Difference between revisions of "2017 AMC 8 Problems/Problem 7"
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==Solution 3== | ==Solution 3== | ||
− | To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, <math>+2-4+7-2+4-7=0</math>. Because the result is 0, the number 247247 is divisible by 11 and so we get <math>\boxed{\textbf{(A)}\ 11}</math>. | + | To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, <math>+2-4+7-2+4-7=0</math>. Because the result is 0, the number 247247 is divisible by 11 and so we get <math>\boxed{\textbf{(A)}\ 11}</math>. ---LarryFlora |
==Video Solution== | ==Video Solution== |
Latest revision as of 13:55, 14 June 2021
Problem
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 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 and so we get .
Solution 3
To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, . Because the result is 0, the number 247247 is divisible by 11 and so we get . ---LarryFlora
Video Solution
https://youtu.be/7an5wU9Q5hk?t=647
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.