Difference between revisions of "2017 AMC 8 Problems/Problem 7"
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-Baolan (hi MVMS) | -Baolan (hi MVMS) | ||
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+ | ==Solution 3== | ||
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+ | The most important step is to realize that any number in the form <math>\overline{ABCABC} = \overline{ABC000}+\overline{ABC} = 1000\overline{ABC}+\overline{ABC} = 1001\overline{ABC}</math>. Thus every number in this form is divisible by <math>1001</math>, and the answer is <math>\textbf{(A) }11</math>, because it is the only choice that is a factor of <math>1001</math>. | ||
==See Also== | ==See Also== |
Revision as of 10:25, 30 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)
Solution 3
The most important step is to realize that any number in the form . Thus every number in this form is divisible by
, and the answer is
, because it is the only choice that is a factor of
.
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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