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Question:

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.

Explanation:

M and N are positive integers such that M > N

∴ M! – N! = abc…999000

∴ [M(M − 1)(M − 2)……N!] – N! = abc…999000

∴ N!{[M(M − 1) (M − 2)……] − 1} = abc…999000

Let the term in the square bracket be x.

Since M is a positive integer, the term in the square brackets i.e., x is also a positive integer, and hence, (x – 1) is also a positive integer.

∴ N!(x – 1) = abc…999000

$∴ (x - 1) = \frac{a b c . . . 999000}{N!}$

Hence, the maximum number of zeroes in N! is 3.

∴ N! ≤ 19 (because from 20! onwards, each factorial has atleast 4 zeroes)

Now, there are 4 possible ranges for N:

N = 0 to 4 (no zeroes in N!)
N = 5 to 9 (1 zero in N!)
N = 10 to 14 (2 zeroes in N!)
N = 15 to 19 (3 zeroes in N!)

Consider case 3, where there are 2 zeroes in N!

Since N!(x – 1) = abc…999000, the third zero on the LHS should come from (x – 1).

For this, x has to be of the form pqrs…1 i.e., x has to be an odd number.

Now, there are two possibilities:

1. M = N + 1

Here, M! – N! = (M × N!) – N! = N!(M – 1)

In this case, M = x

Since N = 10 to 14 and M – 1 should end in 0, M = 11 and N = 10

Hence, M(M – N) = 11(11 – 10) = 11

This does not tally with any of the options.

Hence, M ≠ N + 1

2. There is at least one integer between M and N

Hence, x comprises a series of consecutive integers (at least two as explained above) multiplied with each other. So, there has to be at least one even number in this series.

Hence, x can never be odd.

Hence, the third zero on the LHS can never come from (x – 1).

Hence, there cannot be 2 zeroes in N!. Similarly, it can be proved that there cannot be 1 zero or no zeroes in N!.

Hence, N! has three zeroes i.e. 15 ≤ N ≤ 19

Now, M(M – N) = M² – NM can take four of the five given values.

Hence, there are five possible equations – one for each option.

Consider option 1: M² – NM – 150 = 0

Here, for N = 15 to 19, see if there exists a value of M (> N) that gives positive integral roots in this equation.

For N = 19, the equation becomes

M² – 19M – 150 = 0 i.e. M = 25 or M = −6. Here, M = 25 is valid.

Similarly, consider each option.

Option 3:

For N = 17, the equation becomes

M² – 17M – 200 = 0 i.e. M = 25 or M = −8. Here, M = 25 is valid.

Option 4:

For N = 16, the equation becomes

M² – 16M – 225 = 0 i.e. M = 25 or M = −9. Here, M = 25 is valid.

Option 5:

For N = 17, the equation becomes

M² – 17M – 234 = 0 i.e. M = 26 or M = −9. Here, M = 26 is valid.

However, no value of N from 15 to 19 gives an integral solution for M in M² – NM – 180 = 0

Hence, M(M – N) can never be 180.

Hence, option (b).

Note: There is a logical flaw in this question as the difference of two factorials can never end in 999000. The actual question should have had the last six digits as abc000.