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?

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

Solution 1

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

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

$Thursday$ to $Sunday$ .

Thus $\boxed{\textbf{(A)}\ Monday}$ 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

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2019 AMC 8 Problems/Problem 14

Contents

Problem 14

Solution 1

Solution 2

See Also