pikachu wrote:
If x is a positive integer, is the number of its divisors smaller than 2*<sqrt{x}> - 1 ?
<a> is defined as the smallest integer that is not less than a .
1. x is not a square of an integer
2. x is prime
Source:
m12-72874.htmlBunuel any help on this - how to go about solving this question
We know that \(2*<sqrt{x}> - 1\) will always take on odd-values. Let x be a perfect square and \(x = a^{2k}\), where a is a prime number. For any value of k, x will have an odd number of factors. It is for this very reason that we are given x is not a square of an integer, so that the number of factors of x and any value of \(2*<sqrt{x}> - 1\) don't equal each other, as the latter can only take on odd values. By design from F.S 1, x can not have any even powers and thus, we will always get even no of factors for x. If they had not mentioned this point, we would not be able to give a definitive answer from F.S 1.
So we know that if not anything, \(2*<sqrt{x}> - 1\) and no of divisors for x WILL never be equal.Thus, if we can prove that there is a general rule where either the no of divisors are ALWAYS less than \(2*<sqrt{x}> - 1\) OR are ALWAYS more than \(2*<sqrt{x}> - 1\) , we will get a definitive answer.
No of factors for x can only be even:
Minimum x with 2 factors = 2 Value of \(2*<sqrt{2}> - 1\)= 3.
Minimum x with 4 factors = 6 Value of \(2*<sqrt{6}> - 1\) = 5.
Minimum x with 6 factors = 12 Value of \(2*<sqrt{12}> - 1\) = 7.
Minimum x with 8 factors = 24 Value of \(2*<sqrt{24}> - 1\) = 9. And so on..
What we get is that if for 12, the value of \(2*<sqrt{12}> - 1\) is 9, then for numbers that have 6 factors and are bigger than 12 say 28, the value of \(2*<sqrt{28}> - 1\) will be even more than 9(11). Thus, we can say that the no of factors will always be less than the value of \(2*<sqrt{x}> - 1\) .
If anyone has a doubt as to whether this pattern is uniform, notice that the no of divisors and the value of \(2*<sqrt{x}> - 1\) are even and odd respectively. Thus, assuming that some value of x with say 8 factors has \(2*<sqrt{x}> - 1\) as 7 or 5 or 3,it won't be possible because we have proved that the MINIMUM value of\(2*<sqrt{x}> - 1\) for any x with 8 factors is 9.
From F.S 2, it is very clear that the no of factors is only 2 for any prime value of x and the minimum value for \(2*<sqrt{2}> - 1\) is 3.Sufficient.
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