2017 AMC 8 Problems/Problem 7: Difference between revisions

Revision as of 16:29, 9 November 2020

Let $Z$ 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 $Z$ ?

$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Video Solution

https://youtu.be/7an5wU9Q5hk?t=647

Solution 1

Let $Z = \overline{ABCABC} = 1001 \cdot \overline{ABC} = 7 \cdot 11 \cdot 13 \cdot \overline{ABC}.$ Clearly, $Z$ is divisible by $\boxed{\textbf{(A)}\ 11}$ .

Solution 2

we are given one of the numbers Z can be so we can just try out the options to see which one is divisible by 247247 ans so we get $\boxed{\textbf{(A)}\ 11}$ .

2017 AMC 8 (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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 ==Solution 2==
+we are given one of the numbers Z can be so we can just try out the options to see which one is divisible by 247247 ans so we get <math>\boxed{\textbf{(A)}\ 11}</math>.
-We can see that numbers like <math>247247</math> can be written as <math>ABCABC</math>. We can see that the alternating sum of digits is <math>C-B+A-C+B-A</math>, which is <math>0</math>. Because <math>0</math> is a multiple of <math>11</math>, any number <math>ABCABC</math> is a multiple of <math>11</math>, so the answer is <math>\boxed{\textbf{(A)}\ 11}</math>.
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2017 AMC 8 Problems/Problem 7: Difference between revisions

Revision as of 16:29, 9 November 2020

Contents

Video Solution

Solution 1

Solution 2

See Also