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selvarajan · Jan 13, 2012, 09:54 PM

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

corrigan · Jan 14, 2012, 11:41 PM

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.