Difference between revisions of "2019 AMC 8 Problems/Problem 14"
Ishanthreddy (talk | contribs) (→Solution 1) |
m (→Solution 1) |
||
| Line 5: | Line 5: | ||
==Solution 1== | ==Solution 1== | ||
| − | Let <math>\text{ | + | Let <math>\text{Day |
}1</math> to <math>\text{Day\\ }2</math> denote a day where one coupon is redeemed and the day when the second coupon is redeemed. | }1</math> to <math>\text{Day\\ }2</math> denote a day where one coupon is redeemed and the day when the second coupon is redeemed. | ||
Revision as of 10:33, 1 October 2021
Contents
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
Which clearly 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
Since Sunday is the only day that has not been counted yet. We can just add the 3 days as it will become .
~~ gorefeebuddie
Note: This only works when 7 and 3 are relatively prime.
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 6
Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=MOQj1zxH2gY&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=15
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. 
