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 be the LCM (least common multiple) of the numbers from 2 to 10 -- which is 2520.
And hence, X = 2520-1 = 2519.
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 factors of 2-10, excluding any duplicates. This is equal to 2*2*2*3*3*5*7 = 2520, then just subtract one to get the answer of 2519.
Excuse me Ladies and Gentlemen (and sorry for my poor english): Why do you think the dog have the time to go at 1.5 miles in the time the man go at 3 miles? this is the joke
the man go at 3+1,5 miles in the time the dog go at 1,5 miles. they come back togheter and THE DOG WALK FOR 3 MILES. it's right?