Remainder Remaining

Algebra Level 4

Find the remainder when 1 + x 2 + x 4 + x 6 + x 8 + x 10 + ⋯ + x 22 1+{ x }^{ 2 }+{ x }^{ 4 }+{ x }^{ 6 }+{ x }^{ 8 }+{ x }^{ 10 }+ \cdots+{ x }^{ 22 } is divided by 1 + x 1 + x 2 + x 3 + x 4 + x 5 + ⋯ + x 11 1+{ x }^{ 1 }+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+{ x }^{ 5 }+ \cdots+{ x }^{ 11 } .

Source: KVPY 2016.
1 + x 1 + x 2 + x 3 + ⋯ + x 10 1+{ x }^{ 1 }+{ x }^{ 2 }+{ x }^{ 3 }+\cdots+{ x }^{ 10 } 2 [ 1 + x 1 + x 2 + x 3 + ⋯ + x 10 ] 2[1+{ x }^{ 1 }+{ x }^{ 2 }+{ x }^{ 3 }+\cdots+{ x }^{ 10 }] 2 2 1 1

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Prajwal Krishna by Prajwal Krishna , 22, India

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1 solution

Prajwal Krishna
Nov 8, 2016

1 + x 1 + x 2 + x 3 + x 4 + x 5 . . . . . . . . + x 11 1+{ x }^{ 1 }+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+{ x }^{ 5 }........+{ x }^{ 11 } = 1 − x 12 1 − x \frac { 1-{ x }^{ 12 } }{ 1-x } Similarly 1 + x 2 + x 4 + x 6 + x 8 + x 10 . . . . . . . . + x 22 1+{ x }^{ 2 }+{ x }^{ 4 }+{ x }^{ 6 }+{ x }^{ 8 }+{ x }^{ 10 }........+{ x }^{ 22 } = 1 − x 24 1 − x 2 \frac { 1-{ x }^{ 24} }{ 1-{ x }^{ 2 } }

Now , 1 − x 12 1 − x 2 \frac { 1-{ x }^{ 12 } }{ 1-{ x }^{ 2 } } ÷ \div 1 + x 1 + x 2 + x 3 + x 4 + x 5 . . . . . . . . + x 11 1+{ x }^{ 1 }+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+{ x }^{ 5 }........+{ x }^{ 11 } = 1 + x 12 1 + x \frac { 1+{ x }^{ 12 } }{ 1+x }

Now by factor theorem , remainder when 1 + x 12 1+{ x }^{ 12 } is divided by 1+x = 1 + ( − 1 ) 12 1+{ (-1 )}^{ 12 } = 2

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Can you spot the mistake in this solution?

Hint: Write out the polynomial form as N ( x ) = k ( x ) Q ( x ) + R ( x ) N(x) = k(x) Q(x) + R(x) . See what happens as we proceed with this solution.

Did the same way, please share more KVPY ques.

Md Zuhair - 4 years, 7 months ago

You are right.

1 + x 12 1 + x = x 12 − 1 + 2 x + 1 = ( x + 1 ) ( x 11 − x 10 + x 9 − . . . − 1 ) + 2 x + 1 = ( x 11 − x 10 + x 9 − . . . − 1 ) + 2 x + 1 \begin{aligned} \frac {1+x^{12}}{1+x} & = \frac {x^{12}-1+2}{x+1} \\ & = \frac {(x+1)(x^{11}-x^{10}+x^9-...-1)+2}{x+1} \\ & = (x^{11}-x^{10}+x^9-...-1) + \frac {\boxed{2}}{x+1} \end{aligned}

Chew-Seong Cheong - 4 years, 7 months ago

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Note: What your statement is true, the answer to the problem is not 2.

Calvin Lin Staff - 3 years ago

how can 1 + x 2 + x 4 + x 6 + x 8 + x 10 . . . . . . . . + x 22 1+{ x }^{ 2 }+{ x }^{ 4 }+{ x }^{ 6 }+{ x }^{ 8 }+{ x }^{ 10 }........+{ x }^{ 22 } = 1 − x 12 1 − x 2 \frac { 1-{ x }^{ 12 } }{ 1-{ x }^{ 2 } } this be true it should be 1 − x 24 1 − x 2 \frac { 1-{ x }^{ 24 } }{ 1-{ x }^{ 2 } }

saharsh rathi - 4 years, 7 months ago

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Sry typing mistake

Prajwal Krishna - 4 years, 7 months ago

Does this actually work? In the original polynomials, if we substitute x = 1, the results are the same and so they divide each other for a remainder of zero.

Bryan Hung - 4 years, 7 months ago

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Yes, you can refer to remainder factor theorem .

Chew-Seong Cheong - 4 years, 7 months ago

Yes it is the remainder theorem.

Md Zuhair - 4 years, 7 months ago

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But when I substitute x=1, I don't get a remainder of 2?

Bryan Hung - 4 years, 7 months ago

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@Bryan Hung – If 1 + x = 0 1+x=0 hence x = − 1 x=-1 . Now see x 12 + 1 x^{12} + 1 , we put values , ( − 1 ) 12 = 1 + 1 = 2 (-1)^{12} = 1 + 1 = 2 . So what is your problem?

Md Zuhair - 4 years, 7 months ago

There is a slight error with this exact approach in the final step.

However, the general idea applied here works. It might be easier to think of it as N = k Q + R N = kQ + R , and in polynomial form as N ( x ) = k ( x ) Q ( x ) + R ( x ) N(x) = k(x) Q(x) + R(x) . If so, we're allowed to multiply and divide by non-zero polynomials while the statement stays true.

Calvin Lin Staff - 3 years ago

1 pending report

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