2017 AMC 8 Problems/Problem 7: Difference between revisions
Nukelauncher (talk | contribs) No edit summary |
Nukelauncher (talk | contribs) |
||
| Line 8: | Line 8: | ||
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>. | ||
~nukelauncher | |||
==See Also== | ==See Also== | ||
Revision as of 14:50, 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 
.
~nukelauncher
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
