1000 = 5*5*5*2*2*2

To have a factorial divisible by 1000, it must have 3 factors of 5
and 3 factors of 2.

Every other integer has a factor of 2 so there are plenty of those,
so we must make sure that we have 3 factors of 5 somewhere in the
products.  Let's start buiding our factorial

1*2*3*4*5  <-- that only has 1 factor of 5  
(It already has 3 factors of 2)

So we keep putting on more factors:

1*2*3*4*5*6*7*8*9*10  <-- That only has 2 factors of 5

So we keep going once more:

1*2*3*4*5*6*7*8*9*10*11*12*13*14*15  <-- that's it!

That's 15!. It is the smallest factorial that is divisible by 1000.

Notice that 14! = 87178291200 won't do.

It only has two 0's on the end, so it's only divisible
by 100. But

15! = 1307674368000

is the smallest factorial that has 3 0's on the end, which
a number must have to be divisible by 1000.

Answer: 15! = 1307674368000

Edwin

S(1st 5 terms) = S(next 5 terms) - 90

S(next 5 terms) = S(1st 10 term) - S(1st 5 terms)

Substituting:

S(1st 5 terms) = S(1st 10 terms) - S(1st 5 terms) - 90

2∙S(1st 5 terms) = S(1st 10 terms) - 90




Use the formula for the sum of n terms of an arithmetic
sequence:



The sum of the first five terms of the arithmetic sequence...





The sum of the first ten terms of the arithmetic sequence...





Substituting in:



















Edwin