Difference between revisions of "2017 AMC 8 Problems/Problem 24"
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− | We know that the next year is 2017, and seeing that it's not a leap year, we know that there is a total of <math>365</math> days. If x represents that last time Mrs. Sanders was called by each of her grandchildren, we can get these 3 expressions: <math>x+3, x+4, x+5</math>(in which x represents a different amount each time, as each of her grandchildren call at different times.) Next, we know that if she was last called by all three of them on December 31, 2016, then it will be 2 days that she's not called, and then 3 consecutive days of which she is called. So, we can get the pattern of: 2 days no calls, 3 days with calls. Altogether, you get a period of 5 | + | We know that the next year is 2017, and seeing that it's not a leap year, we know that there is a total of <math>365</math> days. If x represents that last time Mrs. Sanders was called by each of her grandchildren, we can get these 3 expressions: <math>x+3, x+4, x+5</math>(in which x represents a different amount each time, as each of her grandchildren call at different times.) Next, we know that if she was last called by all three of them on December 31, 2016, then it will be 2 days that she's not called, and then 3 consecutive days of which she is called. So, we can get the pattern of: 2 days no calls, 3 days with calls. Altogether, you get a period of 5 days, and then you repeat the pattern. Now we divide <math>{\frac{365}{5}}=73</math>, which means that there will be 73 sets of this pattern in 365 days. Now, since we know that it's every 2 days of nobody calling, and 73 sets of these, so we simply do, <math>{73 \cdot 2}</math>,which equals <math>\boxed{\textbf{(D) } 146}</math>. |
-fn106068 | -fn106068 | ||
Revision as of 20:22, 31 March 2021
Problem
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
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
days without calls. Note that in the last five days of the year, days
and
also do not have any calls, as they are not multiples of
,
, or
. Thus our answer is
.
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
.
Solution 4
We know that the next year is 2017, and seeing that it's not a leap year, we know that there is a total of days. If x represents that last time Mrs. Sanders was called by each of her grandchildren, we can get these 3 expressions:
(in which x represents a different amount each time, as each of her grandchildren call at different times.) Next, we know that if she was last called by all three of them on December 31, 2016, then it will be 2 days that she's not called, and then 3 consecutive days of which she is called. So, we can get the pattern of: 2 days no calls, 3 days with calls. Altogether, you get a period of 5 days, and then you repeat the pattern. Now we divide
, which means that there will be 73 sets of this pattern in 365 days. Now, since we know that it's every 2 days of nobody calling, and 73 sets of these, so we simply do,
,which equals
.
-fn106068
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 |
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