No. And the proof is simple.
Suppose there were such a number, X.
Then X+1 is also a number and X+1 does not divide X.
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No number can satisfy these conditions: To have a remainder of 1 when divided by 6, the number must be odd (as all multiples of 6 are even and an even number plus 1 is odd) To have a remainder of 2 when divided by 8, the number must be even (as all multiples of 8 are even and an even number plus 2 is even) No number is both odd and even. → No number exists that has a remainder of 1 when divided by 6, and 2 when divided by 8.
It means nothing at all. Every whole number, except 1, can be divided evenly by 1. Which leaves 1 which can be divided evenly by -1. Thus every whole number can be divided evenly by another whole number.
There does not exist a number that is divisible by all integers. The opposite is true. The number one can divided into all integers.
1680 can be divided by all three.
A positive number divided by a negative number equals a negative number.