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Difference between revisions of "2017 AMC 10A Problems/Problem 1"

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<math>\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729</math>
 
<math>\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729</math>
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== Solution 1 ==
 
== Solution 1 ==
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Notice this is the term <math>a_6</math> in a recursive sequence, defined recursively as <math>a_1 = 3, a_n = 2a_{n-1} + 1.</math> Thus:
 
Notice this is the term <math>a_6</math> in a recursive sequence, defined recursively as <math>a_1 = 3, a_n = 2a_{n-1} + 1.</math> Thus:
 
<cmath>\begin{split}
 
<cmath>\begin{split}
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a_2 = 3*2 + 1 = 7.\\
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a_2 = 3 \cdot 2 + 1 = 7.\\
−
a_3 = 7 *2 + 1 = 15.\\
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a_3 = 7 \cdot 2 + 1 = 15.\\
−
a_4 = 15*2 + 1 = 31.\\
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a_4 = 15 \cdot 2 + 1 = 31.\\
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a_5 = 31*2 + 1 = 63.\\
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a_5 = 31 \cdot 2 + 1 = 63.\\
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a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127}
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a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127}
−
\end{split}</cmath>.
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\end{split}</cmath>
 +
 
 +
Minor LaTeX edits by fasterthanlight
  
 
== Solution 2 ==
 
== Solution 2 ==
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If you distribute this you get a sum of the powers of <math>2</math>.  The largest power of <math>2</math> in the series is <math>64</math>, so the sum is <math>\boxed{\textbf{(C)}\ 127}</math>.
 
If you distribute this you get a sum of the powers of <math>2</math>.  The largest power of <math>2</math> in the series is <math>64</math>, so the sum is <math>\boxed{\textbf{(C)}\ 127}</math>.
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 +
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==Solution 5==
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<math>(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)</math>
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<math>=(2(2(2(2(2(3)+1)+1)+1)+1)+1)</math>
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<math>=(2(2(2(2(6+1)+1)+1)+1)+1)</math>
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<math>=(2(2(2(2(7)+1)+1)+1)+1)</math>
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<math>=(2(2(2(14+1)+1)+1)+1)</math>
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<math>=(2(2(2(15)+1)+1)+1)</math>
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<math>=(2(2(30+1)+1)+1)</math>
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<math>=(2(2(31)+1)+1)</math>
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<math>=(2(62+1)+1)</math>
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<math>=(2(63)+1)</math>
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<math>=(126+1)</math>
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<math>=127 \Longrightarrow \boxed{\textbf{(C)}\ 127}</math>.
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 +
==Solution 6 (quickest)==
 +
Notice that <math>x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = x (x (x (x (x (x + 1) + 1) + 1) + 1) + 1) + 1</math>. Substituting <math>2</math> for <math>x</math>, we get <cmath>2(2(2(2(2(2+1)+1)+1)+1)+1)+1 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2 + 1 = 2^7 - 1 \Longrightarrow \boxed{\textbf{(C)}\ 127}</cmath>
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==Solution 7 (Fast)==
 +
Notice that there are 5 instances where the sum is multiplied by <math>2</math>. That gives us <math>2^5 = 32</math>. Separate the addition part into <math>3</math> and the <math>1</math>. Multiply the <math>3</math> by <math>32</math>, giving <math>96</math>. Then notice that the <math>1</math> is multiplied by increasing powers of two; therefore, it is equal to <math>2^5-1</math>. Then add these two parts. <math>96+31=\boxed{\textbf{(C)}\ 127}</math>
  
 
==Video Solution==
 
==Video Solution==
−
https://youtu.be/rxeqBnHDQvI
+
https://youtu.be/str7kmcRMY8
 +
 
 +
https://youtu.be/kA6W8SwjitA
 +
 
 +
~savannahsolver
  
 
== See Also ==
 
== See Also ==

Latest revision as of 09:22, 1 November 2024

Problem

What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$?

$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$

Solution 1

Notice this is the term $a_6$ in a recursive sequence, defined recursively as $a_1 = 3, a_n = 2a_{n-1} + 1.$ Thus: \[\begin{split} a_2 = 3 \cdot 2 + 1 = 7.\\ a_3 = 7 \cdot 2 + 1 = 15.\\ a_4 = 15 \cdot 2 + 1 = 31.\\ a_5 = 31 \cdot 2 + 1 = 63.\\ a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127} \end{split}\]

Minor LaTeX edits by fasterthanlight

Solution 2

Starting to compute the inner expressions, we see the results are $1, 3, 7, 15, \ldots$. This is always $1$ less than a power of $2$. The only admissible answer choice by this rule is thus $\boxed{\textbf{(C)}\ 127}$.

Solution 3

Working our way from the innermost parenthesis outwards and directly computing, we have $\boxed{\textbf{(C) } 127}$.

Solution 4

If you distribute this you get a sum of the powers of $2$. The largest power of $2$ in the series is $64$, so the sum is $\boxed{\textbf{(C)}\ 127}$.


Solution 5

$(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$ $=(2(2(2(2(2(3)+1)+1)+1)+1)+1)$ $=(2(2(2(2(6+1)+1)+1)+1)+1)$ $=(2(2(2(2(7)+1)+1)+1)+1)$ $=(2(2(2(14+1)+1)+1)+1)$ $=(2(2(2(15)+1)+1)+1)$ $=(2(2(30+1)+1)+1)$ $=(2(2(31)+1)+1)$ $=(2(62+1)+1)$ $=(2(63)+1)$ $=(126+1)$ $=127 \Longrightarrow \boxed{\textbf{(C)}\ 127}$.

Solution 6 (quickest)

Notice that $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = x (x (x (x (x (x + 1) + 1) + 1) + 1) + 1) + 1$. Substituting $2$ for $x$, we get \[2(2(2(2(2(2+1)+1)+1)+1)+1)+1 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2 + 1 = 2^7 - 1 \Longrightarrow \boxed{\textbf{(C)}\ 127}\]

Solution 7 (Fast)

Notice that there are 5 instances where the sum is multiplied by $2$. That gives us $2^5 = 32$. Separate the addition part into $3$ and the $1$. Multiply the $3$ by $32$, giving $96$. Then notice that the $1$ is multiplied by increasing powers of two; therefore, it is equal to $2^5-1$. Then add these two parts. $96+31=\boxed{\textbf{(C)}\ 127}$

Video Solution

https://youtu.be/str7kmcRMY8

https://youtu.be/kA6W8SwjitA

~savannahsolver

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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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