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

Revision as of 12:29, 16 January 2023

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

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 2 (MAA Original Solution)

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

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

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

Solution 3

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 $2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1$ . Then, we simply do polynomial division, and find that the remainder is $\boxed{\textbf{(D) } 201}$ .

Solution 5 (Modular Arithmetic)

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

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

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

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

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

~ Leo.Euler

~ (edited by asops)

Solution 6(Author: Shiva Kannan)

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

$\frac{2^{202}+202}{2^{101}+2^{51}+1}$ $= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1}$ $-$ $\frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}$

Video Solutions

~ pi_is_3.14

@@ Line 64: / Line 64: @@
 We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:
-<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} </math> - <math>\frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math>
+<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} </math> <math>-</math> <math>\frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math>
 ==Video Solutions==

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

Revision as of 12:29, 16 January 2023

Contents

Problem

Solution 1

Solution 2 (MAA Original Solution)

Solution 3

Solution 4

Solution 5 (Modular Arithmetic)

Solution 6(Author: Shiva Kannan)

Video Solutions

Video Solution 1 by Mathematical Dexterity (2 min)

Video Solution 2 by The Beauty Of Math

Video Solution 3

Video Solution 4 Using Sophie Germain's Identity

See Also