Difference between revisions of "2007 AIME I Problems/Problem 7"
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The change of base formula shows that <math>\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}</math>. For the <math>\log 2</math> term to cancel out, <math>k</math> is a [[exponent|power]] of <math>2</math>. Thus, <math>N</math> is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from <math>2^0 = 1</math> to <math>2^9 = 512</math>. | The change of base formula shows that <math>\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}</math>. For the <math>\log 2</math> term to cancel out, <math>k</math> is a [[exponent|power]] of <math>2</math>. Thus, <math>N</math> is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from <math>2^0 = 1</math> to <math>2^9 = 512</math>. | ||
− | The formula for the sum of an [[arithmetic sequence]] and the sum of a [[geometric sequence]] yields that our answer is <math>[\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9)] \mod{1000}</math>. | + | The formula for the sum of an [[arithmetic sequence]] and the sum of a [[geometric sequence]] yields that our answer is <math>\left[\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9)\right] \mod{1000}</math>. |
Simplifying, we get | Simplifying, we get | ||
− | <math>[1000(\frac{1000+1}{2}) -1023] \mod{1000} \equiv [500-23] \mod{1000} \equiv 477 \mod{1000}.</math> The answer is <math>\boxed{477}</math> | + | <math>\left[1000\left(\frac{1000+1}{2}\right) -1023\right] \mod{1000} \equiv [500-23] \mod{1000} \equiv 477 \mod{1000}.</math> The answer is <math>\boxed{477}</math> |
== See also == | == See also == |
Latest revision as of 20:26, 20 April 2023
Problem
Let
Find the remainder when is divided by 1000. (
is the greatest integer less than or equal to
, and
is the least integer greater than or equal to
.)
Solution
The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer); otherwise, it is equal to 1. Thus, we need to find when or not is an integer.
The change of base formula shows that . For the
term to cancel out,
is a power of
. Thus,
is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from
to
.
The formula for the sum of an arithmetic sequence and the sum of a geometric sequence yields that our answer is .
Simplifying, we get
The answer is
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.