2013 AMC 10B Problem 21

Below is the professionally curated solution for Problem 21 of the 2013 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10B solutions, or check the answer key.

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Concepts:FibonacciDiophantine Equationoptimization

Difficulty rating: 2010

21.

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.N. What is the smallest possible value of NN ?

55 55

89 89

104 104

144 144

273 273

Solution:

A sequence starting with u,vu,v has seventh term 5u+8v5u+8v.

For two sequences (a1,a2)(a_1,a_2) and (b1,b2)(b_1,b_2) with different first terms, assume a1<b1a_1\lt b_1. Then 5a1+8a2=5b1+8b25a_1+8a_2=5b_1+8b_2, so 5(b1a1)=8(a2b2)5(b_1-a_1)=8(a_2-b_2).

Since 55 and 88 are relatively prime, b1a1b_1-a_1 is at least 88, and then nondecreasing order gives b2b1a1+8b_2\ge b_1\ge a_1+8.

The smallest construction is a1=0a_1=0, b1=b2=8b_1=b_2=8, and a2=13a_2=13. This gives N=50+813=104N=5\cdot0+8\cdot13=104.

Thus, the correct answer is C .

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