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

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

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem ==
 What is the value of <math>(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)</math>?
@@ Line 5: / Line 5: @@
 <math>\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729</math>
+== Solution 1 ==
-==Solution==
+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 an arithmetic sequence, such that:
 <cmath>\begin{split}
-a_1 = 3.\\
+a_2 = 3 \cdot 2 + 1 = 7.\\
-a_n = 2a_(n-1) + 1.\\
+a_3 = 7 \cdot 2 + 1 = 15.\\
-It follows that:\\
+a_4 = 15 \cdot 2 + 1 = 31.\\
-a_2 = 3*2 + 1 = 7.\\
+a_5 = 31 \cdot 2 + 1 = 63.\\
-a_3 = 7 *2 + 1 = 15.\\
+a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127}
-a_4 = 15*2 + 1 = 31.\\
-a_5 = 31*2 + 1 = 63.\\
-a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127.}
 \end{split}</cmath>
+Minor LaTeX edits by fasterthanlight
+== Solution 2 ==
+Starting to compute the inner expressions, we see the results are <math>1, 3, 7, 15, \ldots</math>. This is always <math>1</math> less than a power of <math>2</math>. The only admissible answer choice by this rule is thus <math>\boxed{\textbf{(C)}\ 127}</math>.
+== Solution 3 ==
+Working our way from the innermost parenthesis outwards and directly computing, we have <math>\boxed{\textbf{(C) } 127}</math>.
+== Solution 4 ==
+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>.
+==Solution 5==
+<math>(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)</math>
+<math>=(2(2(2(2(2(3)+1)+1)+1)+1)+1)</math>
+<math>=(2(2(2(2(6+1)+1)+1)+1)+1)</math>
+<math>=(2(2(2(2(7)+1)+1)+1)+1)</math>
+<math>=(2(2(2(14+1)+1)+1)+1)</math>
+<math>=(2(2(2(15)+1)+1)+1)</math>
+<math>=(2(2(30+1)+1)+1)</math>
+<math>=(2(2(31)+1)+1)</math>
+<math>=(2(62+1)+1)</math>
+<math>=(2(63)+1)</math>
+<math>=(126+1)</math>
+<math>=127 \Longrightarrow \boxed{\textbf{(C)}\ 127}</math>.
+==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>
+==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==
+https://youtu.be/str7kmcRMY8
+https://youtu.be/kA6W8SwjitA
+~savannahsolver
+== See Also ==
+{{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}}
+{{MAA Notice}}

2017 AMC 10A (Problems • Answer Key • Resources)
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