I came up with a simpler solution after the step at 6:50. Anyone tell me if I am wrong. So basically we have f(n) = h + cg and n + c = f(h) + cg where 0 <= n < c and 0 <= h < c. From this we can conclude f(n) = h (mod c) and f(h) = n (mod c). So for every n < 1987 there is an h that forms a pair (n, h) such that both number is less than c and f maps each one to the other. However since there are an odd number of "n"s (0, 1, 2, ..., 1986), the set cannot be divided evenly into pairs of (n, h)s and so there must be at least two distinct n1, n2 such that f(n1) and f(n2) both map to the same h. Thus f(n1) = f(n2) (mod c). Apply f to both side so n1 + 1987 = n2 + 1987 (mod c) so n1 = n2 (mod c). However n1 and n2 are both less than c so n1 = n2. But we start with two distinct n1, n2 so contradiction.
cr
hi i actually don't understand the part where C is even and then c is even. can anybody explain this to me, please? Thanks a lot.
mangaman
What is the tool used to write and draw? It's really good.
govardhanmucharla
Hello Osman Nal.Your name looks like Turkish's name. Are you from Turkey?
