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Difference between revisions of "1999 AIME Problems/Problem 5"

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== Problem ==
 
== Problem ==
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For any positive integer <math>\displaystyle x_{}</math>, let <math>\displaystyle S(x)</math> be the sum of the digits of <math>\displaystyle x_{}</math>, and let <math>\displaystyle T(x)</math> be <math>\displaystyle |S(x+2)-S(x)|.</math>  For example, <math>\displaystyle T(199)=|S(201)-S(199)|=|3-19|=16.</math>  How many values <math>\displaystyle T(x)</math> do not exceed 1999?
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For any positive integer <math>x_{}</math>, let <math>S(x)</math> be the sum of the digits of <math>x_{}</math>, and let <math>T(x)</math> be <math>|S(x+2)-S(x)|.</math>  For example, <math>T(199)=|S(201)-S(199)|=|3-19|=16.</math>  How many values of <math>T(x)</math> do not exceed 1999?
  
 
== Solution ==
 
== Solution ==
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For most values of <math>x</math>, <math>T(x)</math> will equal |2|. 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,
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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,
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:<math>|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|</math>
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<cmath>|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|</cmath>
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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>\lceil \frac{1999 - 7}{9}\rceil = 222</math> solutions; including <math>2</math> and there are a total of <math>223</math> solutions.
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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>\boxed{223}</math> solutions.
  
 
== See also ==
 
== See also ==
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* [[1999 AIME Problems]]
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{{AIME box|year=1999|num-b=4|num-a=6}}
  
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{{AIME box|year=1999|num-b=4|num-a=6}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 18:40, 4 July 2013

Problem

For any positive integer $x_{}$, let $S(x)$ be the sum of the digits of $x_{}$, and let $T(x)$ be $|S(x+2)-S(x)|.$ For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values of $T(x)$ do not exceed 1999?

Solution

For most values of $x$, $T(x)$ will equal $2$. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take $T(a999)$ as an example, \[|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|\] And in general, the values of $T(x)$ will then be in the form of $|2 - 9n| = 9n - 2$. From $7$ to $1999$, there are $\left\lceil \frac{1999 - 7}{9}\right\rceil = 222$ solutions; including $2$ and there are a total of $\boxed{223}$ 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

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