Find the largest number that will divide exactly the product of four consecutive integers.

Solution:​

Case 1: Let the first number be even = 2n.
The numbers would be 2n, 2n+1, 2n+2, 2n +3.
Product would be 4n(2n+1)(2n+3)(n+1).
Now clearly, there will be at least one multiple of 3 and one more factor 2 coming from n or n + 1 also both 2n + 1 and 2n + 3 will be odd.
So, in this case the largest divisor of the product would be 4 $$\times$$​ 3 $$\times$$​2 = 24
Case 2: Let the first number be odd = 2k + 1.
The product would be (2k + 1)(2k + 2)(2k + 3)(2k + 4) = 4(2k + 1)(k + 1)(2k + 3)(k + 2)
Again, there is a 4, there will be at least one multiple of 3 and one more 2 coming up between k + 1 and k + 2 .
So, the largest divisor here would again be 4 $$\times$$​ 3 $$\times$$​ 2 = 24.
In 4 consecutive Integers there will be 2 even number in which 1 will be a multiple of 4, another number which is not a multiple of 4 but 2 and another number which is a multiple of 3.
So it will be divisible by 4 $$\times$$​ 3 $$\times$$ 2 = 24 

Thus, the largest number which exactly divides the four consecutive number is 24.