Difference between revisions of "2017 AMC 8 Problems/Problem 7"
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| + | ==Problem 7== | ||
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Let <math>Z</math> be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of <math>Z</math>? | Let <math>Z</math> be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of <math>Z</math>? | ||
<math>\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111</math> | <math>\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111</math> | ||
| − | <math>\ | + | ==Solution== |
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| + | Let <math>Z = \overline{ABCABC} = 1001 \cdot \overline{ABC} = 7 \cdot 11 \cdot 13 \cdot \overline{ABC}.</math> Clearly, <math>Z</math> is divisible by <math>\boxed{\textbf{(A)}\ 11}</math>. | ||
Revision as of 14:32, 22 November 2017
Problem 7
Let 
be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of ?
Solution
Let 
Clearly, is divisible by 
.
