Difference between revisions of "2013 AMC 12B Problems/Problem 14"
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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>. | + | 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> | Thus the minimum <math>N=5b_1+8b_2=104\boxed{\mathrm{(C)}}</math> |
Revision as of 16:42, 1 June 2020
- The following problem is from both the 2013 AMC 12B #14 and 2013 AMC 10B #21, so both problems redirect to this page.
Contents
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
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 8. To minimize, let one of them be 0; 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
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.