# Difference between revisions of "2003 AMC 10A Problems/Problem 11"

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== Solution == | == Solution == | ||

− | <math> | + | <math>AMC10+AMC12=123422</math> |

− | <math> | + | <math>AMC00+AMC00=123400</math> |

− | <math> | + | <math>AMC+AMC=1234</math> |

− | <math> | + | <math>2\cdot AMC=1234</math> |

+ | <math>AMC=\frac{1234}{2}=617</math> | ||

+ | |||

+ | Since <math>A</math>, <math>M</math>, and <math>C</math> are digits, <math>A=6</math>, <math>M=1</math>, <math>C=7</math>. | ||

+ | |||

+ | Therefore, <math>A+M+C = 6+1+7 = 14 \Rightarrow E </math>. | ||

== See Also == | == See Also == | ||

− | * [[2003 AMC 10A Problems]] | + | *[[2003 AMC 10A Problems]] |

− | * [[2003 AMC 10A Problems/Problem | + | *[[2003 AMC 10A Problems/Problem 10|Previous Problem]] |

− | * [[2003 AMC 10A Problems/Problem | + | *[[2003 AMC 10A Problems/Problem 12|Next Problem]] |

+ | |||

+ | [[Category:Introductory Algebra Problems]] |

## Revision as of 19:13, 4 November 2006

## Problem

The sum of the two 5-digit numbers and is . What is ?

## Solution

Since , , and are digits, , , .

Therefore, .