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How many numbers that are multiples of 5 divide 1000?

  • #1
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Homework Statement


Okay:

How many numbers divide 1000 that are multiples of 5

I have seen you do 1000/5 = 200

But how does this mean there are 200 numbers that divide 1000 that are multiples of 5?

This just says: 1000 divided into 5 equal pieces, is 200.

So how does this give how many numbers divide 1000 that are multiples of 5?

I just dont get the concept!

Homework Equations




The Attempt at a Solution


[/B]
I dont understand the conceptual part.
 
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Answers and Replies

  • #2
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Basically the question is asking how many divisors of 1000 are multiples of 5.
To make it clear, let me give a simple example - say now the problem is, how many numbers that divide 20 are multiples of 5. Well clearly these numbers are 5, 10 and 20, so the answer is 3. For this small example, of course, you can just eyeball the answer or write down all the divisors. For the problem you have, involving a much bigger number (1000), a useful starting point is to consider its prime factorisation.
 
  • #3
ehild
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200 is not the correct solution. It is true, that starting from zero, every fifth number is divisible by 5. They are 5, 10, 15, 20, 25, 30 35....
But these numbers are not all divisors of 1000. 1000 is not a multiple of 15 or 30 or 35...
 
  • #4
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Amad27,
Your attempt to use LaTeX to write numbers only succeeded in making them harder to read (such as $1000$). There is no real reason to write 1000 or other ordinary numbers using LaTeX markup. I edited your post to remove the $ symbols.
 
  • #5
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How do you suggest finding the answer then? Please suggest?
 
  • #6
ehild
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Fightfish gave you a hint already: consider the prime factorisation of 1000, write up all divisors, and select the ones which are multiple of 5.
 
  • #7
haruspex
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How do you suggest finding the answer then? Please suggest?
Suppose x divides 1000. What can its factorisation look like? What constraint is imposed on the exponents by the requirement that it is a multiple of 5? How many combinations of exponents are thereby allowed?
 
  • #8
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Lets see, this is the fundamental theorem of primes isnt this?

1000 = (2^3)(5^3)

4 ways to choose first constant, (0, 1, 2, 3) and 4 ways to choose second constant (0, 1, 2, 3)

But writing out all divisors would take too long what can be done? On the other hand we could try since 5| 1000

1000 = 0 (mod 5)

But that wont help
 
  • #9
ehild
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Lets see, this is the fundamental theorem of primes isnt this?

1000 = (2^3)(5^3)

4 ways to choose first constant, (0, 1, 2, 3) and 4 ways to choose second constant (0, 1, 2, 3)
You need the product of all possible combinations of 2,2,2, 5,5,5. Among them, at least one has to be 5. So you can choose 5 1, 2, 3 times.
But writing out all divisors would take too long what can be done?
How many are they? You do not really need to write them out, but writing some of them can help.
 
  • #10
haruspex
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4 ways to choose first constant, (0, 1, 2, 3) and 4 ways to choose second constant (0, 1, 2, 3)
How does that change when you include that it must be divisible by 5?
 
  • #11
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  • #12
haruspex
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@haruspex, and @ehild, I am not sure?

Can you help?
If you take any multiple of 5 and factorise it, what is the range of possible values for the exponent of the 5?
 
  • #13
ehild
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@Amad, why not write out a few divisors ? You would understand it at once. Is 5 a divisor? Is 10 a divisor? and 20? ....
 
  • #14
vela
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Lets see, this is the fundamental theorem of primes isnt this?

1000 = (2^3)(5^3)

4 ways to choose first constant, (0, 1, 2, 3) and 4 ways to choose second constant (0, 1, 2, 3)

But writing out all divisors would take too long what can be done?
There are only 16 divisors. It wouldn't take very long to write them out.
 
  • #15
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Keep in mind this is an SAT question. Time is money... Lets take a different problem.

How many numbers less than 1000 are multiples of 30?

1000 = (2^3)(5^3)

The answer is 33

With the factorization method, this simply seems impossible to do then? What to do in a general case?
 
  • #16
haruspex
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How many numbers less than 1000 are multiples of 30?

1000 = (2^3)(5^3)
There's no point in factorising 1000 here since you're not trying count numbers that divide 1000.
What's the largest number less than 1000 that's a multiple of 30?

Back on the original problem, it really is very easy from where you got to in post #8.
4 ways to choose first constant, (0, 1, 2, 3) and 4 ways to choose second constant (0, 1, 2, 3)
You don't need to write any list out. It is a trivial matter to get from that statement to the number of divisors of 1000.
The only other step you need to make is to figure out how those exponent ranges change when you restrict to those numbers which are also multiples of five. Again, it really is very easy, and it's hard to provide a hint that doesn't outright tell you the answer.
I don't know whether you tried to answer my question in post #12:
If you take any multiple of 5 and factorise it, what is the range of possible values for the exponent of the 5?
 
  • #17
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You have 1000=2^3 5^3. So how many combinations of these numbers contain at least a 5?

And about your second problem, you have the inequality 30n<1000. For how many ns this can be true?
 
  • #18
vela
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Keep in mind this is an SAT question. Time is money...
The suggestion to write out the divisors was to get you to realize the pattern that would allow you solve similar problems more efficiently. As haruspex noted, it's hard to give you more of a hint without telling you the answer outright.

For the second problem, you wouldn't use the factorization method to solve that problem.
 
  • #19
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5^3 contains, 5 but that only gives 3 divisors!

For the second one,

30n < 1000

n <= 33


But how do we know 30n will be a multiple of $30$?
Thanks!
 
  • #20
ehild
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30n is 30 multiplied by any integer n. Why do you ask if it is a multiple of 30?
 
  • #21
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@ehild, just checking. so n < 33.

All n less than 33 give a multiple of 30 < 1000 so that is the answer??
 
  • #22
ehild
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edit: all integers less or equal 33 give a multiple of 30 less then 1000. 30 n <1000 if n≤33 .
 
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  • #23
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Almost done, and this comes from the definition that a multiple of 30, $n$ is in the form:

$$30$$?

Thanks!
 
  • #24
ehild
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  • #25
haruspex
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5^3 contains, 5 but that only gives 3 divisors!
I don't understand your logic there. Why are you fixing on 53?
I asked, if you factorise a number that's a multiple of 5, what are the possible values for the exponent on the 5 in the factorisation?
Ignore the exponents on all the other prime factors.
 

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