2017 AMC 8 Problems/Problem 24
Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
Solution 1 (Principle of Inclusion-Exclusion)
We use Principle of Inclusion-Exclusion. There are days in the year, and we subtract the days that she gets at least phone call, which is
To this result we add the number of days where she gets at least phone calls in a day because we double subtracted these days, which is
We now subtract the number of days where she gets three phone calls, which is Therefore, our answer is
Solution 2 (Least Common Multiple)
Note that so there is a cycle every days.
As shown below, all days in a cycle that Mrs. Sanders receives a phone call from any of her grandchildren are colored in red, yellow, or green. The year 2017 has days, or cycles and days.
- For each cycle, there are days that Mrs. Sanders does not receive a phone call, as indicated by the white squares.
- For the last days, there are days that Mrs. Sanders does not receive a phone call, as indicated by the first days in a cycle.
Together, the answer is
Solution 3 (Linearity of Expectation)
For any randomly chosen day, there is a chance the first child does not call her, a chance the second child does not call her and a chance the third child does not call her. So, in a randomly chosen day, there is a chance no child calls her.
In particular, this "unrigorous" reasoning becomes rigorous, by linearity of expectation, when we take a sample of days; over the first days, days will go without a phone call. Now, we simply check the remaining days; on day and , nobody calls her; on day , the child that calls every three days will call her; on day , the child that calls every four days will call her; on day , the child that calls every five days will call her. Thus, we add two more days to , to get our answer of .
|2017 AMC 8 (Problems • Answer Key • Resources)|
|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.