2017 AMC 8 Problems/Problem 7: Difference between revisions
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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>. | 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>. | ||
=Solution 2= | |||
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>A</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>A</math> | ||
Revision as of 11:00, 16 January 2018
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 1
Let 
Clearly, is divisible by 
.
Solution 2
We can see that numbers like 
can be written as 
. We can see that the alternating sum of digits is 
, which is 
. Because is a multiple of 
, any number is a multiple of , so the answer is 
-Baolan (hi MVMS)
See Also
| 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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