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

Revision as of 06:10, 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

@@ Line 29: / Line 29: @@
 <math>\text{Saturday}</math> to <math>\text{Tuesday}</math>.
-<math>\text{Tuesday}</math> to <math>text{Friday}</math>.
+<math>\text{Tuesday}</math> to <math>\text{Friday}</math>.
 <math>\text{Friday}</math> to <math>\text{Monday}</math>.

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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Difference between revisions of "2019 AMC 8 Problems/Problem 14"

Revision as of 06:10, 21 November 2019

Contents

Problem 14

Solution 1

Solution 2

Solution 3

See Also