pure - (simple logic)

1a. Let's assume k is not divisible by 3. Then either k + 1 or k - 1 is divisible by 3.
If k + 1 is divisible by 3, then k + 1 = 3N where N is a natural number.
k = 3N - 1
k2 = 9N2 - 6N + 1 = 3(3N2 - 2N) - 1 => k2 is not divisible by 3 ... (1)
Similarly if k - 1 is divisible by 3, then k - 1 = 3M where M is a natural number
k = 3M + 1
k2 = 9M2 + 6M + 1 = 3(3M2 + 2M) + 1 => k2 is not divisible by 3 ... (2)
(1) and (2) together implies a contradiction of the assumption. Therefore k is divisible by 3.
1b. Let's assume that k = √n is a natural number.
k2 = n is divisible by 3 since the sum of all the digits of n is 2001, and 2001 = 3 x 667
Following 1a, k is also divisible by 3.
Let's put k = 3N where N is a natural number, then k2 = 9N2 => k2 is divisible by 9. However since the sum of all the digits of n is not divisible by 9, n is not divisible by 9. This is a contradiction. Hence n is not a natural number
2. Since x is rational, x = p/q where p and q are integers
Let assume x + y is rational, i.e. x + y = h/k where h and k are integers
y = x + y - x
y = h/k - p/q
y = (qh - pk) / qk which implies that y is rational => contradiction.
So the assumption x + y is rational is wrong and x + y is irrational
3. Let's assume such groupings are possible.
For each if such groups, let's denote the three numbers as ai, bi and ai + bi
So the sum of all these groups are ∑(i=1 to 11)(ai + bi + ai + bi)
= ∑(i=1 to 11)2(ai + bi) is even
Now 1 + 2 + 3 + ... + 33 = (33)(33 + 1)/2 = 561 is odd, which is a contradiction. Hence such groupings are not possible.

1(a) proof by contradiction. If k is not divisible by3. Then k=3n+1 or 3n+2. We can see that (3n+1)^2=9n^2+6n+1 and (3n+2)^2=9n^2+12n+4 which are not divisible by3. A contradiction
(b) if √n is an integer k, then k^2=n. We know that n is divisible by 3=>k^2 is divisible by 3=>k is divisible by 3. Let k=3m. Then n=9m^2 which indicates that n is divisible by 9, which is a contradiction.
2 If x+y is rational
Then (x+y)-x should also be rational which is a contradiction since y is irrational
3 If this can be done then 1+2+3+...+33=33*34/2=561 is even, a contradiction.