. Explain.

2. May 22, 2004

### AKG

What have you done so far, and what are you having trouble with? Here's a few things to consider:

How many multiples of 1 are there between 4 and 20 (including 4 and 20)?
How many multiples of 2 are there between 4 and 20 (you can count them)?
What is the relationship between numbers that are multiples of 11 (or any number for that matter) and numbers that are divisible by 11?
What is an arithmetic sequence and what is a geometric sequence?

For the first problem, consider the fact that the lowest multiple of 7 greater than 30 is 5 x 7 (or 35). The greatest multiple of 7 less than 300 is 42 x 7 (or 294).

For the second problem, realize that the number of 4-digit numbers not divisible by 11 is equal to (the number of 4-digit numbers) - (the number of 4-digit numbers that are divisble by 11).

For the third problem, you know that arithmetic sequences have a general form:

$$t_{n} = a + d(n - 1) \mbox{, where } t_{n} \mbox{ represents the } n^{th} \mbox{ term, } a \mbox{ represents the}$$
$$\mbox{first term, and } d \mbox { represents the difference between}$$
$$\mbox{two consecutive terms.}$$

Therefore, you can write:

$$a = a$$
$$b = a + d$$
$$c = a + 2d$$

So, given that, how can you write the geometric sequence?

3. May 22, 2004

### Mark

problem 3
We have a range between 30, and 300
So lets first try a range of 0 and 300
there are 40 divisibles of 7 to 280, plus an addictional 2 (287,294). Thus 42 multiples.

now lets compensate for the range beggining at 30 instead of 0

there are 4 multiples of 7 from 0 - 30 (7,14,21,28)
thus (multiples of 0-300) - (multiples of 0-30) = correct number of multiples
42 - 4 = 38
Therefore there are 38 multiples between 30 and 300.

problem 4
Again, we will start with a more simple range, lets use 0 - 9999
9999/11 = 909, thus there are 909 numbers divisible by 11 in the 0-9999 range.
Again, we will now compensate for the first chunk of numbers
range (0-1000) --> 999/11 = 90.9
(range 0-9999 multiples) - (range 0-999) = answer
909 = 90.9 = 818.8 multiples

problem 5
substitute for numbers, say a=1, b=2, c=3

1,2,3......arithmatic
2^1 , 2^2 , 2^3.......... geometric

it doesn't matter what numbers we inserted, they will all be evenly spaced, since a,b,c is arithmatic. Therefore 2^a... etc. is geometric.

you mine are still greater."

EDIT: AKG, we posted at the same time...im not gonna bother to see if i was right.....but yea, whatever, sleep time :)
Hope this helps
Mark

4. May 22, 2004

### AKG

Well, seeing as how you've been given the answer already anyways, I guess I'll give you nicer ways to go about it.

For the first example, I said that the lowest multiple of 7 in the given range was 5 x 7, and the greatest was 42 x 7. So, for each number from 5 to 42, there is a multiple of 7 in that range. So, how many numbers are there from 5 to 42, inclusive ("inclusive" means including the end numbers, 5 and 42). Well, that's obvious:
42 - 4 = 38.

Problem 2
How many 4-digit numbers are there? That's the same as the numbers from 1000 to 9999, inclusive. Clearly, there are: 9999 - 999 = 9000. Now, how many multiples of 11 are in that range?

Start by dividing 1000/11. You get 90.909090.... So the first actual multiple of 11 in that range would be 91 x 11. The last number would be found like this: 9999/11 = 909. Since 909 is a nice whole number, we don't have to do any rounding. So, the number of multiples of 11 correspond to the numbers from 91 to 909, which is 909 - 90 = 819. So, the final answer is 9000 - 819 = 8181.

GENERAL APPROACH
The number of multiples of N between A and B, where B > A is:
floor(B/N) - ceil(A/N) + 1.

Problem 3
If the arithmetic sequence progress as follows {a, a + d, a + 2d...}, then the sequence in the form {2^a, 2^b, 2^c...} will progress as follows: {2^a, 2^(a + d), 2^(a + 2d)...}. Now, let r = 2^d, and A = 2^a. We can equivalently describe the sequence as follows: {A, Ar, Ar^2, ...}. In general, we can express this sequence as {Ar^(n-1)}, which is the general form of a geometric series. Therefore, the answer is "true" and we have explained what we set out to explain.

5. May 22, 2004

### Mark

I was given no answers....I saw a post with 0 responses, typed a response, and replied, by the time it posted, yours was before mine, thus we produced two different ways to solve the same problems, thus verifying that correct answers for each of the 3 questions.

But yes, your approach was much cleaner, and makes more sense, good choice.

Anyway, the question is solved lol

Mark