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Re: N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f
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01 Aug 2019, 01:35
N is a positive integer and is divisible by 2,3 and 5
so N is of form = \(2^a*3^b*5^c\)
now given \(\frac{N}{2}\) is a square (that means powers of N are multiple of 2
\(\frac{N}{2}\) = 2^(a-1)*3^b*5^c (when a-1 ,b and c divided by 2 result is integer)
a = 2I+1, b = 2I, c = 2I
a= 3,5,7...., b = 2,4,6...., c = 2,4,6.....
similarly, \(\frac{N}{3}\) is a cube (that means powers of N are multiple of 3)
\(\frac{N}{3}\) = 2^a*3^(b-1)*5^c (when a ,b-1 and c divided by 3 result is integer)
a = 3I, b = 3I+1 , c = 3I
a = 3,6,9..., b = 4,7,10..., c = 3,6,9...
\(\frac{N}{5}\) is a fifth power (that means powers of N are multiple of 5)
\(\frac{N}{5}\) = 2^a*3^b*5^(c-1) (when a ,b and c-1 divided by 5 result is integer)
a = 5I, b = 5I+1 , c = 5I
a= 5,10,15.. b = 6,11,16... c= 5,10,15
taking the LCM of all the 3 cases of value of a, b and c (to satisfy \(\frac{N}{2}\) square, \(\frac{N}{3}\) cube,\(\frac{N}{5}\) fifth power)
a = 15,30,45...., b= 10,20,30...., c = 6,12,18
minimum values of a,b, and c = 15,10, and 6
so minimum value of N = 2^15*3^10*5^6