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Difference between revisions of "2013 AMC 12B Problems/Problem 14"

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==Problem==
 
==Problem==
  
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Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is <math>N</math>. What is the smallest possible value of <math>N</math> ?
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Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is <math>N</math>. What is the smallest possible value of <math>N</math>?
  
 
<math>\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273</math>
 
<math>\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273</math>
  
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==Solution==
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==Solution 1==
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Let the first two terms of the first sequence be <math>x_{1}</math> and <math>x_{2}</math> and the first two of the second sequence be <math>y_{1}</math> and <math>y_{2}</math>.  Computing the seventh term, we see that <math>5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}</math>.  Note that this means that <math>x_{1}</math> and <math>x_{2}</math> must have the same value modulo 8.  To minimize, let one of them be 0; WLOG assume that <math>x_{1} = 0</math>.  Thus, the smallest possible value of <math>y_{1}</math> is <math>8</math>; since the sequences are non-decreasing <math>y_{2} \ge 8</math>.  To minimize, let <math>y_{2} = 8</math>.  Thus, <math>5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}</math>.
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Let the first two terms of the first sequence be <math>x_{1}</math> and <math>x_{2}</math> and the first two of the second sequence be <math>y_{1}</math> and <math>y_{2}</math>.  Computing the seventh term, we see that <math>5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}</math>.  Note that this means that <math>x_{1}</math> and <math>y_{1}</math> must have the same value modulo <math>8</math>.  To minimize, let one of them be <math>0</math>; [[WLOG]], assume that <math>x_{1} = 0</math>.  Thus, the smallest possible value of <math>y_{1}</math> is <math>8</math>; and since the sequences are non-decreasing we get <math>y_{2} \ge 8</math>.  To minimize, let <math>y_{2} = 8</math>.  Thus, <math>5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}</math>.
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== Solution 2 ==
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WLOG, let <math>a_i</math>, <math>b_i</math> be the sequences with <math>a_1<b_1</math>. Then <cmath>N=5a_1+8a_2=5b_1+8b_2</cmath> or <cmath>5a_1+8a_2=5(a_1+c)+8(a_2-d)</cmath> for some natural numbers <math>c</math>, <math>d</math>. Thus <math>5c=8d</math>. To minimize <math>c</math> and <math>d</math>, we have <math>(c,d)=(8,5)</math>, or <cmath>5a_1+8a_2=5(a_1+8)+8(a_2-5).</cmath> To minimize <math>a_1</math> and <math>b_1</math>, we have <math>(a_1,b_1)=(0,0+c)=(0,8)</math>. Using the same method, since <math>b_2\ge b_1</math>, we have <math>b_2\ge8</math>.
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Thus the minimum <math>N=5b_1+8b_2=104\boxed{\mathrm{(C)}}</math>
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~ Nafer
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==Video Solution==
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https://youtu.be/fgQN3e_aa4Y
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~IceMatrix
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==Video Solution 2==
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https://youtu.be/JF3In44sXy0
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~savannahsolver
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2013|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2013|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2013|ab=B|num-b=20|num-a=22}}
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[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:26, 7 August 2023

The following problem is from both the 2013 AMC 12B #14 and 2013 AMC 10B #21, so both problems redirect to this page.

Problem

Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$. What is the smallest possible value of $N$?

$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273$

Solution 1

Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$. Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$. Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$. To minimize, let one of them be $0$; WLOG, assume that $x_{1} = 0$. Thus, the smallest possible value of $y_{1}$ is $8$; and since the sequences are non-decreasing we get $y_{2} \ge 8$. To minimize, let $y_{2} = 8$. Thus, $5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}$.

Solution 2

WLOG, let $a_i$, $b_i$ be the sequences with $a_1<b_1$. Then \[N=5a_1+8a_2=5b_1+8b_2\] or \[5a_1+8a_2=5(a_1+c)+8(a_2-d)\] for some natural numbers $c$, $d$. Thus $5c=8d$. To minimize $c$ and $d$, we have $(c,d)=(8,5)$, or \[5a_1+8a_2=5(a_1+8)+8(a_2-5).\] To minimize $a_1$ and $b_1$, we have $(a_1,b_1)=(0,0+c)=(0,8)$. Using the same method, since $b_2\ge b_1$, we have $b_2\ge8$.

Thus the minimum $N=5b_1+8b_2=104\boxed{\mathrm{(C)}}$

~ Nafer

Video Solution

https://youtu.be/fgQN3e_aa4Y

~IceMatrix

Video Solution 2

https://youtu.be/JF3In44sXy0

~savannahsolver

See also

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 12 Problems and Solutions
2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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