number bases
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HallsofIvy
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Well, first, the base has to be at least 7 since there is the digit "6" in one of the numbers. Let "x" be the base. Then "465" is \(\displaystyle 4x^2+ 6x+ 5\), "24" is \(\displaystyle 2x+ 4\), "225" is \(\displaystyle 2x^2+ 2x+ 5\) and "1050" is \(\displaystyle x^3+ 5x\).
So your equation is \(\displaystyle 4x^2+ 6x+ 5+ 2x+ 4+ 2x^2+ 2x+ 5= x^3+ 5x\) which reduces to \(\displaystyle x^3- 6x^2- 5x- 14= 0\). The fact that x must be an integer makes this especially simple. By the "rational root theorem" any integer root must be a factor of -14 so must be one of 1, 2, 7, 14, -1, -2, -7, or -14. Since the root must be at least 7, it must be either 7 or 14. Try each of those in the polynomial to see which, if any, satisfy it.
So your equation is \(\displaystyle 4x^2+ 6x+ 5+ 2x+ 4+ 2x^2+ 2x+ 5= x^3+ 5x\) which reduces to \(\displaystyle x^3- 6x^2- 5x- 14= 0\). The fact that x must be an integer makes this especially simple. By the "rational root theorem" any integer root must be a factor of -14 so must be one of 1, 2, 7, 14, -1, -2, -7, or -14. Since the root must be at least 7, it must be either 7 or 14. Try each of those in the polynomial to see which, if any, satisfy it.
thanks Bros
Steven G
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While Halls gave a nice mechanical way of solving this problem R.M. nailed the problem by noticing that 5+4+5 must equal 0 which can only be true in base 7 or base 14.
Now one just has to try adding the the numbers in base 7 or 14 to determine the correct base.
Very nicely done.
Now one just has to try adding the the numbers in base 7 or 14 to determine the correct base.
Very nicely done.
What in the world is the point of such problems? They do not represent any question that 99.9% of students will ever face in their future lives. If we ever do have to figure out what is the base of a numeration system using place value and an unknown base (an extremely unlikely event for anyone except possibly a cryptographer), we would count symbols used (the main method to break codes).
Does answering this problem require an understanding of different numeral systems (not different number systems: F in hexadecimal still represents the number fifteen, the product of 5 and 3, the square of 4, 2 to the 4th power, etc) based on place value? Yes. But so would problems involving binary or hexadecimal numerals, all of which have important practical uses that some students will need to understand in their future lives. If you understand binary, decimal, and hexadecimal numeration, you understand bases and place value.
Does this help show the practical uses of the rational root theorem. I doubt whether there is much use for that theorem in an age of graphing calculators.
Does answering this problem require an understanding of different numeral systems (not different number systems: F in hexadecimal still represents the number fifteen, the product of 5 and 3, the square of 4, 2 to the 4th power, etc) based on place value? Yes. But so would problems involving binary or hexadecimal numerals, all of which have important practical uses that some students will need to understand in their future lives. If you understand binary, decimal, and hexadecimal numeration, you understand bases and place value.
Does this help show the practical uses of the rational root theorem. I doubt whether there is much use for that theorem in an age of graphing calculators.
Steven G
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That would be the strangest question I have ever seen if the author of the problem was trying to get the student to use the rational root theorem.
I agree. But seriously when have you ever run across any situation where you were presented with a sum and did not know what numeration system was involved in working the sum.. I used to work a lot with hexadecimal; I never had to guess about whether it was maybe base 7 or 17.
I reiterate: I understand there is value in getting students to understand place value notation. There is practical value in familiarizing students with binary and hexadecimal notation, perhaps even octal. But if you grasp three or four such systems, nothing is added to understanding by adding other systems that are never used.