SOLUTION: Given 60=2^2 x 3 x 5 and 1050=2 x 3x 5^2 x 7, find (a) the smallest integer m such that 60m is a perfect cube (b) the smallest positive integer value of n for which 60n is a mu

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: Given 60=2^2 x 3 x 5 and 1050=2 x 3x 5^2 x 7, find (a) the smallest integer m such that 60m is a perfect cube (b) the smallest positive integer value of n for which 60n is a mu      Log On


   



Question 409629: Given 60=2^2 x 3 x 5 and 1050=2 x 3x 5^2 x 7, find
(a) the smallest integer m such that 60m is a perfect cube
(b) the smallest positive integer value of n for which 60n is a multiple of 1050.

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Since 60+=+2%5E2%2A3%2A5 we can find the smallest (positive) integer such that the prime factors of 60 all have exponents of multiples of 3. Since the exponents of 2, 3, and 5 are 2, 1, 1 respectively, then m can have 2, 3, 5 exponents of 1, 2, 2 respectively. Therefore m+=+2%2A%283%5E2%29%2A%285%5E2%29+=+450.

We want 60n%2F1050 to be an integer. The fraction reduces to 2n%2F35, hence n = 35 is the smallest integer value, as 2 and 35 are relatively prime.