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Subject: "number of dividers"     Previous Topic | Next Topic
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peter (Guest)
guest
Feb-12-01, 03:17 PM (EST)
 
"number of dividers"
 
   hello,

my teacher gave me a task to solve, but I don't know how to start:

"Proove that when watching every positive entire number the number of
dividers ending with 1 or 9 is not smaller than the number of dividers
ending with 3 or 7."

I think the problem is that you also have to take into consideration also the
dividers like 31, 43, 59 and so on.

anyone who got an idea?

greetings,
peter


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  Subject     Author     Message Date     ID  
  RE: number of dividers alexb Feb-12-01 1
     RE: number of dividers peter (Guest) Feb-26-01 2
  RE: number of dividers peter (Guest) Feb-26-01 3
     RE: number of dividers alexb Feb-27-01 4
         RE: number of dividers peter (Guest) Feb-27-01 5
             RE: number of dividers alexb Feb-27-01 6
  RE: number of dividers LilSmall202001 Jul-30-01 7

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alexb
Charter Member
672 posts
Feb-12-01, 04:12 PM (EST)
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1. "RE: number of dividers"
In response to message #0
 
   LAST EDITED ON Feb-12-01 AT 04:17 PM (EST)

Well, you should start somewhere. Forget about unwieldy numbers like 31, 43, etc. Think of smaller ones: 3, 7, 9, 13, 17, 19. The importance of trying by hand some examples can't be overestimated. I'll give one example: 21. 21 has four divisors: 1, 3, 7, 21. Two end in 1 (or 9) and two end in 3 or 7. Take another example: 273. 273 = 3򊐙3. 273 has the following divisors: 1, 3, 7, 13, 21, 39, 91, 273. Of these, 4 end in 1 or 9, another 4 end in 3 or 7.

Let's also try 4641 = 3򊐙317. 4641 has the following factors: 1, 3, 7, 13, 17, 21, 39, 51, 91, 119, 221, 273, 357, 663, 1547, 4641. (Check this. First take factors 3, 7, 13, 17 one by one. Then by twos, then by threes.) Of these 8 end in 1 or 9, and 8 end in 3 or 7.

I suggest you also consider, say, 819 = 3򉁯13.

What's important is to understand that every pair of factors, of which one ends in 3 and the other in 7, produces a factor that ends in 1 - just take their product. But there may be factors that end in 1 that can't be obtained in this manner. 11 is one example.

Let N1, N3, N7, N9 be the number of factors of a given number that end respectively in 1, 3, 7, and 9. Then, because of the previous paragraph

N13種7.

How can you obtain factors that end in 9? You may take by twos either factors that end in 3 or factors that end in 7. There are N3(N3 - 1)/2 of the first kind and N7(N7 - 1)/2 of the second. Thus we obtain that

N9 N3(N3 - 1)/2 + N7(N7 - 1)/2.

We may add up the two inequalities:

N1 + N9 N37 + N3(N3 - 1)/2 + N7(N7 - 1)/2.

Or written a little differently (check this):

N1 + N9 (N3 + N7)(N3 + N7 - 1)/2

Look at the function y = x - 1)/2. Can you compare it to x? I.e., when x - 1)/2 x?. I claim that this happens for x3. You can prove this algebraically or drawing the graph of f. This means that, provided

N3 + N7 3,

we indeed have

N1 + N9

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peter (Guest)
guest
Feb-26-01, 11:46 PM (EST)
 
2. "RE: number of dividers"
In response to message #1
 
   thanks for your immediate help ! My teacher was very impressed.
Your approach seems to be the easiest one.
However, I think that the equation for N1 could even be:

N1 = 1 + N3 + N7

because every number can be divided by 1.

thanks again,
Peter


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peter (Guest)
guest
Feb-26-01, 11:46 PM (EST)
 
3. "RE: number of dividers"
In response to message #0
 
   i think the inequation

N1 >= N3 * N7 is incorrect.

If you take the product of a pair of divisors, you might get a number which is higher than the considered positive integer.

Am I wrong?

Thanks in advance,
peter


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alexb
Charter Member
672 posts
Feb-27-01, 00:03 AM (EST)
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4. "RE: number of dividers"
In response to message #3
 
   LAST EDITED ON Feb-27-01 AT 00:05 AM (EST)

>i think the inequation
>
>N1 >= N3 * N7 is incorrect.
>
>If you take the product of
>a pair of divisors, you
>might get a number which
>is higher than the considered
>positive integer.
>
>Am I wrong?
>

The simplest way to check whether you are or not is to give an example. Do you have one? Do you have an example in which

N1 < N37?

This may be the case when a number N has three divisors a, b, c such that


  1. a穊 ends in 3
  2. a穋 ends in 7
  3. a2穊穋 > N

How this may happen? For example, a may end in 1, b in 3, c in 7. Or, a may end in 7, b in 9, c in 1. Or ...

In all cases, what you lose by having a2穊穋 greater than N, you gain by having one of a, b, or c end in 1. The argument must be made more accurate because this line of reasoning must also account for a possibility that one of a, b, or c will be composite.

But you are right, I have not thought of such a possibility.

The inequality N1 => N37, however, I am sure is correct.


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peter (Guest)
guest
Feb-27-01, 08:06 AM (EST)
 
5. "RE: number of dividers"
In response to message #4
 
   sorry, of course i should have given an example.

it's easy to understand that the product of every pair of factors, of which one ends in 3 and the other in 7, produces a factor that ends in 1.

so you mentioned the following example:

>Let's also try 4641 = 3򊐙317. 4641 has the following >factors: 1, 3, 7, 13, 17, 21, 39, 51, 91, 119, 221, 273, 357, >663, 1547, 4641.

so there are 4 divisors ending in 3 (3, 13, 273, 663), so

N3 = 4,

and other 4 divisors end in 7 (7, 17, 357, 1547), so

N7 = 4. Consequently N3 * N7 = 16, but N1 = 6.

But for example the pair of divisors 663 and 1547 doesn't produce another divisor ending in 1, as their product produces a higher number than the considered integer itself.

Actually i always thought i understood it easily, but then i got a really great black-out that made me wonder...

thank you very much,
peter

P.S.: Was my reflection right that i can add the divisor 1 to the inequation above, because it's a divisor you don't get by the product of N3 * N7 and it's in every integer?


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alexb
Charter Member
672 posts
Feb-27-01, 09:53 PM (EST)
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6. "RE: number of dividers"
In response to message #5
 
   Peter, I appreciate your thoughtfulness. So the hasty reply may also be a wrong one. My apologies. From the example, it is obvious now that one has to consdier N1 and N9 simultaneously.

If anything comes up, I'll continue the thread. If anybody in your class has a solution, I'd like to see it.

>Actually i always thought i understood
>it easily, but then i
>got a really great black-out
>that made me wonder...

Good for you. I also thought it was pretty easy.

>P.S.: Was my reflection right that
>i can add the divisor
>1 to the inequation above,
>because it's a divisor you
>don't get by the product
>of N3 * N7 and
>it's in every integer?

Yes, 1 is one of the factor of any N and should be counted in N1.



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LilSmall202001
Charter Member
2 posts
Jul-30-01, 12:03 PM (EST)
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7. "RE: number of dividers"
In response to message #0
 
   Look I really dont know enough about dividers but look at it like this you need to know what is smaller and what is larger for you to know how to get your answer and also use the greater than and less than signs to figure out which one is greater and which one is less.


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