若2009xM是一個平方數,咁個M係???

The answer of 001 is one possible solution only. There are more than one answer actually. I can derive the solutions for you here.


First of all, 001 is right that 2009 = 7 x 7 x 41, with 7 and 41 relatively prime. Hence, if you multiply a number such that it can form "even" powers of indices of "both 7 and 41". It is thus not limited to 41.


Consider 2009 x M = ( 7 x 41 )^2n where n is a positive integer.


We can then find that, M = 7^( 2n - 2 ) x 41^( 2n - 1 )


Putting different natural no. n results in different M.


Put n = 1: M = 7^0 x 41^1 = 41


Put n = 2: M = 7^2 x 41^3 = 3,377,129


Put n = 3: M = 7^4 x 41^5 = 2.781707386 x 10^11 ...


On the other hand, actually if M is 2009^( 2n - 1 ), where n is an natural number, then 2009M can also be expressed as the square number for 2009M = 2009^2n which is the square number of 2009^n.


Put n = 1: M = 2009


Put n = 2: M = 2009^3 = 8,108,486,729 ...


You can therefore see there are many many possible solutions besides 41, say .2009, 3,377,129, 8,108,486,729 ... can you understand the investigation process above?


Hope I can help you.

先用短除法求得: 2009 = 7 x 7 x 41
而因為 (7 x 41) x (7 x 41) = (7 x 41)^2 =平方數
所以，若 2009 x M 是平方數，則 M = 41