I need to take a look for your first question. Finding difference between the sum of integers and integral divisors is just a bit too complicated for NS, it seems...
AHHHHHH! stupid! stupid me! OK, I'll give you a clue:
28 is a PERFECT #!!!
(Calming down)As for the second one, I'll give you another clue:
TRIANGULAR #!!
Draw each scenario yourself, and find the connection.
Notice how if you list out the numbers that are relatively prime (numbers that share a GCD of 1 with the number in question) they come out in pairs that add up to the number itself. To do this problem quickly, we must use the number sene method we've learned to find how many numbers are relatively prime to the number. To do that, take the prime factorization of the number in exponent form. First, subtract one from the exponents of each factor, and multiplu them together. Next, subtract one from the factors themselves, and multiply them together. Take the two results you worked out and multiply those together. The result is the number of terms that are less than the number and relatively prime to the number. If you want to find the sum of this sequence, divide the number of reletively prime numbers by two and multiply that by the original number. BAM!
ex. What is the sum of the numbers less than and relatively prime to 40?
40 = 2^3 * 5 (prime factorization)
2^(3-1) = 4, 5^(1-1) = 4*1 = 4 (Step 1 result)
2 - 1 = 1, and 5 - 1 = 4. 4*1 = 4 (Step 2 result)
The number of relatively prime numbers is 4*4 = 16.
The sum = 16/2*40 = 320.
On the other one, I think its the same as finding the number of diagonals in a pentagon. You can find that by adding 1+2+3....+S-1, where S is the number of sides in the polygon. In this case, 1+2+3+4 = 5.
Zack described a method for the question about the sum of the numbers that are relatively prime to a number which you posed in another thread.
Also going off what Zack said, I think this works for number of lines determined by n points no 3 of which are collinear
It would be all the diagonals of an n-gon plus the n sides of that n-gon
so: Number of diagonals of n-gon: n(n-3)/2 = (n^2 - 3n)/2 , Number of sides = n
(n^2-3n) / 2 + n = (n^2-3n) /2 + 2n/2 = (n^2 - 3n + 2n) / 2 = (n^2 - n) / 2 = n(n-1)/2 There it is algebraically worked. If 3 are collinear, there would be 2 less than that because
a) a line cannot be drawn between the two endpoints of the segment formed by the 3 collinear points.
b) the two lines connecting those 3 points if they were not collinear would then be 1 line when they are collinear
It amazes me that you can actually pinpoint a post more than sixth months ago.
Yeah, I never did have that shortcut solid in my head, now I hope it has made deep enough an impression. Also thanks for refreshing the 3-pt-noncollinear one.
stupid!
