my first number theory problem....enjoy :D

$(n)(n+1)(n+3)(n+4)(n+5)$ Can be written as $\frac{(n+5)!}{n!}$ Now we have to find an expression which consists of alleast two $5's$ and two $2's$ From the expression the first think which clicks in mind is $n+5$ should be a multiple of 5 as to obtain max no. of $5's$ in the expresssion) The second thing that strikes is that $n+5$ should be an integral power (I think you get it why) So from these two conclusions we notice that $5^{2}$ gives us two $5's$ in the expression . implies $n+5=5^{2}$ $n=20$ Now the no. comes out to be $\frac{(20+5)!}{20!}$ $ans.=6375600$

The number should be divisible by 100, which can be written as 2 * 2 * 5 * 5.when we take 5 consecutive numbers, we can not get two multiples of 5. So, the only possibility is that we include 25, 125, 625, etc. in those five numbers. We require the smallest number so it must be 21 * 22 * 23 * 24 * 25 = 6375600.

$The\quad number\quad can\quad be\quad written\quad in\quad the\quad form:-\\ \\ n(n+1)(n+2)(n+3)(n+4)\\ \\ For\quad this\quad number\quad to\quad be\quad divisible\quad by\quad 100,\\ it\quad must\quad be\quad divisible\quad by\quad { 2 }^{ 2 }\quad and\quad { 5 }^{ 2 }\\ because\quad 100\quad =\quad { 2 }^{ 2 }\quad \times \quad { 5 }^{ 2 }\\ \\ i.e:-\quad it\quad must\quad be\quad divisible\quad by\quad 4\quad and\quad 25.\\ \\ We\quad try\quad to\quad look\quad at\quad the\quad first\quad number\quad which\\ contains\quad 25\quad as\quad a\quad factor\quad as\quad 4\quad is\quad more\quad \\ abundant\quad and\quad will\quad definitely\quad divide\quad one\\ of\\ \\ n(n+1)(n+2)(n+3)(n+4)\\ \\ Now\quad we\quad observe\quad that\quad such\quad a\quad number\quad \\ can\quad only\quad contain\quad one\quad 5\quad as\quad a\quad factor\\ unless\quad 25\quad or\quad any\quad of\quad it's\quad multiples\quad are\\ a\quad factor\quad themself.\\ \\ Thus,\quad the\quad smallest\quad n\quad for\quad which\quad 25\quad is\\ a\quad factor\quad is\quad when\quad n+4\quad =\quad 25\\ \\ Thus\quad n\quad =\quad 21.\\ \\ Thus\quad the\quad number\quad is:-\\ 21\quad \times \quad 22\quad \times \quad 23\quad \times \quad 24\quad \times \quad 25\\ =\quad \boxed { 6375600 } .\\$

I used a wrong approach in getting the answer. I need a better solution.