Difference between revisions of "2013 AMC 12B Problems/Problem 14"
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+ | {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #14]] and [[2013 AMC 10B Problems|2013 AMC 10B #21]]}} | ||
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==Problem== | ==Problem== | ||
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<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> | ||
− | ==Solution== | + | ==Solution 1== |
− | 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> | + | 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>. | ||
+ | |||
+ | Thus the minimum <math>N=5b_1+8b_2=104\boxed{\mathrm{(C)}}</math> | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/fgQN3e_aa4Y | ||
+ | |||
+ | ~IceMatrix | ||
== 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}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:16, 23 January 2021
- 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 . What is the smallest possible value of ?
Solution 1
Let the first two terms of the first sequence be and and the first two of the second sequence be and . Computing the seventh term, we see that . Note that this means that and must have the same value modulo . To minimize, let one of them be ; WLOG assume that . Thus, the smallest possible value of is ; and since the sequences are non-decreasing we get . To minimize, let . Thus, .
Solution 2
WLOG, let , be the sequences with . Then or for some natural numbers , . Thus . To minimize and , we have , or To minimize and , we have . Using the same method, since , we have .
Thus the minimum
~ Nafer
Video Solution
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.