Difference between revisions of "2002 AIME I Problems/Problem 8"
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== Problem == | == Problem == | ||
Find the smallest integer <math>k</math> for which the conditions | Find the smallest integer <math>k</math> for which the conditions | ||
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== Solution == | == Solution == | ||
− | + | From <math>(2)</math>, <math>a_9=</math> <math>a_8+a_7=2a_7+a_6=3a_6+2a_5=5a_5+3a_4=8a_4+5a_3=13a_3+8a_2=21a_2+13a_1</math> <math>=k</math> | |
− | == | + | Suppose that <math>a_1=x_0</math> is the smallest possible value for <math>a_1</math> that yields a good sequence, and <math>a_2=y_0</math> in this sequence. So, <math>13x_0+21y_0=k</math>. |
− | + | ||
+ | Since <math>\gcd(13,21)=1</math>, the next smallest possible value for <math>a_1</math> that yields a good sequence is <math>a_1=x_0+21</math>. Then, <math>a_2=y_0-13</math>. | ||
+ | |||
+ | By <math>(1)</math>, <math>a_2 \ge a_1 \Rightarrow y_0-13 \ge x_0+21 \Rightarrow y_0 \ge x_0+34 \ge 35</math>. So the smallest value of <math>k</math> is attained when <math>(x_0,y_0)=(1,35)</math> which yields <math>(a_1,a_2)=(1,35)</math> or <math>(22,22)</math>. | ||
− | + | Thus, <math>k=13(1)+21(35)=\boxed{748}</math> is the smallest possible value of <math>k</math>. | |
− | + | == See also == | |
+ | {{AIME box|year=2002|n=I|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:54, 4 July 2013
Problem
Find the smallest integer for which the conditions
(1) is a nondecreasing sequence of positive integers
(2) for all
(3)
are satisfied by more than one sequence.
Solution
From ,
Suppose that is the smallest possible value for
that yields a good sequence, and
in this sequence. So,
.
Since , the next smallest possible value for
that yields a good sequence is
. Then,
.
By ,
. So the smallest value of
is attained when
which yields
or
.
Thus, is the smallest possible value of
.
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.