Determine the largest possible value of B such that it is a multiple of all positive whole numbers less than ³√B.

Let M(x) be the LCM of all integers less than or equal to x. M(x+1) will increase any time x+1 introduces a new factor. It will grow most in the cases of prime numbers, but perfect powers will also cause it to grow.

If x+1 is prime, M(x+1)=(x+1)M(x) ³√M(x+1)=³√(x+1)³√M(x) In that case, increasing x by 1 multiplies M(x) by ³√(x+1), which will grow M(x) fairly rapidly even for small x, so we take a straightforward approach.

Several numbers will fit the criteria. In an effort to find the largest overall, we can group numbers which give the same result to floor(³√(number)) because only the largest in each group will be admissable. By the previous paragraph, we have good reason to believe there will be a number such that all groups above it have no member, and thus, it must contain the desired number.

Choosing x as the independent variable, we look for the largest  mutliple of M(x) (called Z) such that x is less than ³√Z but not less than ³√Z-1. By definition, these Z values are multiples of M(x), and by the method of finding them, they are a multiple of all positive whole numbers less than ³√M(x)

x=1, M(x)=1, gives Z=8
x=2, M(x)=2, gives Z=26
x=3, M(x)=6, gives Z=60
x=4, M(x)=12, gives Z=120
x=5, M(x)=60, gives Z=180
x=6, M(x)=60, gives Z=300
x=7, M(x)=420, gives Z=420
x=8, M(x)=840, gives no possible Z.

M(x) will increase too rapidly past this to allow any other possible Z values.

Posted by Gamer on 2007-08-10 16:50:11