document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #698381 by amit5562(1) $\"\"$ $\"About$

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If I am not wrong N will be of the form
\n" ); document.write( "N=2^p*3^q*5^r
\n" ); document.write( "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.
\n" ); document.write( "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.
\n" ); document.write( "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.
\n" ); document.write( "So N=2^15*3^20*5^24.
\n" ); document.write( "Guys i think, there's no number smaller than this number which satisfies the given condition. \n" ); document.write( "

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