https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Thegoodhunter-9115&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-15T23:30:33Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_14&diff=127375 2019 AMC 8 Problems/Problem 14 2020-07-03T16:35:20Z <p>Thegoodhunter-9115: Added Solution 5</p> <hr /> <div>==Problem 14==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; 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 &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; 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?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;\text{Day }1&lt;/math&gt; to &lt;math&gt;\text{Day }2&lt;/math&gt; denote a day where one coupon is redeemed and the day when the second coupon is redeemed. <br /> <br /> If she starts on a &lt;math&gt;\text{Monday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Thursday}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(A)}\ \text{Monday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Tuesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(B)}\ \text{Tuesday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;Wednesday&lt;/math&gt; she redeems her next coupon on &lt;math&gt;Saturday&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Saturday}&lt;/math&gt; to &lt;math&gt;\text{Tuesday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Tuesday}&lt;/math&gt; to &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> And on &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her last coupon. <br /> <br /> <br /> No sunday occured thus &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt; is correct. <br /> <br /> <br /> Checking for the other options, <br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(D)}\ \text{Thursday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Friday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> <br /> Checking for the other options gave us negative results, thus the answer is &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 2==<br /> Let <br /> <br /> &lt;math&gt;Sunday \equiv 0 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Monday \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Tuesday \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Wednesday \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Thursday \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Friday \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Saturday \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;10 \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;20 \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;30 \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;40 \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;50 \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;60 \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> Which clearly indicates if you start form a &lt;math&gt;x \equiv 3 \pmod{7}&lt;/math&gt; you will not get a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Any other starting value may lead to a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Which means our answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 3 ==<br /> Like Solution 2, let the days of the week be numbers&lt;math&gt;\pmod 7&lt;/math&gt;. &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are coprime, so continuously adding &lt;math&gt;3&lt;/math&gt; to a number&lt;math&gt;\pmod 7&lt;/math&gt; will cycle through all numbers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;. If a string of 6 numbers in this cycle does not contain &lt;math&gt;0&lt;/math&gt;, then if you minus 3 from the first number of this cycle, it will always be &lt;math&gt;0&lt;/math&gt;. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. ~~SmileKat32<br /> <br /> == Solution 4 ==<br /> Since Sunday is the only day that has not been counted yet. We can just add the 3 days as it will become &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. <br /> ~~ gorefeebuddie<br /> Note: This only works when 7 and 3 are relatively prime.<br /> <br /> == Solution 5 ==<br /> Let Sunday be Day 0, Monday be Day 1, Tuesday be Day 2, and so forth. We see that Sundays fall on Day &lt;math&gt;n&lt;/math&gt;, 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 &quot;used&quot; 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 &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=13|num-a=15}}<br /> <br /> {{MAA Notice}}</div> Thegoodhunter-9115 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_1&diff=127357 2019 AMC 8 Problems/Problem 1 2020-07-03T15:57:47Z <p>Thegoodhunter-9115: /* Solution 1 */</p> <hr /> <div>==Problem 1==<br /> <br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can and use the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }5\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We maximize the number of sandwiches Mike and Ike can buy by finding the lowest multiple of &lt;math&gt;\$4.50&lt;/math&gt; that is less than &lt;math&gt;\$30.&lt;/math&gt; This number is &lt;math&gt;6.&lt;/math&gt; <br /> <br /> Therefore, they can buy &lt;math&gt;6&lt;/math&gt; sandwiches for &lt;math&gt;\$4.50\cdot6=\$27.&lt;/math&gt; They spend the remaining money on soft drinks, so they buy &lt;math&gt;30-27=3&lt;/math&gt; soft drinks. <br /> <br /> Combining the items, Mike and Ike buy &lt;math&gt;6+3=9&lt;/math&gt; soft drinks.<br /> <br /> The answer is &lt;math&gt;\boxed{\textbf{(D) }9}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|before=First Problem|num-a=2}}<br /> <br /> {{MAA Notice}}</div> Thegoodhunter-9115 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_1&diff=127356 2019 AMC 8 Problems/Problem 1 2020-07-03T15:57:08Z <p>Thegoodhunter-9115: /* Problem 1 */</p> <hr /> <div>==Problem 1==<br /> <br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can and use the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }5\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We maximize the number of sandwiches Mike and Ike can buy by finding the lowest multiple of &lt;math&gt;\$4.50&lt;/math&gt; that is less than &lt;math&gt;\$30.&lt;/math&gt; This number is &lt;math&gt;6.&lt;/math&gt; <br /> <br /> Therefore, they can buy &lt;math&gt;6&lt;/math&gt; sandwiches for &lt;math&gt;\$4.50\cdot6=\$27.&lt;/math&gt; They spend the remaining money on soft drinks, so they buy &lt;math&gt;30-27=3&lt;/math&gt; soft drinks. <br /> <br /> Combining the items, Mike and Ike buy &lt;math&gt;3+3=6&lt;/math&gt; soft drinks.<br /> <br /> The answer is &lt;math&gt;\boxed{\textbf{(A) }6}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|before=First Problem|num-a=2}}<br /> <br /> {{MAA Notice}}</div> Thegoodhunter-9115