https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Vaishnavim&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T00:30:10ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_14&diff=1352032019 AMC 8 Problems/Problem 142020-10-17T21:25:28Z<p>Vaishnavim: /* Solution 1 */</p>
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<div>==Problem 14==<br />
Isabella has <math>6</math> 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 <math>10</math> days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the <math>6</math> 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 />
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<math>\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}</math><br />
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==Solution 1==<br />
Let <math>\text{Day }1</math> to <math>\text{Day }2</math> denote a day where one coupon is redeemed and the day when the second coupon is redeemed. <br />
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If she starts on a <math>\text{Monday}</math> she redeems her next coupon on <math>\text{Thursday}</math>. <br />
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<math>\text{Thursday}</math> to <math>\text{Sunday}</math>.<br />
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Thus <math>\textbf{(A)}\ \text{Monday}</math> is incorrect.<br />
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If she starts on a <math>\text{Tuesday}</math> she redeems her next coupon on <math>\text{Friday}</math>.<br />
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<math>\text{Friday}</math> to <math>\text{Monday}</math>.<br />
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<math>\text{Monday}</math> to <math>\text{Thursday}</math>.<br />
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<math>\text{Thursday}</math> to <math>\text{Sunday}</math>.<br />
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Thus <math>\textbf{(B)}\ \text{Tuesday}</math> is incorrect.<br />
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If she starts on a <math>\text{Wednesday}</math> she redeems her next coupon on <math>\text{Saturday}</math>.<br />
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<math>\text{Saturday}</math> to <math>\text{Tuesday}</math>.<br />
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<math>\text{Tuesday}</math> to <math>\text{Friday}</math>.<br />
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<math>\text{Friday}</math> to <math>\text{Monday}</math>.<br />
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<math>\text{Monday}</math> to <math>\text{Thursday}</math>.<br />
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And on <math>\text{Thursday}</math> she redeems her last coupon. <br />
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No sunday occured thus <math>\boxed{\textbf{(C)}\ \text{Wednesday}}</math> is correct. <br />
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Checking for the other options, <br />
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If she starts on a <math>\text{Thursday}</math> she redeems her next coupon on <math>\text{Sunday}</math>.<br />
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Thus <math>\textbf{(D)}\ \text{Thursday}</math> is incorrect.<br />
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If she starts on a <math>\text{Friday}</math> she redeems her next coupon on <math>\text{Monday}</math>.<br />
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<math>\text{Monday}</math> to <math>\text{Thursday}</math>.<br />
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<math>\text{Thursday}</math> to <math>\text{Sunday}</math>.<br />
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Checking for the other options gave us negative results, thus the answer is <math>\boxed{\textbf{(C)}\ \text{Wednesday}}</math>.<br />
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== Solution 2==<br />
Let <br />
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<math>Sunday \equiv 0 \pmod{7}</math><br />
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<math>Monday \equiv 1 \pmod{7}</math><br />
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<math>Tuesday \equiv 2 \pmod{7}</math><br />
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<math>Wednesday \equiv 3 \pmod{7}</math><br />
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<math>Thursday \equiv 4 \pmod{7}</math><br />
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<math>Friday \equiv 5 \pmod{7}</math><br />
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<math>Saturday \equiv 6 \pmod{7}</math><br />
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<math>10 \equiv 3 \pmod{7}</math><br />
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<math>20 \equiv 6 \pmod{7}</math><br />
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<math>30 \equiv 2 \pmod{7}</math><br />
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<math>40 \equiv 5 \pmod{7}</math><br />
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<math>50 \equiv 1 \pmod{7}</math><br />
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<math>60 \equiv 4 \pmod{7}</math><br />
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Which clearly indicates if you start form a <math>x \equiv 3 \pmod{7}</math> you will not get a <math>y \equiv 0 \pmod{7}</math>.<br />
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Any other starting value may lead to a <math>y \equiv 0 \pmod{7}</math>.<br />
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Which means our answer is <math>\boxed{\textbf{(C)}\ Wednesday}</math>.<br />
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~phoenixfire<br />
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== Solution 3 ==<br />
Like Solution 2, let the days of the week be numbers<math>\pmod 7</math>. <math>3</math> and <math>7</math> are coprime, so continuously adding <math>3</math> to a number<math>\pmod 7</math> will cycle through all numbers from <math>0</math> to <math>6</math>. If a string of 6 numbers in this cycle does not contain <math>0</math>, then if you minus 3 from the first number of this cycle, it will always be <math>0</math>. So, the answer is <math>\boxed{\textbf{(C)}\ Wednesday}</math>. ~~SmileKat32<br />
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== 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 <math>\boxed{\textbf{(C)}\ Wednesday}</math>. <br />
~~ gorefeebuddie<br />
Note: This only works when 7 and 3 are relatively prime.<br />
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== 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 <math>n</math>, 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 <math>\boxed{\textbf{(C)}\text{ Wednesday}}</math>.<br />
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== Solution 6 ==<br />
Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E<br />
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Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)<br />
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==See Also==<br />
{{AMC8 box|year=2019|num-b=13|num-a=15}}<br />
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{{MAA Notice}}</div>Vaishnavimhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_13&diff=1351792015 AMC 8 Problems/Problem 132020-10-17T14:49:52Z<p>Vaishnavim: /* Solution 2 */</p>
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<div>How many subsets of two elements can be removed from the set <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}</math> so that the mean (average) of the remaining numbers is 6?<br />
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<math>\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}</math><br />
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==Solution 1==<br />
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Since there will be <math>9</math> elements after removal, and their mean is <math>6</math>, we know their sum is <math>54</math>. We also know that the sum of the set pre-removal is <math>66</math>. Thus, the sum of the <math>2</math> elements removed is <math>66-54=12</math>. There are only <math>\boxed{\textbf{(D)}~5}</math> subsets of <math>2</math> elements that sum to <math>12</math>: <math>\{1,11\}, \{2,10\}, \{3, 9\}, \{4, 8\}, \{5, 7\}</math>.<br />
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==Solution 2== <br />
We can simply remove <math>5</math> subsets of <math>2</math> numbers, while leaving only <math>6</math> behind. The average of this one-number set is still <math>6</math>, so the answer is <math>\boxed{\textbf{(D)}~5}</math>.<br />
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==See Also==<br />
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{{AMC8 box|year=2015|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Vaishnavim