document.write( "Question 1043962: If 360=2x×3y×5z ;than find x+y+z \n" ); document.write( "

Algebra.Com's Answer #659191 by ikleyn(52788) $\"\"$ $\"About$

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document.write( "1.  The condition is incorrect.\r\n" );
document.write( "    The correct condition must say that x, y and z are positive integers.\r\n" );
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document.write( "    Otherwise the number of solution is more than infinite.\r\n" );
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document.write( "2.  OK. Let us add it into the condition.\r\n" );
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document.write( "    Then, if  360=2x×3y×5z,  then xyz = 12,   and it is much easier to solve.\r\n" );
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document.write( "    The solutions are  (\"by the modulus of permutations\", which doesn't matter, because the question is about the sum x+y+z invariant to permutations)\r\n" );
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document.write( "    a)  1, 1, 12  --->  x + y + z = 1 + 1 + 12 = 14;\r\n" );
document.write( "    b)  1, 2,  6  --->  x + y + z = 1 + 2 +  6 =  9;\r\n" );
document.write( "    c)  1, 3, 4  --->   x + y + z = 1 + 3 +  4 =  8;\r\n" );
document.write( "    d)  2, 2, 3  --->   x + y + z = 2 + 2 + 3 =   7.\r\n" );
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document.write( "Answer.  x + y + z = 14, 9, 8, 7.\r\n" );
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