My answer is below, is this the answer you were looking for or is there a lower one?
Just because your voice reaches halfway around the world doesn't mean you are wiser than when it reached o

If we're looking for n where n divided by x (x being each number 2-10) leaves a remainder of x-1, n+1 divided by each number 2-10 leaves no remainder. To find n+1, we must simply multiply all prime fa

Let X be the number. From the above properties of X, one can see that,
"X+1" is a multiple of the numbers from 2 to 10. As Bonanova is asking for the smallest number with such properties, "X+1" must

## Question

## bonanova 85

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.

When I divide it by 3, the remainder is 2.

When I divide it by 4, the remainder is 3.

When I divide it by 5, the remainder is 4.

When I divide it by 6, the remainder is 5.

When I divide it by 7, the remainder is 6.

When I divide it by 8, the remainder is 7.

When I divide it by 9, the remainder is 8.

When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.

When I looked for a smaller number with this property I couldn't find one.

Can you find it?

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## Guest

My answer is below, is this the answer you were looking for or is there a lower one? Just because your voice reaches halfway around the world doesn't mean you are wiser than when it reached o

## Guest

If we're looking for n where n divided by x (x being each number 2-10) leaves a remainder of x-1, n+1 divided by each number 2-10 leaves no remainder. To find n+1, we must simply multiply all prime fa

## Guest

Let X be the number. From the above properties of X, one can see that, "X+1" is a multiple of the numbers from 2 to 10. As Bonanova is asking for the smallest number with such properties, "X+1" must

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