2019 AMC 8

Complete problem set with solutions and individual problem pages

Problem 14 Medium

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?

  • A.

    \text{Monday}

  • B.

    \text{Tuesday}

  • C.

    \text{Wednesday}

  • D.

    \text{Thursday}

  • E.

    \text{Friday}

Answer:C

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}}.