alexb
Charter Member
672 posts 
Feb1201, 04:12 PM (EST) 

1. "RE: number of dividers"
In response to message #0

LAST EDITED ON Feb1201 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 N_{1}, N_{3}, N_{7}, N_{9} be the number of factors of a given number that end respectively in 1, 3, 7, and 9. Then, because of the previous paragraph N_{1}3·N_{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 N_{3}·(N_{3}  1)/2 of the first kind and N_{7}·(N_{7}  1)/2 of the second. Thus we obtain that N_{9} N_{3}·(N_{3}  1)/2 + N_{7}·(N_{7}  1)/2. We may add up the two inequalities: N_{1} + N_{9} N_{3}·N_{7} + N_{3}·(N_{3}  1)/2 + N_{7}·(N_{7}  1)/2. Or written a little differently (check this): N_{1} + N_{9} (N_{3} + N_{7})·(N_{3} + N_{7}  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 N_{3} + N_{7} 3, we indeed have N_{1} + N_{9} 
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peter (Guest)
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Feb2601, 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 N_{1} could even be:N_{1} = 1 + N_{3} + N_{7} because every number can be divided by 1. thanks again, Peter 

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peter (Guest)
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Feb2601, 11:46 PM (EST) 

3. "RE: number of dividers"
In response to message #0

i think the inequation N_{1} >= N_{3} * N_{7} 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 
Feb2701, 00:03 AM (EST) 

4. "RE: number of dividers"
In response to message #3

LAST EDITED ON Feb2701 AT 00:05 AM (EST) >i think the inequation > >N_{1} >= N_{3} * N_{7} 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 N_{1} < N_{3}·N_{7}? 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
 a^{2}·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 a^{2}·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 N_{1} => N_{3}·N_{7}, however, I am sure is correct. 

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peter (Guest)
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Feb2701, 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 = 3·7·13·17. 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 N_{3} = 4, and other 4 divisors end in 7 (7, 17, 357, 1547), so N_{7} = 4. Consequently N_{3} * N_{7} = 16, but N_{1} = 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 blackout 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 N_{3} * N_{7} and it's in every integer? 

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alexb
Charter Member
672 posts 
Feb2701, 09:53 PM (EST) 

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 N_{1} and N_{9} 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 blackout >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 N_{3} * N_{7} and >it's in every integer? Yes, 1 is one of the factor of any N and should be counted in N_{1}.


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