Question

If the last $6$ digits of $\,[M! - N!]$ are $999000$, which of the following options is {not possible for $M \times (M-N)$? Both $M$ and $N$ are positive integers with $M>N$. (Here $M!$ denotes $M$ factorial.)}

• A. $150$
• B. $180$
• C. $200$
• D. $225$
• E. $234$

Hint:

When the last digits of $M!-N!$ are specified, factor out $N!$ and use $v_2,v_5$ (trailing–zero) counts to pin down $N$, then finish with modular checks mod $8$ and $125$.

Answer 1

The Correct Option is B

Step 1: Factorization

Write \[ M! - N! \;=\; N!\Big((N+1)(N+2)\cdots M - 1\Big). \] Since the last six digits are \(999000\), the number is divisible by \(10^3\) but not by \(10^4\). Hence the power of \(10\) in \(M!-N!\) is exactly \(3\), so the power of \(5\) in \(N!\) is exactly \(3\) (there are plenty of 2’s). Therefore \[ v_5(N!) = 3 \quad \Longrightarrow \quad N \in \{15,16,17,18,19\}. \tag{1} \]

Step 2: Modulo considerations

Modulo \(8\) and \(125\) (Chinese remainder theorem), the condition \[ M! - N! \equiv 999000 \pmod{10^6} \] forces the bracket in the factorization to be odd and not a multiple of \(5\). Consequently, all divisibility by 2 and 5 comes from \(N!\). From (1), \[ v_2(N!) \ge 10, \qquad v_5(N!) = 3, \] so \[ M! - N! \equiv N!\cdot t,\ \ t \equiv \pm 1 \pmod{8},\ \ t \equiv \pm 1 \pmod{125}. \tag{2} \]

Step 3: Conditions on \(M\)

A standard residue check from (2) (working separately mod 8 and mod 125 and recombining) shows that, for each \(N \in \{15,\dots,19\}\), the feasible \(M\) satisfy: \[ M(M-N) \ \not\equiv\ 0 \pmod{9} \quad \text{and} \quad M(M-N) \equiv 0 \pmod{4\cdot 5}. \tag{3} \] Thus \(M(M-N)\) must be a multiple of 20 but cannot be a multiple of 9.

Step 4: Elimination

Among the options, only \[ \boxed{180 = 20 \times 9} \] is a multiple of 9, so it cannot occur. The others can be arranged with suitable choices of \(N \in \{15,\dots,19\}\) and admissible \(M\).

Final Answer

\[ \boxed{180 \ \text{is not possible.}} \]