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
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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 = 3·7·13. 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 = 3·7·13·17. 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 = 3·3·7·13. 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·N7. 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 N3·N7 + 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  that the inequality remains true.<P>P.S. I wonder whether your teacher had in mind anything of that sort, or I missed a simpler approach. If and when you learn that the problem may be solved in a simpler way, I would appreciate your posting it to the forum. Thank you.<p></font></td></tr></table>
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peter (Guest)
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Feb-26-01, 11:46 PM (EST) |
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2. "RE: number of dividers"
In response to message #1
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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)
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Feb-26-01, 11:46 PM (EST) |
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3. "RE: number of dividers"
In response to message #0
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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
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672 posts |
Feb-27-01, 00:03 AM (EST) |
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4. "RE: number of dividers"
In response to message #3
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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 < N3·N7? This may be the case when a number N has three divisors a, b, c such that
- a·b ends in 3
- a·c ends in 7
- a2·b·c > 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·b·c 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 => N3·N7, however, I am sure is correct. |
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peter (Guest)
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Feb-27-01, 08:06 AM (EST) |
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5. "RE: number of dividers"
In response to message #4
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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 = 3·7·13·17. 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
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672 posts |
Feb-27-01, 09:53 PM (EST) |
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6. "RE: number of dividers"
In response to message #5
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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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