Perfect Square, Cube, Fourth PowerDate: 01/25/2002 at 19:05:54 From: James A. Hopton Subject: Perfect square, cube, fourth power My wife is an 8th Grade Math Teacher, and in the Bonus Section in one of the chapters of the current math book is this problem: Find the least integer greater than 1 that is a perfect square, a perfect cube, and a perfect fourth power. There was no equation with the problem. We have come up with the number 4096, (do not know if this is correct), by pushing and shoving numbers around. Is there an equation that will enable someone to solve this problem in a reasonable amount of time? Thank you for your help, and we anxiously await your answer. Date: 01/25/2002 at 20:06:08 From: Doctor Paul Subject: Re: Perfect square, cube, fourth power Every number can be written as a product of powers of primes - this is the Fundamental Theorem of Arithmetic. A number is a perfect square if, when raised to the 1/2 power, it still has an integer for all of the exponents in its prime factorization. For example: 3136 = 2^6*7^2 is a perfect square because 3136^(1/2) = [2^6*7^2]^(1/2) = 2^3*7 = 56. Similarly, a number will be a cube root if, when raised to the 1/3 power, it still has an integer for all of the exponents in its prime factorization. A number will be both a perfect square and a perfect cube if, when raised to both the 1/2 and the 1/3 powers, it still has an integer for all of the exponents in its prime factorization. otice that successively raising a number to the 1/2 and then the 1/3 power is the same as taking the 6th root of a number. So the numbers that are going to be perfect squares and perfect cubes are precisely the 6th powers of the integers: 1^6 = 1 2^6 = 64 3^6 = 729 4^6 = 4096 5^6 = 15625 etc... A similar computation shows that for a number to be a perfect square, a perfect cube, and the fourth power of some integer, it must be the 12th power of some integer. Note that we don't require the 24th power. The 24th power will work, but it's overkill. Essentially, you want to compute lcm(2,3,4) = lcm(2, lcm(3,4) ) = lcm(2,12) = 12. Does this make sense? Notice that the square root of x^12 = x^6, the cube root of x^12 is x^4, and the 4th root of x^12 is x^3. So the numbers that will be perfect squares, cubes, and fourth powers are the powers of 12: 1^12 = 1 2^12 = 4096 3^12 = 531441 etc... You have correctly computed (I guess by trial and error) that 4096 is the right answer. Perhaps now you understand why. I hope this helps. Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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