Question 32179: Find the smallest positive integer n such that 2n is a perfect square, 3n is a perfect cube, and 5n is a perfect fifth power.
Found 4 solutions by mszlmb, Prithwis, lyra, amit5562:
Answer by mszlmb(115)   (Show Source):
You can put this solution on YOUR website! duh dude its one
>>oops!
how about 2*3*5?
30(2) doesn't work
how about 4*9*25?
900.
900*2=1800 sqrt=2*3*3*2*2*5*5 that's 900sqrt(2) :PPPPP
ok so the problem is much harder than i thot, i apologize a lot!
Answer by Prithwis(166)  (Show Source):
Answer by lyra(94)  (Show Source):
You can put this solution on YOUR website! The definition of a perfect sqaure is "A number whose square root is an integer". We can test a few small integers, lets start with one, the smallest.  which is not a perfect square since  is not an integer. Thus it cannot be 1. Lets try 2,  ,  which is an integer. So 2 works for 2n, lets try it with 3n:  the cube root of 6 is not an integer, so it cannot be 2. Lets try 3, ,  is not an integer, so it cannot be 3. As we keep going down, we find a pattern, for 2n to work 2(some even number n)=perfect square. We can skip to 8=n, so  which is a perfect square, but 3 is not a perfect cube. This seems to be going pretty slowly, so lets try a different technique (don't be afraid in math to start over, and go down a different path, sometimes you just have to reorganize your thinking.) Lets try a different aproach:
lets go by cubes.
 so n=9 might work,  oops, that wont work.  non-integer.  =non-integer.   okay two down, one to go,   darn, that won't work!  non-integer.  non-integer.   non-integer. non-integer. Now that you see the pattern you can keep on testing,
Good luck!
Hope this helps, this is a very difficult problem,
lyra
Answer by amit5562(1)  (Show Source):
You can put this solution on YOUR website! If I am not wrong N will be of the form
N=2^p*3^q*5^r
Now,p should be divisible by 3&5(as while taking 3rd root or 5th root power of 2 remains unchanged)so p=15,30,45... Etc but we need the least value of p which when multiplied by 2(as question says 2N=2^(p+1)) becomes an even power(to become a perfect sqaure). So least possible value of p=15.
Similarly, q should be divisible by 2&5(as while taking square root or 5th root,its power remains unchanged). So, q=10,20,30.... But we need that value into which 1 added will give multiple of 3. Here it is 20.
Similarly for r, it should be multiple of 2&3. So, r=6,12,18,24,30... But we need that value which when increased by 1 (5N=5^r+1) becomes a multiple of 5. Here it is 24.
So N=2^15*3^20*5^24.
Guys i think, there's no number smaller than this number which satisfies the given condition.
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