Difference between revisions of "2010 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
− | For each positive integer n, let <math>f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor</math>. Find the largest value of n for which <math>f(n) \le 300</math>. | + | For each positive integer <math>n</math>, let <math>f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor</math>. Find the largest value of <math>n</math> for which <math>f(n) \le 300</math>. |
'''Note:''' <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>. | '''Note:''' <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>. | ||
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Observe that <math>f</math> is strictly increasing in <math>n</math>. We realize that we need <math>100</math> terms to add up to around <math>300</math>, so we need some sequence of <math>2</math>s, <math>3</math>s, and then <math>4</math>s. | Observe that <math>f</math> is strictly increasing in <math>n</math>. We realize that we need <math>100</math> terms to add up to around <math>300</math>, so we need some sequence of <math>2</math>s, <math>3</math>s, and then <math>4</math>s. | ||
− | It follows that <math>n \approx 100</math>. Manually checking shows that <math>f(109) = 300</math> and <math>f(110) > 300</math>. Thus, our answer is <math>\boxed{109}</math>. | + | It follows that <math>n \approx 100</math> (alternatively, use binary search to get to this, with <math>n\le 1000</math>). Manually checking shows that <math>f(109) = 300</math> and <math>f(110) > 300</math>. Thus, our answer is <math>\boxed{109}</math>. |
== Solution 2 == | == Solution 2 == | ||
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Therefore, <cmath>M + \left\lfloor \frac{M}{10} \right\rfloor \ge 100 \implies M \ge \left\lceil\frac{1000}{11}\right\rceil = 91 \implies n \le \left\lfloor\frac{9999}{91}\right\rfloor = 109.</cmath> Since we want the largest possible <math>n</math>, the answer is <math>\boxed{109}</math>. | Therefore, <cmath>M + \left\lfloor \frac{M}{10} \right\rfloor \ge 100 \implies M \ge \left\lceil\frac{1000}{11}\right\rceil = 91 \implies n \le \left\lfloor\frac{9999}{91}\right\rfloor = 109.</cmath> Since we want the largest possible <math>n</math>, the answer is <math>\boxed{109}</math>. | ||
− | == See | + | == Solution 4 (similar to Solution 1, but with more detail) == |
+ | |||
+ | Since we're working with base-<math>10</math> logarithms, we can start by testing out <math>n</math>'s that are powers of <math>10</math>. For <math>n = 1</math>, the terms in the sum are <math>\lfloor \log_{10} (1)\rfloor, \lfloor \log_{10} (2)\rfloor, \lfloor \log_{10} (3) \rfloor , . . . , \lfloor \log_{10} (100) \rfloor</math>. For numbers <math>1</math>-<math>9</math>, <math>\lfloor \log_{10} (kn) \rfloor = 0</math>. Then we have <math>90</math> numbers, namely <math>10</math>-<math>99</math>, for which <math>\lfloor \log_{10} (kn) \rfloor = 1</math>. The last number we have is <math>100</math>, which gives us <math>\lfloor \log_{10} (kn) \rfloor = 2</math>. This sum gives us only <math>90 + 2 = 92</math>, which is much too low. However, applying the same counting technique for <math>n = 10</math>, our sum comes out to be <math>9 + 180 + 3 = 192</math>, since there are <math>9</math> terms for which <math>\lfloor \log_{10} (kn) \rfloor = 1</math>, <math>90</math> terms for which <math>\lfloor \log_{10} (kn) \rfloor = 2</math>, and one term for which <math>\lfloor \log_{10} (kn) \rfloor = 3</math>. So we go up one more power of <math>10</math> and get <math>18 + 270 + 4 = 292</math>, which is very close to what we are looking for. | ||
+ | |||
+ | Now we only have to bump up the value of <math>n</math> a bit and check our sum. Each increase in <math>n</math> by <math>1</math> actually increases the value of our sum by <math>1</math> as well (except for <math>n = 101</math>), because whenever a <math>4</math> is added to the sum, a <math>3</math> is taken away. It doesn't take long to check and see that the value of <math>n</math> we're looking for is <math>\boxed{109}</math> , which corresponds to a sum of exactly <math>300</math>. | ||
+ | |||
+ | ~ anellipticcurveoverq | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/K2xPdIti_AI | ||
+ | |||
+ | == See Also == | ||
{{AIME box|year=2010|num-b=13|num-a=15|n=I}} | {{AIME box|year=2010|num-b=13|num-a=15|n=I}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:09, 25 November 2023
Contents
Problem
For each positive integer , let
. Find the largest value of
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 (alternatively, use binary search to get to this, with
). 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
For any where the sum is close to
, all the terms in the sum must be equal to
,
or
. 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,
Therefore, Since we want the largest possible
, the answer is
.
Solution 4 (similar to Solution 1, but with more detail)
Since we're working with base- logarithms, we can start by testing out
's that are powers of
. For
, the terms in the sum are
. For numbers
-
,
. Then we have
numbers, namely
-
, for which
. The last number we have is
, which gives us
. This sum gives us only
, which is much too low. However, applying the same counting technique for
, our sum comes out to be
, since there are
terms for which
,
terms for which
, and one term for which
. So we go up one more power of
and get
, which is very close to what we are looking for.
Now we only have to bump up the value of a bit and check our sum. Each increase in
by
actually increases the value of our sum by
as well (except for
), because whenever a
is added to the sum, a
is taken away. It doesn't take long to check and see that the value of
we're looking for is
, which corresponds to a sum of exactly
.
~ anellipticcurveoverq
Video Solution
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 |
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