Difference between revisions of "2010 AIME I Problems/Problem 14"
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== Solution 3 == | == Solution 3 == | ||
− | The sum will contain some 4's, 3's and 2's. Let <math>M</math> be the number of terms less than or equal to <math>3</math> and <math>N</math> be the number of terms equal to <math>2</math> (also counted in <math>M</math>). With this definition of <math>M</math> and <math>N</math> the total will be <math>400 - M - N = 300</math>, from which <math>M + N = 100</math>. Now <math>M+1</math> is the smallest integer <math>k</math> for which <math>log(kn) >= 4</math> or <math>kn >= 10000</math>, thus < | + | The sum will contain some 4's, 3's and 2's. Let <math>M</math> be the number of terms less than or equal to <math>3</math> and <math>N</math> be the number of terms equal to <math>2</math> (also counted in <math>M</math>). With this definition of <math>M</math> and <math>N</math> the total will be <math>400 - M - N = 300</math>, from which <math>M + N = 100</math>. Now <math>M+1</math> is the smallest integer <math>k</math> for which <math>\log(kn) >= 4</math> or <math>kn >= 10000</math>, thus <cmath>M = \left\lceil\frac{10000}{n}\right\rceil - 1</cmath>. Similarly, <cmath>N = \left\lceil\frac{1000}{n}\right\rceil - 1</cmath>. Unless <math>n</math> is a divisor of <math>10000</math>, we have <math></math>M = \left\lfloor\frac{10000}{n}\right\rfloor<math> and <cmath>N = \left\lfloor\frac{1000}{n}\right\rfloor = \left\lfloor \frac{M}{10} \right\rfloor</cmath>. Under the same assumption </math>M + \left\lfloor \frac{M}{10} \right\rfloor = 100<math>, and so </math>M = 91<math>. Dividing </math>10000<math> by </math>91<math> we see that </math>n<math> is </math>109<math>. Since </math>n = 110<math> yields </math>M + N < 100<math>, and the sum is monotone in </math>n<math>, the largest value is </math>109$. |
== See also == | == See also == |
Revision as of 18:11, 21 March 2010
Contents
[hide]Problem
For each positive integer n, let . Find the largest value of n for which
.
Note: is the greatest integer less than or equal to
.
Solution 1
Observe that is strictly increasing in
. We realize that we need
terms to add up to around
, so we need some sequence of
s,
s, and then
s.
It follows that . Manually checking shows that
and
. Thus, our answer is
.
Solution 2
Because we want the value for which , the average value of the 100 terms of the sequence should be around
. For the value of
to be
,
. We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let
, so
, and
.
, so we want to lower
. Testing
yields
, so our answer is still
.
Solution 3
The sum will contain some 4's, 3's and 2's. Let be the number of terms less than or equal to
and
be the number of terms equal to
(also counted in
). With this definition of
and
the total will be
, from which
. Now
is the smallest integer
for which
or
, thus
. Similarly,
. Unless
is a divisor of
, we have $$ (Error compiling LaTeX. Unknown error_msg)M = \left\lfloor\frac{10000}{n}\right\rfloor
M + \left\lfloor \frac{M}{10} \right\rfloor = 100
M = 91
10000
91
n
109
n = 110
M + N < 100
n
109$.
See also
2010 AIME I (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 | ||
All AIME Problems and Solutions |