2019 AMC 8 Problems/Problem 14
Contents
- 1 Problem 14
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution Explained
- 8 Video Solution
- 9 Video Solution by Math-X (Extremely simple approach!!!)
- 10 Solution 6
- 11 Video Solution
- 12 Video Solution (CREATIVE THINKING + MOST EFFICIENT!!!)
- 13 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 14 See also
Problem 14
Isabella has coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?
Solution 1
Let to denote a day where one coupon is redeemed and the day when the second coupon is redeemed.
If she starts on a , she redeems her next coupon on .
to .
Thus, is incorrect.
If she starts on a , she redeems her next coupon on .
to .
to .
to .
Thus is incorrect.
If she starts on a , she redeems her next coupon on .
to .
to .
to .
to .
And on , she redeems her last coupon.
No Sunday occured; thus, is correct.
Checking for the other options,
If she starts on a , she redeems her next coupon on .
Thus, is incorrect.
If she starts on a , she redeems her next coupon on .
to .
to .
Checking for the other options gave us negative results; thus, the answer is .
Solution 2
Let
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Which indicates if you start from a you will not get a .
Any other starting value may lead to a .
Which means our answer is .
~phoenixfire
Solution 3
Like Solution 2, let the days of the week be numbers. and are coprime, so continuously adding to a number will cycle through all numbers from to . If a string of 6 numbers in this cycle does not contain , then if you minus 3 from the first number of this cycle, it will always be . So, the answer is .
~~SmileKat32
Solution 4
Start counting on Sunday, to maximize the number of 10-day jumps before reaching Sunday again. Add the 10 () days to reach the first coupon day on .
~~ gorefeebuddie
Note: This only works when 7 and 3 are relatively prime. Otherwise, you’d prefer to start on a day whose distance from Sunday is relatively prime to the jump size.
Solution 5
Let Sunday be Day 0, Monday be Day 1, Tuesday be Day 2, and so forth. We see that Sundays fall on Day , where n is a multiple of seven. If Isabella starts using her coupons on Monday (Day 1), she will fall on a Day that is a multiple of seven, a Sunday (her third coupon will be "used" on Day 21). Similarly, if she starts using her coupons on Tuesday (Day 2), Isabella will fall on a Day that is a multiple of seven (Day 42). Repeating this process, if she starts on Wednesday (Day 3), Isabella will first fall on a Day that is a multiple of seven, Day 63 (13, 23, 33, 43, 53 are not multiples of seven), but on her seventh coupon, of which she only has six. So, the answer is .
Solution Explained
https://youtu.be/gOZOCFNXMhE ~ The Learning Royal
Video Solution
Video Solution by Math-X (Extremely simple approach!!!)
https://youtu.be/IgpayYB48C4?si=jnBHI2Gbvdu-cCyR&t=4331
~Math-X
Solution 6
Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=MOQj1zxH2gY&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=15
Video Solution
~savannahsolver
Video Solution (CREATIVE THINKING + MOST EFFICIENT!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
See also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.