2019 AMC 8 Problems/Problem 14

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.

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 eleventh coupon, of which she only has ten. So, the answer is $\boxed{\textbf{(C)}\ Wednesday}$ .

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

Page

Toolbox

Search

2019 AMC 8 Problems/Problem 14

Contents

Problem 14

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

See Also