2011 AMC 12A Problems/Problem 19
Problem
At a competition with players, the number of players given elite status is equal to
. Suppose that
players are given elite status. What is the sum of the two smallest possible values of
?
Solution 1
We start with . After rearranging, we get
.
Since is a positive integer,
must be in the form of
for some positive integer
. From this fact, we get
.
If we now check integer values of N that satisfy this condition, starting from , we quickly see that the first values that work for
are
and
, that is,
and
, giving values of
and
for
, respectively. Adding up these two values for
, we get
Solution 2
We examine the value that takes over various intervals. The
means it changes on each multiple of 2, like so:
2 --> 1
3 - 4 --> 2
5 - 8 --> 3
9 - 16 --> 4
From this, we see that is the difference between the next power of 2 above
and
. We are looking for
such that this difference is 19. The first two
that satisfy this are
and
for a final answer of
Solution 3 (using the answer choices)
Note that each is
less than a power of
. So, the answer will be
less than the sum of
powers of
. Adding
to each answer, we get
,
,
,
, and
. Obviously we can take out
and
. Also,
will not work because two powers of two will never sum to another power of
(unless they are equal, which is a contradiction to the question). So, we have
and
. Note that
, etc. We quickly see that
will not work, leaving
which corresponds to
. We can also confirm that this works because
.
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |
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