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

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== Solution 3 ==
 
== Solution 3 ==
Like Solution 2, let the days of the week be numbers <math>\pmod 7</math>. <math>3</math> and <math>7</math> are coprime, so continuously adding <math>3</math> to a number <math>\pmod 7</math> will cycle through all numbers from <math>0</math> to <math>6</math>. If a string of 6 numbers in this cycle does not contain <math>0</math>, then if you minus 3 from the first number of this cycle, it will always be <math>0</math>. So, the answer is <math>\boxed{\textbf{(C)}\ Wednesday}</math>.
+
Like Solution 2, let the days of the week be numbers<math>\pmod 7</math>. <math>3</math> and <math>7</math> are coprime, so continuously adding <math>3</math> to a number<math>\pmod 7</math> will cycle through all numbers from <math>0</math> to <math>6</math>. If a string of 6 numbers in this cycle does not contain <math>0</math>, then if you minus 3 from the first number of this cycle, it will always be <math>0</math>. So, the answer is <math>\boxed{\textbf{(C)}\ Wednesday}</math>. ~~SmileKat32
  
 
==See Also==
 
==See Also==

Revision as of 22:10, 20 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?

$\textbf{(A) }$Monday$\qquad\textbf{(B) }$Tuesday$\qquad\textbf{(C) }$Wednesday$\qquad\textbf{(D) }$Thursday$\qquad\textbf{(E) }$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 $\text{Sunday$ (Error compiling LaTeX. Unknown error_msg)}.

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


If she starts on a $Tuesday$ she redeems her next coupon on $Friday$.

$Friday$ to $Monday$.

$Monday$ to $Thursday$.

$Thursday$ to $Sunday$.

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


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

$Saturday$ to $Tuesday$.

$Tuesday$ to $Friday$.

$Friday$ to $Monday$.

$Monday$ to $Thursday$.

And on $Thursday$ she redeems her last coupon.


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


Checking for the other options,


If she starts on a $Thursday$ she redeems her next coupon on $Sunday$.

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


If she starts on a $Friday$ she redeems her next coupon on $Monday$.

$Monday$ to $Thursday$.

$Thursday$ to $Sunday$.


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

~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

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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