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

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==Problem 14==
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==Problem==
 
Isabella has <math>6</math> 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 <math>10</math> days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the <math>6</math> 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?
 
Isabella has <math>6</math> 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 <math>10</math> days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the <math>6</math> 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?
  
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Let  
 
Let  
  
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<math>Sunday \equiv 0 \pmod{7}</math>
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Sunday <math>\equiv 0 \pmod{7}</math>
  
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<math>Monday \equiv 1 \pmod{7}</math>
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Monday <math>\equiv 1 \pmod{7}</math>
  
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<math>Tuesday \equiv 2 \pmod{7}</math>
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Tuesday <math>\equiv 2 \pmod{7}</math>
  
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<math>Wednesday \equiv 3 \pmod{7}</math>
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Wednesday <math>\equiv 3 \pmod{7}</math>
  
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<math>Thursday \equiv 4 \pmod{7}</math>
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Thursday <math>\equiv 4 \pmod{7}</math>
  
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<math>Friday \equiv 5 \pmod{7}</math>
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Friday <math>\equiv 5 \pmod{7}</math>
  
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<math>Saturday \equiv 6 \pmod{7}</math>
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Saturday <math>\equiv 6 \pmod{7}</math>
  
  
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Which clearly indicates if you start from a <math>x \equiv 3 \pmod{7}</math> you will not get a <math>y \equiv 0 \pmod{7}</math>.
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Which indicates if you start from a <math>x \equiv 3 \pmod{7}</math> you will not get a <math>y \equiv 0 \pmod{7}</math>.
  
 
Any other starting value may lead to a <math>y \equiv 0 \pmod{7}</math>.
 
Any other starting value may lead to a <math>y \equiv 0 \pmod{7}</math>.
  
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Which means our answer is <math>\boxed{\textbf{(C)}\ Wednesday}</math>.
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Which means our answer is <math>\boxed{\textbf{(C)\ Wednesday}}</math>.
  
 
~phoenixfire
 
~phoenixfire
  
 
== Solution 3 ==
 
== Solution 3 ==
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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
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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
  
 
== Solution 4 ==
 
== Solution 4 ==
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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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Start counting on Sunday, to maximize the number of 10-day jumps before reaching Sunday again. Add the 10 (<math>\equiv 3 \pmod 7</math>) days to reach the first coupon day on <math>\boxed{\textbf{(C)\ Wednesday}}</math>.
 +
 
 
~~ gorefeebuddie
 
~~ gorefeebuddie
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Note: This only works when 7 and 3 are relatively prime.
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 +
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 ==
 
== Solution 5 ==
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==Solution Explained==
 
==Solution Explained==
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https://youtu.be/gOZOCFNXMhE ~ The Learning Royal
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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 ==
 
== Solution 6 ==
 
Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E
 
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
 
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=MOQj1zxH2gY&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=15
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~savannahsolver
 
~savannahsolver
 +
 +
==Video Solution (CREATIVE THINKING + MOST EFFICIENT!!!)==
 +
https://youtu.be/9NNJE3pkmVc
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 +
~Hayabusa1
  
 
==See also==
 
==See also==

Latest revision as of 09:32, 9 November 2024

Problem

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 $\text{Wednesday}$, she redeems her next coupon on $\text{Saturday}$.

$\text{Saturday}$ 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 $\boxed{\textbf{(C)}\ \text{Wednesday}}$.

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 indicates if you start from 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

Start counting on Sunday, to maximize the number of 10-day jumps before reaching Sunday again. Add the 10 ($\equiv 3 \pmod 7$) days to reach the first coupon day on $\boxed{\textbf{(C)\ Wednesday}}$.

~~ 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 $n$, 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 $\boxed{\textbf{(C)}\text{ Wednesday}}$.

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

https://youtu.be/8VQc6fbZMvg

~savannahsolver

Video Solution (CREATIVE THINKING + MOST EFFICIENT!!!)

https://youtu.be/9NNJE3pkmVc

~Education, the Study of Everything

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~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

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