Difference between revisions of "1994 AIME Problems/Problem 4"
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Through computation: <math>\sum_{j=0}^{7}(j\cdot2^j)=1538<1994</math> and <math>\sum_{j=0}^{8}(j\cdot2^j)=3586>1994</math>. Thus, <math>k=8</math>. | Through computation: <math>\sum_{j=0}^{7}(j\cdot2^j)=1538<1994</math> and <math>\sum_{j=0}^{8}(j\cdot2^j)=3586>1994</math>. Thus, <math>k=8</math>. | ||
− | So, <math>\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}</math>. | + | So, <math>\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}</math>. |
− | + | ||
+ | Alternatively, one could notice this is an [[arithmetico-geometric series]] and avoid a lot of computation. | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=1994|num-b=3|num-a=5}} | {{AIME box|year=1994|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:33, 11 August 2018
Problem
Find the positive integer for which
(For real
,
is the greatest integer
)
Solution
Note that if for some
, then
.
Thus, there are integers
such that
. So the sum of
for all such
is
.
Let be the integer such that
. So for each integer
, there are
integers
such that
, and there are
such integers such that
.
Therefore, .
Through computation: and
. Thus,
.
So, .
Alternatively, one could notice this is an arithmetico-geometric series and avoid a lot of computation.
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.