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

Latest revision as of 10:22, 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 1==
 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}
-a_2 = 3*2 + 1 = 7.\\
+a_2 = 3 \cdot 2 + 1 = 7.\\
-a_3 = 7 *2 + 1 = 15.\\
+a_3 = 7 \cdot 2 + 1 = 15.\\
-a_4 = 15*2 + 1 = 31.\\
+a_4 = 15 \cdot 2 + 1 = 31.\\
-a_5 = 31*2 + 1 = 63.\\
+a_5 = 31 \cdot 2 + 1 = 63.\\
-a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127}
+a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127}
 \end{split}</cmath>
-==Solution 2==
+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==
+== Solution 3 ==
 Working our way from the innermost parenthesis outwards and directly computing, we have <math>\boxed{\textbf{(C) } 127}</math>.
-==Solution 4==
+== 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: Guessing Strategies for Lazy People==
-"I'm too lazy, so I'll try to guess this," says the lazy person. Well then this is the solution for you! (note that in general however, these guessing strategies are just meant to be confidence boosters for your numerical thinking). If you look up a probability table for each of the answer choices on the AMC, you will get that <math>\boxed{\text{C}}</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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