How many Fibonacci numbers F n divide F 1 0 0 , where n is a positive integer greater than 1?
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The number of Fibonacci factors of F k being equal to the number of factors of k is a corollary of an even more interesting theorem that m ∣ n iff F m ∣ F n ! Nice problem Prashant Ramnani, although the number isn't that well chosen.
The question is ill formed. The question asks for number of fibonacci numbers that divide 354...75 and not its number of factors.
{1, 3, 5, 55, 6765, 75025, 12586269025, 354224848179261915075}
I found 8 only. I don't know which one that I missed out. I did this carefully. Anyone who is willing to do hard onto this question won't be wrong with information obtainable.
You are absolutely right. There are only eight distinct factors. F 1 0 0 divides F i for i = 1 , 2 , 4 , 5 , 1 0 , 2 0 , 2 5 , 5 0 , 1 0 0 . But, F 1 = F 2 = 1 . Hence, only 8 distinct factors.
Exactly. Count the Fibonacci number 1 only once.
Use the well known fact that g cd ( F m , F n ) = F g cd ( m , n )
Let m = 1 0 0 . Then it is easy to see that n ∣ 1 0 0 .There are only 9 factors of 1 0 0 . Hence the answer is 9 .
Note that F 1 = F 2 = 1 . Hence, only 8 distinct factors.
Correct.My solution is slightly wrong then.
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Let F ( k ) be a k t h fibonacci number then no. of factors of F ( k ) = no. of factors of k
Which gives us that no. of factors of F 1 0 0 =No. of factors of 1 0 0
as 1 0 0 = 2 2 × 5 2 ∴ no. of factors of 1 0 0 = ( 2 + 1 ) × ( 2 + 1 ) ⇒ 9
hence our answer is ⇒ 9
For list of fibonacci nos. click here