Q1} How many ordered triples (a,b,c) of positive integers are there such that none of a,b,c exceeds 2010 and each of a,b,c divides a+b+c ?
Q2} Let n be a positive integer. Prove that there are no positive integers x and y such as
√n + √n+1 < √x + √y < √4n+2
211.Ans 2010+670=2680
Because the triplets will be of the form (c,c,c) or (3c,2c,c) only.
11Effectively, b+c=ak, a+c=bl, a+b=cm......be the 3 equations.
Adding them gives 2(a+b+c)=ak+bl+cm
Implies a(2-k)+b(2-l)+c(2-m)=0. That gives at least 1 of k,l,m=1....let k=1
what follows is b|2c and c|2b
Now lets have 2c=bg, giving c=bg2 .....2b=4cg, which means g|4, implying =1,2,4....
Work out the cases to get the above forms of #2 & #3
341The only triplets that satisfy the given conditions are of the form (2k,k,k) or (3k,2k,k) or (k,k,k).
The first gives 1005 unordered triplets and 3015 ordered triplets
The second gives 670 unordered triplets = 670X6 = 4020 ordered triplets
The third gives 2010 ordered triplets
Hence we have 9045 unordered triplets
341The proof of the first part:
Assume a \ge b \ge c
Now a|(b+c) \Rightarrow a \le b+c \le 2b
Also b|a+c and a+c \le 3b means a+c=2b or a+c=3b
Now a+c = 3b \Leftrightarrow c = 3b-a \ge 3b-2b=b \Rightarrow c=b
so that a=2b
So we have (2k,k,k) as one family of solutions.
Now if a+c = 2b, a|b+c \Rightarrow a|(a+b+c) \Rightarrow a|3b
So either a = b or 2a=3b (since a≤2b)
If a=b, then a=b=c so that (k,k,k) is another family of solutions.
If 2a=3b, we see that 3|a. So let a=3k, then b = 2k and c=k
giving (3k,2k,k) is the last family of solutions.
11I've done the same thing in another way
Referring to #4 - the very fact that g=1,2,4 gives these forms explicitly
