Difference between revisions of "2019 AMC 8 Problems/Problem 14"

(Solution 4)
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== Solution 4 ==
 
== 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 <math>\boxed{\textbf{(C)}\ Wednesday}</math>. ~~ gorefeebuddie
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Since Sunday is the only day that has not been counted yet. We can just add the 3 days as it will become <math>\boxed{\textbf{(C)}\ Wednesday}</math>.  
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~~ gorefeebuddie
 
Note: This only works when 7 and 3 are relatively prime.
 
Note: This only works when 7 and 3 are relatively prime.
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==See Also==
 
==See Also==
 
{{AMC8 box|year=2019|num-b=13|num-a=15}}
 
{{AMC8 box|year=2019|num-b=13|num-a=15}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:26, 24 November 2019

Problem 14

Isabella has $6$ 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 $10$ days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the $6$ 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?

$\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}$

Solution 1

Let $\text{Day }1$ to $\text{Day }2$ denote a day where one coupon is redeemed and the day when the second coupon is redeemed.

If she starts on a $\text{Monday}$ she redeems her next coupon on $\text{Thursday}$.

$\text{Thursday}$ to $\text{Sunday}$.

Thus $\textbf{(A)}\ \text{Monday}$ is incorrect.


If she starts on a $\text{Tuesday}$ she redeems her next coupon on $\text{Friday}$.

$\text{Friday}$ to $\text{Monday}$.

$\text{Monday}$ to $\text{Thursday}$.

$\text{Thursday}$ to $\text{Sunday}$.

Thus $\textbf{(B)}\ \text{Tuesday}$ is incorrect.


If she starts on a $Wednesday$ she redeems her next coupon on $Saturday$.

$\text{Saturday}$ to $\text{Tuesday}$.

$\text{Tuesday}$ to $\text{Friday}$.

$\text{Friday}$ to $\text{Monday}$.

$\text{Monday}$ to $\text{Thursday}$.

And on $\text{Thursday}$ she redeems her last coupon.


No sunday occured thus $\boxed{\textbf{(C)}\ \text{Wednesday}}$ is correct.


Checking for the other options,


If she starts on a $\text{Thursday}$ she redeems her next coupon on $\text{Sunday}$.

Thus $\textbf{(D)}\ \text{Thursday}$ is incorrect.


If she starts on a $\text{Friday}$ she redeems her next coupon on $\text{Monday}$.

$\text{Monday}$ to $\text{Thursday}$.

$\text{Thursday}$ to $\text{Sunday}$.


Checking for the other options gave us negative results, thus the answer is $\boxed{\textbf{(C)}\ \text{Wednesday}}$.

~phoenixfire

Solution 2

Let

$Sunday \equiv 0 \pmod{7}$

$Monday \equiv 1 \pmod{7}$

$Tuesday \equiv 2 \pmod{7}$

$Wednesday \equiv 3 \pmod{7}$

$Thursday \equiv 4 \pmod{7}$

$Friday \equiv 5 \pmod{7}$

$Saturday \equiv 6 \pmod{7}$


$10 \equiv 3 \pmod{7}$

$20 \equiv 6 \pmod{7}$

$30 \equiv 2 \pmod{7}$

$40 \equiv 5 \pmod{7}$

$50 \equiv 1 \pmod{7}$

$60 \equiv 4 \pmod{7}$


Which clearly indicates if you start form a $x \equiv 3 \pmod{7}$ you will not get a $y \equiv 0 \pmod{7}$.

Any other starting value may lead to a $y \equiv 0 \pmod{7}$.

Which means our answer is $\boxed{\textbf{(C)}\ Wednesday}$.

~phoenixfire

Solution 3

Like Solution 2, let the days of the week be numbers$\pmod 7$. $3$ and $7$ are coprime, so continuously adding $3$ to a number$\pmod 7$ will cycle through all numbers from $0$ to $6$. If a string of 6 numbers in this cycle does not contain $0$, then if you minus 3 from the first number of this cycle, it will always be $0$. So, the answer is $\boxed{\textbf{(C)}\ Wednesday}$. ~~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 $\boxed{\textbf{(C)}\ Wednesday}$. ~~ gorefeebuddie Note: This only works when 7 and 3 are relatively prime.

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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