Difference between revisions of "2017 AMC 8 Problems/Problem 24"
Vaathsalya (talk | contribs) (→Solution 2) |
|||
Line 10: | Line 10: | ||
==Solution 1== | ==Solution 1== | ||
− | + | For this problem, we use the Principle of Inclusion and Exclusion(PIE). Notice every <math>3</math> day a day would be a multiple of <math>3</math>, every <math>4</math> day a day is a multiple of 4, and every 5 days it's a multiple of <math> 144</math><math> days without calls. Note that in the last five days of the year, days </math>361<math> and </math>362<math> also do not have any calls, as they are not multiples of </math>3<math>, </math>4<math>, or </math>5<math>. Thus our answer is </math>144+2 = \boxed{\textbf{(D)}\ 146}$. | |
<cmath></cmath> | <cmath></cmath> |
Revision as of 20:24, 13 December 2020
Contents
Problem 24
Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
Video Solution
https://youtu.be/Zhsb5lv6jCI?t=2797
Solution 1
For this problem, we use the Principle of Inclusion and Exclusion(PIE). Notice every day a day would be a multiple of
, every
day a day is a multiple of 4, and every 5 days it's a multiple of
361
362
3
4
5
144+2 = \boxed{\textbf{(D)}\ 146}$.
Solution 2
Since there are 365 days in the year, days without calls.
~Gr8
Solution 3
We use Principle of Inclusion and 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. This number is
We now subtract the number of days where she gets three phone calls, which is . Therefore, our answer is
.
Video Solution
https://youtu.be/a3rGDEmrxC0 - Happytwin
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.