**Solution**

1. If x, y is a solution then so is y-x, -x. Hence also -y, x-y. If the first
two are the same, then y = -x, and x = y-x = -2x, so x = y = 0, which is impossible, since n > 0. Similarly, if any other
pair are the same.

2891 = 2 (mod 9) and there is no solution to x^{3} - 3xy^{2}
+ y^{3} = 2 (mod 9). The two cubes are each -1, 0 or 1, and the other term is 0, 3 or 6, so the only solution is to
have the cubes congruent to 1 and -1 and the other term congruent to 0. But the other term cannot be congruent to 0, unless
one of x, y is a multiple of 3, in which case its cube is congruent to 0, not 1 or -1.

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2. We show that f(n) = [n/3] for n <= 9999, where [ ] denotes the integral
part.

We show first that f(3) = 1. f(1) must be 0, otherwise f(2) - f(1) - f(1)
would be negative. Hence f(3) = f(2) + f(1) + 0 or 1 = 0 or 1. But we are told f(3) > 0, so f(3) = 1. It follows by induction
that f(3n) ≥ n. For f(3n+3) = f(3) + f(3n) + 0 or 1 = f(3n) + 1 or 2. Moreover if we ever get f(3n) > n, then the
same argument shows that f(3m) > m for all m > n. But f(3.3333) = 3333, so f(3n) = n for all n <= 3333.

Now f(3n+1) = f(3n) + f(1) + 0 or 1 = n or n + 1. But 3n+1 = f(9n+3) ≥
f(6n+2) + f(3n+1) ≥ 3f(3n+1), so f(3n+1) < n+1. Hence f(3n+1) = n. Similarly, f(3n+2) = n. In particular f(1982)
= 660.

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4. a^{2 }+ b^{2}
– 8c = 6

or, a^{2} + b^{2}
= 6 + 8c

Right Hand Side is an even number.

__Case I__

If both a & b are even,

Let, a = 2m & b = 2n

4m^{2} + 4n^{2}
= 8c + 6

2m^{2} + 2n^{2}
= 4c + 3

2(m^{2} + n^{2})
= 4c + 3

LHS is an even number, but RHS
is an odd number.

Hence, both cannot be even.

__Case II__

If both a & b are odd,

Let, a = 2m+1 & b = 2n+1

(2m+1)^{2} + (2n+1)^{2}
= 8c+6

m(m+1) + n(n+1) = 2c+1

The product of consecutive number
is even

RHS is odd

Therefore both are not
possible

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5. n.n! + (n-1).(n-1)! + (n-2)(n-2)! + ………3.3! + 2.2! + 1.1!

= (n+1-1).n! + (n-1).(n-1)! + (n-1-1).(n-2)! + ………(4-1).3!
+ (3-1).2! + (2-1).1!

= [(n+1).n! – n!] + [n.(n-1)! – (n-1)!] + [(n-1).(n-2)!
– (n-2)!] + ………..[4.3! – 3!] + [3.2! – 2!] + [2.1! – 1!]

= {(n+1).n! + n.(n-1)! + (n-1).(n-2)! + …..4.3! + 3.2! +
2.1!} – {n! + (n-1)! + (n+2)! + …..3! + 2! + 1!}

= {(n+1)! + n! + (n-1)! + ……4! + 3!+ 2!} - {n! + (n-1)!
+ (n+2)! + …..3! + 2! + 1!}

= (n+1)! – 1!

= **(n+1)! - 1**

Solution by Kapil Goel