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標題:
若2009xM是一個平方數,咁個M係???
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最佳解答:
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.
其他解答:
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
若2009xM是一個平方數,咁個M係???
發問:
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若2009xM是一個平方數,咁個M係??? + 原因!!!最佳解答:
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.
其他解答:
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
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