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

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If she starts on a <math>Tuesday</math> she redeems her next coupon on <math>Friday</math>.
 
If she starts on a <math>Tuesday</math> she redeems her next coupon on <math>Friday</math>.
  
−
<math>Friday</math> to <math>Monday</math>.
+
<math>\text{Friday}</math> to <math>\text{Monday}</math>.
  
−
<math>Monday</math> to <math>Thursday</math>.
+
<math>\text{Monday}</math> to <math>\text{Thursday}</math>.
  
−
<math>Thursday</math> to <math>Sunday</math>.
+
<math>\text{Thursday}</math> to <math>\text{Sunday}</math>.
  
 
Thus <math>\boxed{\textbf{(B)}\ \text{Tuesday}}</math> is incorrect.
 
Thus <math>\boxed{\textbf{(B)}\ \text{Tuesday}}</math> is incorrect.
Line 27: Line 27:
 
If she starts on a <math>Wednesday</math> she redeems her next coupon on <math>Saturday</math>.
 
If she starts on a <math>Wednesday</math> she redeems her next coupon on <math>Saturday</math>.
  
−
<math>Saturday</math> to <math>Tuesday</math>.
+
<math>\text{Saturday}</math> to <math>\text{Tuesday}</math>.
  
−
<math>Tuesday</math> to <math>Friday</math>.
+
<math>\text{Tuesday}</math> to <math>text{Friday}</math>.
  
−
<math>Friday</math> to <math>Monday</math>.
+
<math>\text{Friday}</math> to <math>\text{Monday}</math>.
  
−
<math>Monday</math> to <math>Thursday</math>.
+
<math>\text{Monday}</math> to <math>\text{Thursday}</math>.
  
−
And on <math>Thursday</math> she redeems her last coupon.  
+
And on <math>\text{Thursday}</math> she redeems her last coupon.  
  
  
Line 49: Line 49:
  
  
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If she starts on a <math>Friday</math> she redeems her next coupon on <math>Monday</math>.
+
If she starts on a <math>\text{Friday}</math> she redeems her next coupon on <math>\text{Monday}</math>.
  
−
<math>Monday</math> to <math>Thursday</math>.
+
<math>\text{Monday}</math> to <math>\text{Thursday}</math>.
  
−
<math>Thursday</math> to <math>Sunday</math>.
+
<math>\text{Thursday}</math> to <math>\text{Sunday}</math>.
  
  

Revision as of 06:09, 21 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 $\text{Sunday}$.

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


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

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

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

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

Thus $\boxed{\textbf{(B)}\ \text{Tuesday}}$ 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 $Thursday$ she redeems her next coupon on $Sunday$.

Thus $\boxed{\textbf{(D)}\ \text{Thursday}}$ 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

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

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