Difference between revisions of "1999 AIME Problems/Problem 5"
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== Solution == | == Solution == | ||
For most values of <math>x</math>, <math>T(x)</math> will equal <math>2</math>. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take <math>T(a999)</math> as an example, | For most values of <math>x</math>, <math>T(x)</math> will equal <math>2</math>. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take <math>T(a999)</math> as an example, | ||
− | + | <cmath>|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|</cmath> | |
− | And in general, the values of <math>T(x)</math> will then be in the form of <math>|2 - 9n| = 9n - 2</math>. From 7 to 1999, there are <math>\left\lceil \frac{1999 - 7}{9}\right\rceil = 222</math> solutions; including <math>2</math> and there are a total of <math>223</math> solutions. | + | And in general, the values of <math>T(x)</math> will then be in the form of <math>|2 - 9n| = 9n - 2</math>. From <math>7</math> to <math>1999</math>, there are <math>\left\lceil \frac{1999 - 7}{9}\right\rceil = 222</math> solutions; including <math>2</math> and there are a total of <math>223</math> solutions. |
== See also == | == See also == |
Revision as of 13:13, 26 April 2008
Problem
For any positive integer , let
be the sum of the digits of
, and let
be
For example,
How many values of
do not exceed 1999?
Solution
For most values of ,
will equal
. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take
as an example,
And in general, the values of
will then be in the form of
. From
to
, there are
solutions; including
and there are a total of
solutions.
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |