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Difference between revisions of "2020 AMC 10B Problems/Problem 22"

m (→‎Solution 3 (MAA Original Solution): add alignment)
(→‎Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation))
 
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<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math>
 
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math>
  
−
==Solution 1==
+
==Solution 1 (MAA Original Solution)==
 +
Completing the square, then difference of squares:
 +
 
 +
<cmath>
 +
\begin{align*}
 +
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\
 +
&= (2^{101} + 1)^2 - 2^{102} + 201\\
 +
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.
 +
\end{align*}
 +
</cmath>
 +
 
 +
 
 +
Thus, we see that the remainder is <math>\boxed{\textbf{(D) } 201}</math>
 +
 
 +
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)
 +
 
 +
==Solution 2==
  
 
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.
 
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.
  
−
We could proceed with polynomial division, but the denominator looks awfully similar to the [[Sophie Germain Identity]],  which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath>
+
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]],  which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath>
  
 
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have
 
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have
Line 21: Line 37:
 
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath>
 
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath>
  
−
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88
+
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math>  
  
−
==Solution 2==
+
~quacker88
−
Similar to Solution 1, let <math>x=2^{50}</math>. It suffices to find remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. Dividing polynomials results in a remainder of <math>\boxed{\textbf{(D) } 201}</math>.
 
  
−
==Solution 3 (MAA Original Solution)==
 
  
−
<cmath>
 
−
\begin{align*}
 
−
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\
 
−
&= (2^{101} + 1)^2 - 2^{102} + 201\\
 
−
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.
 
−
\end{align*}
 
−
</cmath>
 
  
−
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math>
+
==Solution 3 (Same As Solution 2)==
−
 
 
−
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)
 
−
 
 
−
==Solution 4==
 
 
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.
 
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.
 
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.
 
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.
 
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math>
 
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math>
  
−
==Solution 5==
+
==Solution 4 ==
−
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.
+
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+2^{0.5}x+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.
 +
 
 +
The long division is essentially the same if you work with <math>x=2</math>, or do repeated multiplication and subtraction using the original expression.
 +
 
 +
==Solution 5 (Modular Arithmetic)==
 +
Let <math>n=2^{101}+2^{51}+1</math>. Then, mod <math>n</math>:
 +
 
 +
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 </math>
 +
 
 +
<math>\equiv 2^{102}+2^{52}+203 </math>
 +
 
 +
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>.
 +
 
 +
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.
 +
 
 +
~ Leo.Euler
 +
 
 +
~ (edited by asops)
 +
 
 +
==Solution 6(Author: Shiva Kumar Kannan - Least insightful & very straightforward + Manipulation)==
 +
 
 +
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:
 +
 
 +
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} </cmath> <cmath>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>=2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</cmath>
 +
 
 +
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is    <math>\boxed{\textbf{(D) } 201}</math>.
  
 
==Video Solutions==
 
==Video Solutions==

Latest revision as of 16:00, 9 October 2024

Problem

What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$?

$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$

Solution 1 (MAA Original Solution)

Completing the square, then difference of squares:

\begin{align*} 2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\ &= (2^{101} + 1)^2 - 2^{102} + 201\\ &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201. \end{align*}


Thus, we see that the remainder is $\boxed{\textbf{(D) } 201}$

(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)

Solution 2

Let $x=2^{50}$. We are now looking for the remainder of $\frac{4x^4+202}{2x^2+2x+1}$.

We could proceed with polynomial division, but the numerator looks awfully similar to the Sophie Germain Identity, which states that \[a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)\]

Let's use the identity, with $a=1$ and $b=x$, so we have

\[1+4x^4=(1+2x^2+2x)(1+2x^2-2x)\]

Rearranging, we can see that this is exactly what we need:

\[\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1\]

So \[\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1}\]

Since the first half divides cleanly as shown earlier, the remainder must be $\boxed{\textbf{(D) }201}$

~quacker88


Solution 3 (Same As Solution 2)

We let \[x = 2^{50}\] and \[2^{202} + 202 = 4x^{4} + 202\]. Next we write \[2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1\]. We know that \[4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)\] by the Sophie Germain identity so to find \[4x^{4} + 202,\] we find that \[4x^{4} + 202 = 4x^{4} + 201 + 1\] which shows that the remainder is $\boxed{\textbf{(D) } 201}$

Solution 4

We let $x=2^{50.5}$. That means $2^{202}+202=x^{4}+202$ and $2^{101}+2^{51}+1=x^{2}+2^{0.5}x+1$. Then, we simply do polynomial division, and find that the remainder is $\boxed{\textbf{(D) } 201}$.

The long division is essentially the same if you work with $x=2$, or do repeated multiplication and subtraction using the original expression.

Solution 5 (Modular Arithmetic)

Let $n=2^{101}+2^{51}+1$. Then, mod $n$:

$2^{202}+202 \equiv (-2^{51}-1)^2 + 202$

$\equiv 2^{102}+2^{52}+203$

$= 2(n-1)+203 \equiv 201 \pmod{n}$.

Thus, the remainder is $\boxed{\textbf{(D) } 201}$.

~ Leo.Euler

~ (edited by asops)

Solution 6(Author: Shiva Kumar Kannan - Least insightful & very straightforward + Manipulation)

We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:

\[\frac{2^{202}+202}{2^{101}+2^{51}+1}\] \[= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\] \[= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\] \[=2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}\] \[=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1}\] \[= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}\] \[= 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.\]

Clearly, $201 < 2^{201} + 2^{51} + 1$, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is $\boxed{\textbf{(D) } 201}$.

Video Solutions

Video Solution 1 by Mathematical Dexterity (2 min)

https://www.youtube.com/watch?v=lLWURnmpPQA

Video Solution 2 by The Beauty Of Math

https://youtu.be/gPqd-yKQdFg

Video Solution 3

https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx

Video Solution 4 Using Sophie Germain's Identity

https://youtu.be/ba6w1OhXqOQ?t=5155

~ pi_is_3.14

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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