Product Of Consecutives

Note that in order to be divisible by 100, the product of these 2 numbers is divisible by both 25 and 4. So from this we can deduce that one of the numbers is a multiple of 25, otherwise they would not be consecutive.

25 - 1 = 24 which is a multiple of 4. Therefore the 2 numbers are 24 and 25, because 25 is the smallest multiple of 25. $24\times25=\boxed{600}$

This is my correctly done $3000^{th}$ problem, so I must provide the solution, which is same as Eamon Gupta 's.

Let $m$ and $n$ be positive integers such that:

$100n = m(m+1) \quad \Rightarrow n = \dfrac {m(m+1)}{100} = \dfrac {m(m+1)}{100} = \dfrac {m(m+1)}{2^2\times 5^2}$

We note that the nominator of RHS $m(m+1)$ is a product of an odd and an even numbers. For $n$ to be an integer, the odd number in the nominator of RHS must divisible by $25$ and the even one, $4$ . The smallest $100n$ is when the odd number of nominator is $25$ and since $25-1=24$ is a multiple of $4$ , it meets the bill. Therefore the smallest multiple of $100$ which can be expressed as the product of two consecutive numbers is $24\times 25 = \boxed{600}$ .

Brute force

     boolean found = false;
     int x = 11;
     while(!found){             
         if( x * (x-1) % 100 == 0){
             found = true;
         }else{
             x +=1;
         }

     }
     System.out.println(x * (x - 1));