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2013 AMC 12B Problems/Problem 14

Revision as of 12:23, 7 August 2019 by Nafer (talk | contribs) (→‎Solution 2)
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

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 $N=5b_1+8b_2=104\boxed{\mathrm{(B)}}$

~ Nafer

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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