A number is divided by 10 the reminder is 9--the same number is divided by 9 the reminder is 8--the same number is divided by 8 the reminder is 7 ---like this the same number is divided by 7 the reminder is 6--divided by 6 reminder 5--divided by 5 reminder 4--divided by 4 reminder 3--divided by 3 reminder 2 --divided by 2 reminder 1--What is the Number

Piece of cake. Let x be the number. Since x divided by 10 gives a remainder of 9, then x+1 gives a remainder of zero when divided by 10. So x+1 is a multiple of ten. Similarly, since x divided by 9 gives a remainder of 8, then x+1 gives a remainder of zero when divided by 9. And we can continue this for all the numbers from one to ten. So we get:

x +1= 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1
\Rightarrow x = 3, 628, 800 -1 = 3, 628, 799

Now that I think about it, there are an infinite number of possible answers.
(10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)^2 -1 = (3, 628, 800)^2 -1
would also give a remainder of 9 when divided by ten, etc. In fact, (3, 628, 800)^k -1 for any positive integer k would give the remainders that you are looking for. I could prove this by induction, but I don't think you need that much.