Challenge Accepted

Among ten consecutive integers that divide n, there must exist numbers divisible by $2^3, 3^2, 5, 7,11$ (since we are to find the smallest 10 consecutive divisors :P).

Thus the desired number has the form n = $2^{α_1}\times 3^{α_2} \times 5^{α_3} \times 7^{α_4} \times 11^{α_5}$ , where $α_1 ≥ 3; α_2 ≥ 2; α_3 ≥ α_4\geq a_5\geq 1$ .

Since n has $(α1 + 1)(α2 + 1)(α3 + 1)...$ distinct factors, and $(α_1 + 1)(α_2 + 1)(α_3 + 1)(α_4 + 1) ≥ 48$ , we must have $(α_5 + 1) · · · ≤ 3$ . Hence at most one $a_{j>4}$ , Is positive, and in the minimal n this must be $a_5$ .

Checking through logic and some bashing, the possible combinations satisfying $(α_1 + 1)(α_2 + 1)...(α_5 + 1) = 144$ one finds that the minimal n is $2^5 · 3^2 · 5 · 7 · 11 = 110880$ .

$Simple\quad Solution\quad For:\quad X\\ From\quad (ii):\\ Since\quad there\quad are\quad 10\quad consecutive\quad integers\quad in\quad divisors,\quad the\quad number\\ should\quad be\quad divisible\quad by\quad 2,3,5,7\quad (primes\quad whose\quad multiples\quad present\quad in\quad any\quad 10\quad consecutive\quad no.)\quad \\ so\quad X\quad should\quad be\quad something\quad like\quad { 2 }^{ p }\times { 3 }^{ q }\times { 5 }^{ r }\times { 7 }^{ s }\quad -\quad least\quad no\quad of\quad this\quad kind\quad is\quad 2520\quad (has\quad 10\quad consecutive\quad no's\quad as\quad divisors\quad 1-10)\\ 2520\quad =\quad { 2 }^{ 3 }\times { 3 }^{ 2 }\times { 5 }\times { 7 }\quad \quad \Rightarrow \quad Count(Divisors)\quad =\quad (3+1)\times (2+1)\times (1+1)\times (1+1)\quad =\quad 4\times 3\times 2\times 2\quad \neq \quad 144\quad \quad --\quad (I)\\ From\quad (i):\quad \\ No\quad of\quad divisors\quad =\quad 144\quad =\quad 3\times 3\times 4\times 4\\ We\quad can\quad multiply\quad 2520\quad with\quad least\quad number\quad \& \quad also\quad Count(Divisors)=144\\ Two\quad ways\quad to\quad do\quad that\quad \\ 1.\quad Using\quad same\quad primes\quad { 2 }^{ 3 }\times { 3 }^{ 3 }\times { { 5 }^{ 2 } }\times { { 7 }^{ 2 } }\quad =\quad 262400\quad (with\quad 144\quad divisors)\quad [Least\quad possible\quad no.]\quad \vdots \quad Multiplied\quad 2520\quad with\quad 35\\ 2.\quad Use\quad one\quad more\quad prime\quad { 2 }^{ 5 }\times { 3 }^{ 2 }\times { 5 }\times { 7 }\times 11\quad =\quad 110880\quad (with\quad 144\quad divisors)\quad [Least\quad possible\quad no.]\quad \vdots \quad Multiplied\quad with\quad 44\\ Least\quad Possible\quad X\quad =\quad 110880\quad (with\quad 144\quad divisors)$