If the last 6 digits of [(M)! – (N)!] are 999000, which of the following option is not possible for (M) × (M – N) ? Both (M) and (N) are positive integers and M > N. (M)! is factorial M.
(b) M! – N! = M (M – 1) (M – 2) ... (N + 1) N! – N! = {M (M – 1) (M – 2) .... (N + 1) – 1} N! = (Z – 1) N! where Z = M (M – 1) (M - 2) ... (N + 1) Last 3 digits are '000' and 4th from last is 9 (i.e. not zero). So, what would be the value of N? To get three zeros, it requires 5 (along with an even nmber), 10, and 15 (along with an even number) and NOT 20 (then it would be 'XX0000'). 4th to 6th places of this number is '999' (i.e. Z –1 = 1000× 10– 1). So, Z should be more than 19 and it would have at least three zeros → Z should include both 20 and 25 (20 would give one zero and 25 would give two zeros) → M t25 Option A : 150 = 25 × 6 = 25 × (25 – 19) Option C : 200 = 25 × 8 = 25 × (25 – 17) Option D : 225 = 25 × 9 = 25 × (25 – 16) Option E : 234 = 26 × 9 = 26 × (26 – 17) All are possible solution, however, 180 = 2 × 2 × 3 × 3 × 5, cannot be broken to fulfil the requirements.