MHB What is the last odd digit in the factorial sequence?

[ittalo25]

Hi guys, my english is very bad but let me try translate the question:

Let $ a_n=\frac{(n+9)!}{(n-1)!} $ . Let k the lesser natural number since that the first digit (on the right side) after all the zeros of $ a_k $ is odd.

Example: $ a_k =$ 4230000000 or $ a_k =$ 62345000

This odd digit number is:

a) 1
b) 3
c) 5
d) 7
e) 9$ a_k = \frac{(k+9)!}{(k-1)!} = k \cdot (k+1)\cdot(k+2)\cdot (k+3)\cdot ... \cdot (k+9) $

We have ten consecutives numbers, which are 5 odds and 5 evens.

By the conditions, we need to have $ a_k = 2^{x}\cdot 5^{y}\cdot... $ with $y \geq x$

But I don't know how continue.

 

[kaliprasad]

Hi guys, my english is very bad but let me try translate the question:

Let $ a_n=\frac{(n+9)!}{(n-1)!} $ . Let k the lesser natural number since that the first digit (on the right side) after all the zeros of $ a_k $ is odd.

Example: $ a_k =$ 4230000000 or $ a_k =$ 62345000

This odd digit number is:

a) 1
b) 3
c) 5
d) 7
e) 9$ a_k = \frac{(k+9)!}{(k-1)!} = k \cdot (k+1)\cdot(k+2)\cdot (k+3)\cdot ... \cdot (k+9) $

We have ten consecutives numbers, which are 5 odds and 5 evens.

By the conditions, we need to have $ a_k = 2^{x}\cdot 5^{y}\cdot... $ with $y \geq x$

But I don't know how continue.

There are more 5's than 2 . after 5's get paired with 2 we get zeros at the end and only od numberer remains along with 5. so last non zero digit is 5.