Remainder Theorem Advanced Question #1

Algebra Level 5

Let f ( x ) f(x) be a polynomial such that

  • f ( x ) ÷ ( x 3 + x + 1 ) f(x) \div (x^3+x+1) has a remainder of 3 3 ;
  • f ( x ) ÷ ( x 3 − x + 1 ) f(x) \div (x^3-x+1) has a remainder of x + 1 x+1 .

Also, let R ( x ) R(x) be the remainder of the division f ( x ) ÷ ( x 6 + 2 x 3 − x 2 + 1 ) f(x) \div (x^6+2x^3-x^2+1) and let R ( 2 ) − R ( − 2 ) = k R(2)-R(-2)=k .

Now, for all positive integers n , n, if g ( n ) g(n) is the number of factors of n n that are positive integers, what is the value of k − g ( k ) ? k-g(k)?


The answer is 62.

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Boi (보이) by Boi (보이) , 19, South Korea

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2 solutions

Atomsky Jahid
Jun 9, 2017

We can write the polynomial f ( x ) f(x) in the following way, f ( x ) = Q ( x ) ( x 3 + x + 1 ) ( x 3 − x + 1 ) + R ( x ) f(x)=Q(x)(x^3+x+1)(x^3-x+1)+R(x) Now, R ( x ) R(x) is a polynomial of degree 5 at most. And, it leaves a remainder of 3 3 and x + 1 x+1 when divided by x 3 + x + 1 x^3+x+1 and x 3 − x + 1 x^3-x+1 respectively. So, R ( x ) R(x) can be written as, R ( x ) = ( A x 2 + B x + C ) ( x 3 + x + 1 ) + 3 R(x)=(Ax^2+Bx+C)(x^3+x+1)+3 and, R ( x ) = ( D x 2 + E x + F ) ( x 3 − x + 1 ) + x + 1 R(x)=(Dx^2+Ex+F)(x^3-x+1)+x+1 Equating the above two equations we get, A = D = 1 A=D=1 ; B = E = 0 B=E=0 ; C = − 1 2 C=-\frac{1}{2} and, F = 3 2 F=\frac{3}{2} So, R ( x ) = ( x 2 − 1 2 ) ( x 3 + x + 1 ) + 3 R(x)=(x^2-\frac{1}{2})(x^3+x+1)+3 or, R ( x ) = x 5 + 1 2 x 3 + x 2 − 1 2 x + 5 2 R(x)=x^5+\frac{1}{2}x^3+x^2-\frac{1}{2}x+\frac{5}{2} Now, R ( 2 ) − R ( − 2 ) = 39 + 5 2 − ( − 31 + 5 2 ) = 70 R(2)-R(-2)=39+\frac{5}{2}-(-31+\frac{5}{2})=70 and, 70 = 2 × 5 × 7 70=2 \times 5 \times 7 . Hence, g ( 70 ) = 2 3 = 8 g(70)=2^3=8 . Therefore, the answer is 70 − g ( 70 ) = 70 − 8 = 62 70-g(70)=70-8=\boxed{62} .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Ohhh that's really neat!

Didn't think about dividing R ( x ) R(x) .

'Brilliant' solution you got there!

Boi (보이) - 4 years ago

Impressed! ! First tell me how these ideas comes into mind ? And how many days did it take to solve it ?

Amit Kumar - 3 years, 12 months ago

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The necessary parts of this solution didn't take much time. But, there was another issue which did cause enough trouble for me. You may want to look at this discussion thread. Stuck in Remainders!

Atomsky Jahid - 3 years, 12 months ago
Boi (보이)
Jun 6, 2017

From f ( x ) = ( x 3 + x + 1 ) Q 1 ( x ) + 3 f(x)=(x^3+x+1)Q_{1}(x)+3 , we can get

( x 3 − x + 1 ) f ( x ) = ( x 3 + x + 1 ) ( x 3 − x + 1 ) Q 1 ( x ) + 3 ( x 3 − x + 1 ) (x^3-x+1)f(x)=(x^3+x+1)(x^3-x+1)Q_{1}(x)+3(x^3-x+1) , which would be simplified as:

( x 3 − x + 1 ) f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) Q 1 ( x ) + 3 x 3 − 3 x + 3 ⋯ A (x^3-x+1)f(x)=(x^6+2x^3-x^2+1)Q_{1}(x)+3x^3-3x+3\quad\cdots A

.

And from f ( x ) = ( x 3 − x + 1 ) Q 2 ( x ) + x + 1 f(x)=(x^3-x+1)Q_{2}(x)+x+1 , we can get

( x 3 + x + 1 ) f ( x ) = ( x 3 + x + 1 ) ( x 3 − x + 1 ) Q 2 ( x ) + ( x + 1 ) ( x 3 + x + 1 ) (x^3+x+1)f(x)=(x^3+x+1)(x^3-x+1)Q_{2}(x)+(x+1)(x^3+x+1) , which would be simplified as:

( x 3 + x + 1 ) f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) Q 2 ( x ) + x 4 + x 3 + x 2 + 2 x + 1 ⋯ B (x^3+x+1)f(x)=(x^6+2x^3-x^2+1)Q_{2}(x)+x^4+x^3+x^2+2x+1\quad\cdots B

.

Subtract equation A A from equation B B .

2 x f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) { Q 2 ( x ) − Q 1 ( x ) } + x 4 − 2 x 3 + x 2 + 5 x − 2 2xf(x)=(x^6+2x^3-x^2+1)\{Q_{2}(x)-Q_{1}(x)\}+x^4-2x^3+x^2+5x-2 .

Let Q 3 ( x ) = Q 2 ( x ) − Q 1 ( x ) Q_{3}(x)=Q_{2}(x)-Q_{1}(x) , and then substitute x = 0 x=0 to the above equation.

2 x f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) Q 3 ( x ) + x 4 − 2 x 3 + x 2 + 5 x − 2 ⋯ C 2xf(x)=(x^6+2x^3-x^2+1)Q_{3}(x)+x^4-2x^3+x^2+5x-2\quad\cdots C

0 = Q 3 ( 0 ) − 2 0=Q_{3}(0)-2

∴ Q 3 ( 0 ) = 2 \therefore Q_{3}(0)=2 .

.

Then, according to the remainder theorem,

Q 3 ( x ) = x Q 4 ( x ) + 2 Q_{3}(x)=xQ_{4}(x)+2 .

Substitute that to equation C C .

2 x f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) { x Q 4 ( x ) + 2 } + x 4 − 2 x 3 + x 2 + 5 x − 2 ; 2xf(x)=(x^6+2x^3-x^2+1)\{xQ_{4}(x)+2\}+x^4-2x^3+x^2+5x-2;

2 x f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) ⋅ x Q 4 ( x ) + 2 x 6 + 4 x 3 − 2 x 2 + 2 + x 4 − 2 x 3 + x 2 + 5 x − 2 ; 2xf(x)=(x^6+2x^3-x^2+1)\cdot xQ_{4}(x)+2x^6+4x^3-2x^2+2+x^4-2x^3+x^2+5x-2;

2 x f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) ⋅ x Q 4 ( x ) + 2 x 6 + x 4 + 2 x 3 − x 2 + 5 x 2xf(x)=(x^6+2x^3-x^2+1)\cdot xQ_{4}(x)+2x^6+x^4+2x^3-x^2+5x .

.

Now let's divide both sides by 2 x 2x .

f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) ⋅ Q 4 ( x ) 2 + 1 2 ( 2 x 5 + x 3 + 2 x 2 − x + 5 ) f(x)=(x^6+2x^3-x^2+1)\cdot\frac{Q_{4}(x)}{2}+\frac{1}{2}\left(2x^5+x^3+2x^2-x+5\right) .

∴ R ( x ) = 1 2 ( 2 x 5 + x 3 + 2 x 2 − x + 5 ) ; \therefore R(x)=\frac{1}{2}\left(2x^5+x^3+2x^2-x+5\right);

2 R ( x ) = 2 x 5 + x 3 + 2 x 2 − x + 5 2R(x)=2x^5+x^3+2x^2-x+5

.

2 R ( 2 ) = 83 , 2 R ( − 2 ) = − 57 2R(2)=83,\quad 2R(-2)=-57

2 R ( 2 ) − 2 R ( − 2 ) = 83 − ( − 57 ) = 140 2R(2)-2R(-2)=83-(-57)=140

∴ k = R ( 2 ) − R ( − 2 ) = 70 \therefore k=R(2)-R(-2)=70

.

70 = 2 × 5 × 7 70=2\times5\times7 . So, there are 8 8 factors of 70 70 that are natural numbers.

∴ k − g ( k ) = 70 − 8 = 62 \therefore k-g(k)=70-8=\boxed{62}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

x x cancellation is awesome!

I use this 1 x 6 + 2 x 3 − x 2 + 1 = 1 2 ( x 2 + 1 x 3 + x + 1 − x 2 − 1 x 3 − x + 1 ) \frac{1}{x^6+2x^3-x^2+1}=\frac{1}{2}(\frac{x^2+1}{x^3+x+1}-\frac{x^2-1}{x^3-x+1})

Kelvin Hong - 4 years ago

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Hmmm.

Care to elaborate more, please?

Boi (보이) - 4 years ago

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Sure,

f ( x ) = ( x 3 + x + 1 ) Q 1 ( x ) + 3 f(x)=(x^3+x+1)Q_1 (x)+3

f ( x ) = ( x 3 − x + 1 ) Q 2 ( x ) + x + 1 f(x)=(x^3-x+1)Q_2 (x)+x+1

f ( x ) = ( x 6 + 2 x 3 − x 2 + 1 ) Q 3 ( x ) + R ( x ) f(x)=(x^6+2x^3-x^2+1)Q_3(x)+R(x)

Using equation above,

2 f ( x ) = ( x 2 + 1 ) ( x 3 − x + 1 ) f ( x ) − ( x 2 − 1 ) ( x 3 + x + 1 ) f ( x ) 2f(x)=(x^2+1)(x^3-x+1)f(x)-(x^2-1)(x^3+x+1)f(x)

2 f ( x ) = ( x 2 + 1 ) ( x 3 − x + 1 ) ( x 3 + x + 1 ) Q 1 ( x ) + 3 ( x 2 + 1 ) ( x 3 − x + 1 ) − [ ( x 2 − 1 ) ( x 3 + x + 1 ) ( x 3 − x + 1 ) Q 2 ( x ) + ( x + 1 ) ( x 2 − 1 ) ( x 3 + x + 1 ) ] 2f(x)=(x^2+1)(x^3-x+1)(x^3+x+1)Q_1(x)+3(x^2+1)(x^3-x+1)-[(x^2-1)(x^3+x+1)(x^3-x+1)Q_2(x)+(x+1)(x^2-1)(x^3+x+1)]

Simplified gets

f ( x ) = 1 2 ( ( x 2 + 1 ) Q 1 ( x ) + ( x 2 − 1 ) Q 2 ( x ) − 1 ) ( x 6 + 2 x 3 − x 2 + 1 ) + x 5 + 1 2 x 3 + x 2 − 1 2 x + 5 2 f(x)=\frac{1}{2} ((x^2+1)Q_1(x)+(x^2-1)Q_2(x)-1)(x^6+2x^3-x^2+1)+x^5+\frac{1}{2}x^3+x^2-\frac{1}{2}x+\frac{5}{2}

So let

Q 3 ( x ) = 1 2 ( ( x 2 + 1 ) Q 1 ( x ) + ( x 2 − 1 ) Q 2 ( x ) − 1 ) Q_3(x)=\frac{1}{2}((x^2+1)Q_1(x)+(x^2-1)Q_2(x)-1)

Then gets,

R ( x ) = x 5 + 1 2 x 3 + x 2 − 1 2 x + 5 2 R(x)=x^5+\frac{1}{2}x^3+x^2-\frac{1}{2}x+\frac{5}{2}

After as the same as yours.

Kelvin Hong - 4 years ago

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@Kelvin Hong – Ooooh, nice! That's really neat!

Boi (보이) - 4 years ago

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@Boi (보이) – Thanks~ but I did a really trivial mistake so I can just post my solution here...

Kelvin Hong - 4 years ago

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@Kelvin Hong – Awh, that's bad :c

Boi (보이) - 4 years ago

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