What is the smallest positive integer ending in 1986 which is divisible by 1987?

Question

What is the least positive integer n that can be placed in the following expression:

n!(n+1)!(2n+1)! - 1

and yields a number ending with thirty digits of 9's.

Answer 1

[1987x- 1986) mod 1000 = 0: gets us the smallest number that has the required property.
So the smallest number to make this possible is 1078.
1987 × 1078 = 2,141,986.

Comments

How to find out solution

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The objective here was to find out the number 1078 in { 1987 x 1078 }
Here the last digit 8 (in 1078) is obvious as we had to get 6 as the last digit of the product (2,141,986). This would only happen when 8x7 = 56.
Now 1987 x 8 = 15896. This will help us find the tens place of the answer
We now know that 9 (from 15896) and A x 7 ( A is the first unknown) must add up to 8 (tens place of 1986) ie.,
9 + 7A = 8. This can happen when A = 7 or when A = 4 (By trial and error eliminate 4)
Now we have 15896 (1987*8), 13909 (1987*8) and 7 x B (B is the next unknown)
8 + 0 + 7B +1(carry from previous operation) = 9 (2nd number in 1986)
7B = 0 { Happens when B = 10}

This way I ended up with 1078 as a number that could yield 1986 as the last 4 digits of the product of 1987 x 1078.