document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1171176>Question 1171176</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If a^2 x b^3 x c^4 = 648 000, where a, b, and c are distinct positive integers greater than 1, what is the least possible value of a + b + c? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #796095 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=796095\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Find the prime factorization<BR>\n" );
document.write( "648,000 = 2^6 * 3^4 * 5^3\r<BR>\n" );
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document.write( "Compare this to a^2 * b^3 * c^4\r<BR>\n" );
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document.write( "Note how 2^6 * 3^4 * 5^3 doesn't have a squared factor term, but it has cubic and fourth power terms.\r<BR>\n" );
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document.write( "We can fix this by pulling 2^4 out of 2^6 and having it pair with the 3^4 like so:<BR>\n" );
document.write( "a^2 * b^3 * c^4 = 2^6 * 5^3 * 3^4<BR>\n" );
document.write( "a^2 * b^3 * c^4 = (2^6) * 5^3 * 3^4<BR>\n" );
document.write( "a^2 * b^3 * c^4 = (2^2*2^4) * 5^3 * 3^4<BR>\n" );
document.write( "a^2 * b^3 * c^4 = 2^2 * 5^3 * (2^4*3^4)<BR>\n" );
document.write( "a^2 * b^3 * c^4 = 2^2 * 5^3 * (2*3)^4<BR>\n" );
document.write( "a^2 * b^3 * c^4 = 2^2 * 5^3 * 6^4\r<BR>\n" );
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document.write( "At this point, we can see<BR>\n" );
document.write( "a^2 = 2^2<BR>\n" );
document.write( "b^3 = 5^3<BR>\n" );
document.write( "c^4 = 6^4<BR>\n" );
document.write( "which leads to<BR>\n" );
document.write( "a = 2<BR>\n" );
document.write( "b = 5<BR>\n" );
document.write( "c = 6<BR>\n" );
document.write( "and finally,<BR>\n" );
document.write( "a+b+c = 2+5+6 = 13\r<BR>\n" );
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document.write( "This is the smallest value of a+b+c possible. The proof is given in the next section. \r<BR>\n" );
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document.write( "Here's a way to help prove we have the smallest a,b,c triple.\r<BR>\n" );
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document.write( "We know that a > 1 is some integer. Let's say 'a' is the smallest possible integer given that condition, so let's make a = 2.\r<BR>\n" );
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document.write( "a = 2<BR>\n" );
document.write( "a^2 = 4<BR>\n" );
document.write( "Divide like so<BR>\n" );
document.write( "(648,000)/4 = 162,000\r<BR>\n" );
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document.write( "This must mean<BR>\n" );
document.write( "b^3*c^4 = 162,000<BR>\n" );
document.write( "c^4 = 162000/(b^3)<BR>\n" );
document.write( "c = (162000/(b^3))^(1/4)\r<BR>\n" );
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document.write( "Let<BR>\n" );
document.write( "c = (162000/(x^3))^(1/4)<BR>\n" );
document.write( "so that implies that b = x\r<BR>\n" );
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document.write( "We can construct the function<BR>\n" );
document.write( "f(x) = a+b+c<BR>\n" );
document.write( "f(x) = 2+x+(162000/(x^3))^(1/4)<BR>\n" );
document.write( "in which we want to minimize, such that x > 0 and x is an integer.\r<BR>\n" );
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document.write( "The use of differential calculus or a graphing calculator will find that the min of f(x) is located at approximately (4.71, 12.98)\r<BR>\n" );
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document.write( "But we can't have non-integer x values, so we could try f(4) and f(5) since 4.71 is between 4 and 5.<BR>\n" );
document.write( "It turns out that f(4) = 13.09 approximately and f(5) = 13 exactly. Any other x value will lead to f(x) being larger. So f(5) = 13 leads to x = 5 and b = 5 and c = 6.\r<BR>\n" );
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document.write( "This fully proves that we have the smallest a,b,c triple possible. I skipped over the actual steps of using differential calculus, so let me know if you need me to go over that. The use of a graphing calculator is the preferred, quickest, and most efficient way to do this problem in my opinion.\r<BR>\n" );
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document.write( "Answers:<BR>\n" );
document.write( "a = 2<BR>\n" );
document.write( "b = 5<BR>\n" );
document.write( "c = 6<BR>\n" );
document.write( "a+b+c = 13 is the smallest possible sum.  \r<BR>\n" );
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