Difference between revisions of "2008 iTest Problems/Problem 80"
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Rockmanex3 (talk | contribs) (Alternate Solution to Problem 80) |
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the remainder when <math>|r(2008)|</math> is divided by <math>1000</math>. | the remainder when <math>|r(2008)|</math> is divided by <math>1000</math>. | ||
− | == Solution == | + | == Solutions == |
+ | |||
+ | ===Solution 1=== | ||
+ | |||
<math>x^4+x^3+2x^2+x+1 = (x^2 + 1)(x^2 + x + 1)</math>. We apply the polynomial generalization of the [[Chinese Remainder Theorem]]. | <math>x^4+x^3+2x^2+x+1 = (x^2 + 1)(x^2 + x + 1)</math>. We apply the polynomial generalization of the [[Chinese Remainder Theorem]]. | ||
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using similar reasoning. Hence <math>p(x) \equiv x+1 \pmod{x^2 + x + 1}, p(x) \equiv 1 \pmod{x^2 + 1}</math>, and by CRT we have <math>p(x) \equiv -x^2 \pmod{x^4+x^3+2x^2+x+1}</math>. | using similar reasoning. Hence <math>p(x) \equiv x+1 \pmod{x^2 + x + 1}, p(x) \equiv 1 \pmod{x^2 + 1}</math>, and by CRT we have <math>p(x) \equiv -x^2 \pmod{x^4+x^3+2x^2+x+1}</math>. | ||
− | Then <math>|r(2008)| \equiv 2008^2 \equiv 64 \pmod{1000}</math>. | + | Then <math>|r(2008)| \equiv 2008^2 \equiv \boxed{64} \pmod{1000}</math>. |
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | The given information can be represented as | ||
+ | <cmath>x^{2008} + x^{2007} + \cdots + x^2 + x + 1 = q(x) \cdot (x^4 + x^3 + 2x^2 + x + 1) + r(x)</cmath> | ||
+ | where <math>r(x)</math> is the remainder and <math>q(x)</math> is the quotient. Multiplying both sides by <math>x-1</math> would make computation easier; doing so results in | ||
+ | <cmath>\begin{align*} | ||
+ | x^{2009} - 1 &= q(x) \cdot (x^4 + x^3 + 2x^2 + x + 1)(x-1) + r(x) \cdot (x-1) \\ | ||
+ | &= q(x) \cdot (x^2 + 1)(x^2 + x + 1)(x-1) + r(x) \cdot (x-1) | ||
+ | \end{align*}</cmath> | ||
+ | Now plug in values of <math>x</math> not equal to 1 that make <math>q(x) \cdot (x^2 + 1)(x^2 + x + 1)(x-1) = 0</math>. By the Zero Product Property, the values of <math>x</math> that make the expression equal to 0 are <math>\pm i</math> and <math>\tfrac{-1 \pm i\sqrt{3}}{2}</math>. | ||
+ | |||
+ | <br> | ||
+ | By plugging in <math>\pm i</math> and <math>\tfrac{-1 \pm i\sqrt{3}}{2}</math> into the equation and solving for the remainder function, we have <math>r(\pm i) = 1, r(\tfrac{-1 + i\sqrt{3}}{2}) = \tfrac{1 + i\sqrt{3}}{2}, r(\tfrac{-1 - i\sqrt{3}}{2}) = \tfrac{1 - i\sqrt{3}}{2}</math>. Since the remainder function's degree is at most 3 and we know four points, we can construct the unique remainder function. | ||
+ | |||
+ | <br> | ||
+ | Let <math>r(x) = ax^3 + bx^2 + cx + d</math>. Plugging in <math>i</math> results in <math>-ai - b + ci + d = 1</math>, and plugging in <math>-i</math> results in <math>ai - b - ci + d = 1</math>. Thus, <math>a = c</math> and <math>d-b = 1</math>. Plugging in <math>\tfrac{-1 + i\sqrt{3}}{2}</math> results in <math>a + \tfrac{-1 - i\sqrt{3}}{2}b + \tfrac{-1 + i\sqrt{3}}{2}a + d = \tfrac{1 + i\sqrt{3}}{2}</math>. Plugging in <math>\tfrac{-1 - i\sqrt{3}}{2}</math> results in <math>a + \tfrac{-1 + i\sqrt{3}}{2} + \tfrac{-1 - i\sqrt{3}}{2}a + d = \tfrac{1 - i\sqrt{3}}{2}</math>. Thus, <math>-b + a = 1</math> and <math>\tfrac12 a - \tfrac12 b + d = \tfrac12</math>, so | ||
+ | <cmath>\begin{align*} | ||
+ | a - b + 2d &= 1 \\ | ||
+ | 1 + 2d &= 1 \\ | ||
+ | d &= 0 | ||
+ | \end{align*}</cmath> | ||
+ | Substituting <math>d = 0</math> results in <math>b = -1</math> and <math>a = c = 0</math>. Therefore, <math>r(x) = -x^2</math>, so <math>|r(2008)| = 4032064</math>, and the remainder when divided by 1000 is <math>\boxed{64}</math>. | ||
+ | |||
+ | == See Also == | ||
+ | {{2008 iTest box|num-b=79|num-a=81}} | ||
− | |||
[[Category:Intermediate Algebra Problems]][[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Algebra Problems]][[Category:Intermediate Number Theory Problems]] |
Latest revision as of 19:49, 23 November 2018
Problem
Let
and let be the polynomial remainder when is divided by . Find the remainder when is divided by .
Solutions
Solution 1
. We apply the polynomial generalization of the Chinese Remainder Theorem.
Indeed,
since . Also,
using similar reasoning. Hence , and by CRT we have .
Then .
Solution 2
The given information can be represented as where is the remainder and is the quotient. Multiplying both sides by would make computation easier; doing so results in Now plug in values of not equal to 1 that make . By the Zero Product Property, the values of that make the expression equal to 0 are and .
By plugging in and into the equation and solving for the remainder function, we have . Since the remainder function's degree is at most 3 and we know four points, we can construct the unique remainder function.
Let . Plugging in results in , and plugging in results in . Thus, and . Plugging in results in . Plugging in results in . Thus, and , so
Substituting results in and . Therefore, , so , and the remainder when divided by 1000 is .
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 79 |
Followed by: Problem 81 | |
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