A number theory problem by Ankit Kumar Jain

Here's a straightforward solution using some trick on order. We first notice that this equation can be written as $\frac{c!-a!}{b!} = 2001.$ Of course, $a!, b!, c!$ are all positive, hence $c>a$ . Next we find the place for $b$ . Obviously, it cannot exceed $c$ , since the left hand side of equation would be fraction. Also, it cannot lie between $a$ and $c$ , using the same reasoning. Therefore, $c > a \ge b$ . From here, we can rewrite the equation as $\left( \frac{a!}{b!} \right) \left( \frac{c!}{a!}-1 \right) = 2001,$ and both inside those brackets are integers. In fact, $\frac{a!}{b!}$ and $\frac{c!}{a!}$ are products of some consecutive numbers. The factorization of $2001$ is $3 \times 23 \times 29$ , and its divisors are $1,3,23,29,69,87,667,2001$ .

Now, we look at $\frac{a!}{b!}$ . Notice that all of divisors of $2001$ except $1$ is not a product of two or more consecutive numbers, so we have that the possible solutions are $(a,b)=(3,2),(23,22),(29,28),(69,68),(87,86),(667,666),(2001,2000)$ . However, if we put these numbers into equation, we would have, from $c!=a! \left( \frac{2001}{a!/b!}+1\right)$ , that $c!=668 \cdot 3!, 88 \cdot 23!, 70 \cdot 29!, 30 \cdot 69!, 24 \cdot 87!, 4 \cdot 667!, 2 \cdot 2001!,$ which are all impossible for integer $c$ . Therefore, there is only one case left, that is when $\frac{a!}{b!}=1$ , or $a=b$ .

If $a=b$ , then we have that $\frac{c!}{a!}=2002$ . The divisors of $2002=2 \times 7 \times 11 \times 13$ are $1,2,7,11,13,14,22,26,77,91,143,154,182,286,1001,2002$ . Observe that we cannot write $2002$ as product of two or more consecutive numbers again, so we must have that $c=2002$ and $a=b=2001$ . The solution is thus $2001+2001+2002=\boxed{6004}$ .

When a = b , we have

2002b ! = c !

c(c - 1)(c - 2)...(b + 1) = 2002

2002 = 2 $\times$ 7 $\times$ 11 $\times$ 13 so there is no way for 2002 to be expressed as a product of 2 or more consecutive integers. Thus when a = b the only solution for (a,b,c) is (2001, 2001, 2002).

For the case of b > a we consider the largest power of 3 that divides a, b and c. Let them be $a_{1}$ , $b_{1}$ and $c_{1}$ respectively. Since b > a ,

$a_{1}$ $\leqslant$ $b_{1}$

Therefore we can say that 2001b! = a!(3k) for some integer k since 3 divides 2001.

c! = 2001b! + a! = a!(3k + 1)

Since 3 does not divide 3k + 1 , $c_{1}$ = $a_{1}$ , which implies c $\leqslant$ a + 2 . Since none of a, b and c are equal and c > b > a , It then follows that c = a + 2, b = a+1 . The original condition then simplifies to

a!(1 + 2001(a + 1)) = (a + 2)(a + 1)a!

2001a + 2002 = $a^{2}$ + 3a + 2

$a^{2}$ - 1998a - 2000 = 0

Since the quadratic has no integer solutions, there are no solutions where b > a .

The final case is when b < a . Consider the largest power of 2 that divides a, b and c, letting them be $a_{2}$ , $b_{2}$ and $c_{2}$ respectively.

c > a > b

if $a_{2}$ > $b_{2}$ ,

c! = a! + 2001b! = b!(2k + 2001) for some integer k

$c_{2}$ = $b_{2}$ < $a_{2}$ , a contradiction.

Therefore $a_{2}$ = $b_{2}$ , i.e. a = b + 1 and b is even.

c! = (b+1)b! + 2001b! = (b + 2002)b!

(b +1)(b + 2)...(c) = b + 2002

(b + 1)(b + 2) | b + 2002 because c $\geqslant$ b + 2

b + 1 | 2001

b + 2 | 2000

Given that 2001 = 3 $\times$ 23 $\times$ 29, by considering all possible values of b + 1 we find that the only possible value of b is 2, which does not have a solution for a and c.

In conclusion, the only solution for (a,b,c) is (2001, 2001, 2002), so the answer is 2001 + 2001 + 2002 = $\boxed{6004}$