MATHCOUNTS
https://www.mathcounts.org
enThirst Quencher
https://www.mathcounts.org/resources/problem-of-the-week/thirst-quencher
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-06-17T06:15:00-07:00">June 17, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/algebraic-expressions-equations">Algebraic Expressions & Equations</a>, <a href="/problem-topic-content-area/measurement">Measurement</a>, <a href="/problem-topic-content-area/proportional-reasoning">Proportional Reasoning</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>In the summer months, a common problem people suffer from is dehydration – not enough water in their body. Waiting until you feel thirsty is not a good indicator that it’s time to drink water. In fact, once thirst has settled in, you are already dehydrated! A general guideline is that people should drink eight 8-ounce glasses of water per day. According to this guideline, how many gallons of water should a person drink in a day, given that 1 gallon = 4 quarts and 1 quart = 32 ounces? Express your answer as a decimal to the nearest tenth.</p>
<p> </p>
<p>When dehydrated, your body needs water. Juices and sodas are not the optimal beverages for replenishing your body with the water it needs. And if you take cost into account, you’ll definitely want to grab a glass of water! It is estimated that 4000 glasses of tap water cost the same as a six-pack of soda. If a six-pack of soda costs $2.99, how many glasses of water would have a cost of 10¢? Express your answer to the nearest whole number.</p>
<p> </p>
<p>Lake Tahoe is the second deepest lake in the U.S. and it holds 40 trillion gallons of water – enough to cover the state of California to a depth of 14 inches! Given that 1 ft<sup>3</sup> = 7.48 gallons and 1 mile = 5280 ft, how many square miles are in the area of California? Express your answer to the nearest thousand.</p>
<p style="text-align: center;"><strong>CHECK THE <a href="https://www.mathcounts.org/resources/problem-archive"> PROBLEM OF THE WEEK ARCHIVE </a><br>
FOR SOLUTIONS TO PREVIOUS PROBLEMS </strong></p>
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Mon, 17 Jun 2019 15:33:24 +0000kera.johnson242227 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/thirst-quencher#commentsFather's Day
https://www.mathcounts.org/resources/problem-of-the-week/fathers-day
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-06-10T06:15:00-07:00">June 10, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/algebraic-expressions-equations">Algebraic Expressions & Equations</a>, <a href="/problem-topic-content-area/plane-geometry">Plane Geometry</a>, <a href="/problem-topic-content-area/proportional-reasoning">Proportional Reasoning</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>For Father’s Day, Martin’s family is having a barbeque. Martin is making the lemonade. If the directions calls for 2 tablespoons of mix per quart of water, how many tablespoons of mix are needed to make 1.5 gallons of lemonade?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Since there are 4 quarts in a gallon, it follows that 4 <em>×</em> 2 = 8 tablespoons of mix are required per gallon of water, and half that amount, or 8/2 = 4 tablespoons of mix are required per 0.5 gallon of water. Therefore, to make 1.5 gallons of lemonade, Martin needs to use 8 + 4 = </em></span><span style="color:#272561;"><em><strong>12</strong> </em></span><span style="color:#19a2bf;"><em>tablespoons of mix.</em></span></p>
<p>Martin’s brother is grilling the food. He puts 2 more burgers than hotdogs on the grill and 3 more ears of corn than hotdogs. In total, there are of 20 items on the grill. How many ears of corn are on the grill?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Let b, h and c represent the numbers of burgers, hot dogs and ears of corn, respectively. Based on the information provided, we can set up the following equations: b = h + 2, c = h + 3 and b + h + c = 20. We now have a system of three equations in three variables. We can solve by substituting h + 2 and h + 3 for b and c, respectively in the equation b + h + c = 20. Doing so, we get (h + 2) + h + (h + 3) = 20 <span style="color:#19a2bf;"><em>→</em></span> 3h + 5 = 20 <em><em>→</em></em> h = 5 hot dogs. Since c = h + 3 and h = 5, we substitute and find that there are c = 5 + 3 = </em></span><span style="color:#272561;"><em><strong>8</strong> </em></span><span style="color:#19a2bf;"><em>ears of corn on the grill.</em></span></p>
<p>Martin’s dad is 6’2” tall, but while playing catch with the football, Martin notices that his dad’s shadow is only 2 feet long. If one of the trees in their backyard casts a shadow that was 11 feet long at the same time, how tall is the tree? Express your answer to the nearest whole number.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Since there are 12 inches in a foot, Martin’s dad is 6 + 2/12 = 6 1/6 = 37/6 feet tall. Now we can set the proportion between the ratio of the shadow lengths of Martin’s dad and the tree and the ratio of the heights of Martin’s dad and the tree. We have 2/11 = (37/6)/t, where t represents the height of the tree. Cross-multiplying, we get 2 <em><em>×</em></em> t = 11 <em><em>×</em></em> (37/6) <em><em>→</em></em> 2t = 407/6 <em><em>→</em></em> t = (407/6)(1/2) = 407/12 =33.916666… ≈ </em></span><span style="color:#272561;"><em><strong>34</strong> </em></span><span style="color:#19a2bf;"><em>feet.</em></span></p>
<p style="text-align: center;">
<strong>
♦
<span style="color:#f47321;">
Page 1
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of the linked PDF contains PROBLEMS & SOLUTIONS.<br>
<span style="color:#f47321;">
Page 2
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Mon, 10 Jun 2019 12:02:04 +0000kera.johnson242219 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/fathers-day#commentsSchool's Out for Summer
https://www.mathcounts.org/resources/problem-of-the-week/schools-out-summer
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-06-03T06:15:00-07:00">June 3, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/number-theory">Number Theory</a>, <a href="/problem-topic-content-area/probability-counting-combinatorics">Probability, Counting & Combinatorics</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>At Walter Whitman Middle School, to celebrate the last day of school, Mrs. Hanks brought in treats for her class. She has chocolate bars, lollipops and gumballs. Mrs. Hanks tells each student they can choose any combination of three treats from what she has. How many different combinations of three treats can one of her students chose?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Let’s call the three types of treats A, B and C. There is 1 combination that has one of each type: ABC. There are 3 combinations that each have three of the same type: AAA, BBB and CCC. There are 6 combinations that have two of one type and one of another: AAB, AAC, BBA, BBC, CCA and CCB. These are the only 1 + 3 + 6 = </em></span><span style="color:#272561;"><em><strong>10 </strong></em></span><span style="color:#19a2bf;"><em>possible combinations of three treats.</em></span></p>
<p>At the end-of-year assembly, WWMS holds a raffle, in which several lucky students will each win a gift certificate to the famous Giorgio’s Gelato. There are 10 gift certificates to be raffled to the 6th graders, 15 gift certificates to be raffled to the 7th graders and 20 gift certificates to be raffled to the 8th graders. Each grade has 300 students. Junie is in 8th grade at WWMS. Junie’s sister and brother also attend WWMS and are in 7th and 6th grade, respectively. What is the probability that at least one of the three siblings wins a gift certificate? Express your answer as a common fraction.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>For each of the gift certificates, each sibling has a 1/300 chance of winning. Junie’s brother in 6th grade as a 1/300 chance at each of the 10 gift certificates, so 10/300 = 1/30 chance of winning a gift certificate. Junie’s sister in 7th grade, similarly, has a 15/300 = 1/20 chance of winning. Junie has a 20/300 = 1/15 chance of winning. The probability that at least one of them wins a gift certificate is, therefore, 1/30 + 1/20 + 1/15 = </em></span><span style="color:#272561;"><em><strong>3/20</strong></em></span><span style="color:#19a2bf;"><em>.</em></span></p>
<div style="float:right;"><img alt="" src="/sites/default/files/u5328/lockers.jpg" style="width: 150px; height: 217px; margin: 10px; float: right;"></div>
<p>Before WWMS students were officially released for summer vacation, the staff had to conduct an official check of all 900 lockers, numbered from 1 to 900. Once the locker check was completed, all 900 lockers remained open. As students exited in an orderly fashion, walking down the hallway single file, the 1st student closed all the lockers. The 2nd student then opened every even numbered locker. The 3rd student changed the status of every locker whose number was a multiple of three – opened it if closed or closed it if opened. The 4th student then did the same for every locker whose number was a multiple of four, and so on through the 900th student who changed the status of locker number 900. When all of the students had left the school, how many lockers were closed?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>The lockers that will be open after the 900th student leaves, are all of the lockers whose numbers have an even number of factors. Take 6, for example, its factors are 1, 2, 3 and 6. Student 1 closes the locker, student 2 opens it, student 3 closes it and then student 6 opens it again. A locker number with an odd number of factors will end up closed after the 900th student leaves. Which numbers have an odd number of factors? Only perfect squares have an odd number of factors. The number 1 only has itself as a factor; 4 has 1, 2 and 4 as factors; 9 has 1, 3 and 9 as factors; and so on. This is because factors come in pairs and only perfect squares have 1 pair consisting of the same integer repeated. For example, the factor pairs of 9 are 1 × 9 and 3 × 3. The last perfect square locker number is 30<sup>2</sup> = 900, so since there are 30 perfect square locker numbers, 1<sup>2</sup> to 30<sup>2</sup>, after all the students left the school, there were </em></span><span style="color:#272561;"><em><strong>30</strong></em></span><span style="color:#19a2bf;"><em> closed lockers.</em></span></p>
<p style="text-align: center;">
<strong>
♦
<span style="color:#f47321;">
Page 1
</span>
of the linked PDF contains PROBLEMS & SOLUTIONS.<br>
<span style="color:#f47321;">
Page 2
</span>
contains ONLY PROBLEMS. ♦
</strong>
</p>
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Mon, 03 Jun 2019 15:15:11 +0000kera.johnson242214 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/schools-out-summer#commentsMemorial Day
https://www.mathcounts.org/resources/problem-of-the-week/memorial-day-1
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-05-27T06:00:00-07:00">May 27, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/general-math">General Math</a>, <a href="/problem-topic-content-area/number-theory">Number Theory</a>, <a href="/problem-topic-content-area/problem-solving-misc">Problem Solving (Misc.)</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>The distinct letters from the word MEMORIAL are used to create set Z, such that Z = {M, E, O, R, I, A, L}. The letters from the word DAY are used to create set Y, such that Y = {D, A, Y}. If set X is the intersection of sets Z and Y, what are the letters in set X?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>The intersection of two sets includes all the elements (or members) that are in both sets. So set X = </em></span><span style="color:#272561;"><em><strong>{A}</strong></em></span><span style="color:#19a2bf;"><em>.</em></span></p>
<p>If set V is the union of sets Z and Y from the previous problem, what letters are in set V?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>The union of two sets contains all the elements in both sets combined, thus set V = </em></span><span style="color:#272561;"><em><strong>{M, E, O, R, I, A, L, D, Y}</strong></em></span><span style="color:#19a2bf;"><em>.</em></span></p>
<p>If set S is the intersection of sets X and V from the previous problems and set T is the union of sets X and V, what letters are in sets S and T?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Based on the definitions of intersection and union previously given, we see that S = </em></span><span style="color:#272561;"><em><strong>{A} </strong></em></span><span style="color:#19a2bf;"><em>and T = </em></span><span style="color:#272561;"><em><strong>{M, E, O, R, I, A, L, D, Y}</strong></em></span><strong><span style="color:#19a2bf;"><em>.</em></span></strong></p>
<p style="text-align: center;"><strong><strong>♦ <span style="color:#f47321;"> Page 1 </span> of the linked PDF contains PROBLEMS & SOLUTIONS.<br>
<span style="color:#f47321;">Page 2 </span> contains ONLY PROBLEMS. ♦ </strong> </strong></p>
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Tue, 28 May 2019 14:20:55 +0000kera.johnson242212 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/memorial-day-1#comments2019 National Champion
https://www.mathcounts.org/resources/problem-of-the-week/2019-national-champion
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-05-20T06:00:00-07:00">May 20, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/coordinate-geometry">Coordinate Geometry</a>, <a href="/problem-topic-content-area/number-theory">Number Theory</a>, <a href="/problem-topic-content-area/probability-counting-combinatorics">Probability, Counting & Combinatorics</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>Last week the national competition concluded, and Daniel Mai from Massachusetts earned the title of MATHCOUNTS National Champion. Let’s look at some of the problems he had to solve on the way to the top!</p>
<p><strong>Sprint #14</strong><br>
Two opposite vertices of a certain square are located at (1, 6) and (−3, 1). If the line <em>y</em> = <em>mx</em> divides this square into two regions of equal area, what is the value of <em>m</em>? Express your answer as a common fraction.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Because of symmetry, any line that separates the square into two regions of equal area must intersect the center point of the square. Since the segment with endpoints (−3, 1) and (1, 6) is a diagonal of the square, we know that the center point of the square is at the midpoint of this segment, which is ((−3 + 1)/2, (6 + 1)/2) = (−1, 7/2). The line given by y = mx intersects the point (0, 0). So, the line in question passes through (0, 0) and (−1, 7/2) and has slope m = (7/2 − 0)/(−1 − 0) = </em></span><span style="color:#272561;"><em><strong>−7/2</strong></em></span><span style="color:#19a2bf;"><em>.</em></span></p>
<p><strong>Sprint #20</strong><br>
A certain digital lock has a keypad with five buttons labeled 1 to 5. To activate the locking mechanism, a secret code is set with these restrictions: the code must contain three parts, each part will consist of either pressing one button or pressing two buttons simultaneously, and no button may be pressed more than once. For example [3][4][1], [1][ 2 & 4][5] and [1 & 2][4][3 & 5] are three possible codes. Also, note that [1][2 & 4][5] and [1][4 & 2][5] are indistinguishable. How many distinct secret codes can be set for this lock?</p>
<div>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>A code for this lock can consist of three parts that are three presses of a single button, or two presses of a single button and one of two buttons pressed simultaneously, or one press of a single button and two of two buttons pressed simultaneously. Let’s examine these three cases. </em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><strong><em>Case 1</em></strong><em>: The number of codes of the form [a][b][c] is 5 × 4 × 3 = <u>60</u> codes. </em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><strong><em>Case 2</em></strong><em>: For codes of the form [a], [b], [c&d], there are 5 × 4 = 20 options for [a] and [b], leaving <sub>3</sub>C<sub>2</sub> =3 options for [c&d]. That gives us 20 × 3 = 60 combinations of the three parts, each of which can be arranged in 3 orders, for a total of 60 × 3 = <u>180</u> codes. </em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><strong><em>Case 3</em></strong><em>: For codes of the form [a][b&c][d&e], there are 5 options for [a], leaving <sub>4</sub>C<sub>2</sub> = 6 options for [b&c], which then leaves 1 option for [[d&e], for a total of 5 × 6 × 1 = 30 combinations of the three parts, each of which can be arranged in 3 orders, for a total of 30 × 3 = <u>90</u> codes. </em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Therefore, the total number of distinct codes possible is 60 + 180 + 90 = </em></span><span style="color:#272561;"><strong><em>330</em></strong><em> </em></span><span style="color:#19a2bf;"><em>codes.</em></span></p>
</div>
<p><strong>Sprint #29</strong><br>
How many of the first 100,000 positive integers have no single-digit prime factors?</p>
<div>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Here is what we get counting multiples of 2, 3, 5 and 7. We must be careful, though, as some integers are counted two, three, even four times:</em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em> 2: 50,000 2 × 3: 16,666 2 × 3 × 5: 3333 2 × 3 × 5 × 7: 476<br>
3: 33,333 2 × 5: 10,000 2 × 3 × 7: 2380<br>
5: 20,000 2 × 7: 7142 2 × 5 × 7: 1428<br>
7: <u>14,285</u> 3 × 5: 6666 3 × 5 × 7: <u> 952</u><br>
117,618 3 × 7: 4761 8093<br>
5 × 7: <u> 2857</u><br>
48,092</em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>First, we see that the 48,092 multiples of 6, 10, 14, 15, 21 and 35 were counted twice among the 117,618 multiples of 2, 3, 5 and 7. For example, multiples of 2 × 3 = 6 were counted with multiples of 2 and with multiples of 3. Since these integers have been double counted, we subtract and get 117,618 − 48,092 = 69,526 integers. </em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Next, the 8093 multiples of 30, 42, 70 and 105 were counted three times among the 48,092 multiples of 6, 10, 14, 15, 21 and 35. For example, multiples of 2 × 3 × 5 = 30 were counted with multiples of 2 × 3 = 6, with multiples of 2 × 5 = 10 and with multiples of 3 × 5 = 15. We just removed all three counts of these 8093 integers when we subtracted the 48,092 integers to resolve the double counting. So, we add these 8093 integers to our previous total and get 69,526 + 8093 = 77,619 integers. </em></span></p>
</div>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Finally, the 476 multiples of 210 were counted four times among the 8093 multiples of 30, 42, 70 and 105. For example, multiples of 2 × 3 × 5 × 7 = 210 were counted with multiples of 2 × 3 × 5 = 30, with multiples of 2 × 3 × 7 = 42, with multiples of 2 × 5 × 7 = 70 and with multiples of 3 × 5 × 7 = 105. We resolved three of the counts of these 476 integers when we initially subtracted 48,092 to resolve double counting, but we brought one of the counts back when we just added 8093 to our previous total. Now, we see that, 476 of those 77,619 integers are counted twice and subtracting gives us a total of 77,619 − 476 = 77,143 of the first 100,000 positive integers that have a single-digit prime factor. Therefore, the total number of integers that do not have a single-digit prime factor is 100,000 − 77,143 = </em></span><span style="color:#272561;"><em><strong>22,857</strong></em></span><span style="color:#19a2bf;"><em> </em><em>integers.</em></span></p>
<p style="text-align: center;"><strong>♦ <span style="color:#f47321;"> Page 1 </span> of the linked PDF contains PROBLEMS & SOLUTIONS.<br>
<span style="color:#f47321;">Page 2 </span> contains ONLY PROBLEMS. ♦ </strong></p>
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Mon, 20 May 2019 14:47:24 +0000kera.johnson242206 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/2019-national-champion#commentsFinal Countdown
https://www.mathcounts.org/resources/problem-of-the-week/final-countdown-0
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-05-13T06:00:00-07:00">May 13, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/coordinate-geometry">Coordinate Geometry</a>, <a href="/problem-topic-content-area/number-theory">Number Theory</a>, <a href="/problem-topic-content-area/problem-solving-misc">Problem Solving (Misc.)</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>On Sunday, May 12th, 224 middle-school math students participated in the written rounds of the 2019 Raytheon MATHCOUNTS National Competition. On Monday, May 13th, the top 12 competitors will go head to head in the National Countdown Round to determine the 2019 MATHCOUNTS National Champion. Let’s solve a few problems from the 2018 National Countdown Round.</p>
<p><strong>2018 National Countdown #21</strong></p>
<p>In square units, what is the area of the triangle with vertices P(−2, 1), Q(3, 8) and R(9, 3)? Express your answer as a decimal to the nearest tenth.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>One way to solve this problem is by graphing the triangle with vertices P(−2, 1), Q(3, 8) and R(9, 3) on a coordinate grid. Then, we construct a rectangle with vertices P(−2, 1), O(−2, 8), N(9, 8) and M(9, 3) that circumscribes this triangle. We see that rectangle MNOP is composed of triangle PQR and right triangles POQ, RNQ and PMR, and that the area of triangle PQR is the area rectangle MNOP minus the areas of right triangles POQ, RNQ and PMR. Therefore, the area of triangle PQR is (7)(11) − (1/2)(5)(7) − (1/2)(5)(6) − (1/2)(2)(11) = 77 − 35/2 − 15 − 11 = 51 − 35/2 = (102 − 51)/2 = 67/2 = </em></span><span style="color:#272561;"><em><strong>33.5</strong></em></span><span style="color:#19a2bf;"><em> units<sup>2</sup>. Another way to solve this problem is by using the shoelace method, which says that the are of a closed figure with vertices A(x<sub>1</sub>, y<sub>1</sub>), B(x<sub>2</sub>, y<sub>2</sub>) and C(x<sub>3</sub>, y<sub>3</sub> ) is |(x<sub>1</sub>y<sub>2</sub> + x<sub>2</sub>y<sub>3</sub> + x<sub>3</sub>y<sub>1</sub>) − (x<sub>2</sub>y<sub>1</sub> + x<sub>3</sub>y<sub>2</sub> + x<sub>1</sub>y<sub>3</sub>)|/2. So, for the area of triangle PQR, we have |((−2)(8) + (3)(3) + (9)(1)) − ((1)(3) + (8)(9) + (3)(−2))|/2 = |(−16 + 9 + 9) − (3 + 72 − 6)|/2 = |2 − 69|/2 = 67/2 = </em></span><span style="color:#272561;"><em><strong>33.5</strong> </em></span><span style="color:#19a2bf;"><em>units<sup>2</sup>. </em></span></p>
<p><strong>2018 National Countdown #16</strong></p>
<p>What is the greatest prime factor of 2<sup>13</sup> + 2<sup>11</sup> − 10?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>The expression 2<sup>13</sup> + 2<sup>11</sup> − 10 can be rewritten (2<sup>11</sup>)(4 + 1) − 10 = (2<sup>10</sup>)(10) − 10 = (10)(2<sup>10</sup> − 1) = (10)((2<sup>5</sup>)<sup>2</sup> − 1) = (10)(2<sup>5</sup> + 1)(2<sup>5</sup> − 1) = 10(33)(31) = 2 × 5 × 3 × 11 × 31. Thus, the greatest prime factor of this expression is </em></span><span style="color:#272561;"><em><strong>31</strong></em></span><span style="color:#19a2bf;"><em>. </em></span></p>
<p><strong>2018 National Countdown #13</strong></p>
<p>The four largest numbers in a set of seven numbers have a mean of 10. The four smallest numbers in the same set have a mean of 5. What is the least possible sum of the seven numbers?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Let a, b, c, d, e, f and g represent the seven numbers in ascending order. We are told that (d + e + f + g)/4 = 10, so d + e + f + g = 40. We also are told that (a + b + c + d)/4 = 20, so a + b + c + d = 80. That means a + b + c + d + d + e + f + g = 40 + 20, and a + b + c + d + e + f + g = 60 − d. In order to minimize the sum of these seven numbers, which is 60 − d, we need to maximize the value of d. The greatest possible value of d occurs when d = e = f = g = 10. Therefore, the least possible value of the sum of these seven numbers is 60 − d = 60 − 10 = </em></span><span style="color:#272561;"><em><strong>50</strong></em></span><span style="color:#19a2bf;"><em>.</em></span></p>
<p style="text-align: center;">
<strong>
♦
<span style="color:#f47321;">
Page 1
</span>
of the linked PDF contains PROBLEMS & SOLUTIONS.<br>
<span style="color:#f47321;">
Page 2
</span>
contains ONLY PROBLEMS. ♦
</strong>
</p>
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Mon, 13 May 2019 12:41:27 +0000kera.johnson242199 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/final-countdown-0#commentsNational Competition
https://www.mathcounts.org/resources/problem-of-the-week/national-competition-8
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-05-06T06:00:00-07:00">May 6, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/algebraic-expressions-equations">Algebraic Expressions & Equations</a>, <a href="/problem-topic-content-area/number-theory">Number Theory</a>, <a href="/problem-topic-content-area/probability-counting-combinatorics">Probability, Counting & Combinatorics</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>On Sunday, May 12th, 224 of the nation’s most talented middle-school math minds will be in Orlando, FL for the 2019 Raytheon MATHCOUNTS National Competition. The stakes are high, and the problems will be tough. Here are a few problems national competitors solved in 2018.</p>
<p><strong><u>Sprint #17</u></strong><br>
In a card game, Nora draws cards at random, without replacement, from a deck of 21 cards. Twenty of the cards are numbered 1 through 20, and the other card is marked “Joker.” Nora keeps all of the cards she draws before she draws the Joker. What is the probability that the cards Nora keeps include exactly four prime-numbered cards? Express your answer as a common fraction.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>The eight prime-numbered cards among the cards in this game are 2, 3, 5, 7, 11, 13, 17 and 19. We are interested in the outcomes that result in four of these eight cards being included in the cards Nora keeps. Determining the number of favorable outcomes and the number of total outcomes with permutations of prime- and composite-numbered cards would be too time consuming. Instead we can simply focus on where the Joker is drawn relative to the eight prime-numbered cards. There are nine possible positions in which the Joker can be drawn (JPPPPPPPP, PJPPPPPPP, PPJPPPPPP, ...). There is only one favorable outcome, when it is drawn between the fourth and fifth prime-numbered cards (PPPPJPPPP). So, the probability of that outcome is </em></span><span style="color:#272561;"><em><strong><em>1/9</em></strong></em></span><span style="color:#19a2bf;"><em>.</em></span></p>
<p><strong><u>Target #7</u></strong><br>
A <em>4-up number </em>is defined as a positive integer that is divisible by neither 2 nor 3 and does not have 2 or 3 as any of its digits. How many numbers from 400 to 600, inclusive, are 4-up numbers?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Since 600 is divisible by both 2 and 3, it is not a 4-up number. Thus, we need only consider three-digit numbers of the forms 4BC and 5BC, where B and C represent the tens and units digit of each, respectively. Since 4-up numbers are not divisible by 2, the possible values for the units digit of each three-digit number are 1, 5, 7 and 9. We now see that possible 4-up numbers from 400 to 600, inclusive are of the forms 4B1, 4B5, 4B7, 4B9, 5B1, 5B5, 5B7 and 5B9. For a number to be divisible by 3, the sum of its digits must be divisible by three. So, we need to determine the number of possible values of B so that the sums 4 + B + 1, 4 + B + 5, …, 5 + B + 7, 5 + B + 9 are not divisible by 3. The possible values of B for each are listed below:</em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>4B1: 0, 5, 6, 8 and 9 are the <u>5</u> possible values of B.<br>
4B5: 1, 4, 5, 7 and 8 are the <u>5</u> possible values of B.<br>
4B7: 0, 5, 6, 8 and 9 are the <u>5</u> possible values of B.<br>
4B9: 0, 1, 4, 6, 7 and 9 are the <u>6</u> possible values of B.<br>
5B1: 1, 4, 5, 7 and 8 are the <u>5</u> possible values of B.<br>
5B5: 0, 1, 4, 6, 7 and 9 are the <u>6</u> possible values of B.<br>
5B7: 1, 4, 5, 7 and 8 are the <u>5</u> possible values of B.<br>
5B9: 0, 5, 6, 8 and 9 are the <u>5</u> possible values of B.</em></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>That’s a total of 5 + 5 + 5 + 6 + 5 + 6 + 5 + 5 = </em></span><span style="color:#272561;"><em><strong><em>42</em></strong> </em></span><span style="color:#19a2bf;"><em>numbers.</em></span></p>
<p><strong><u>Team #3</u></strong><br>
Captain Greenbeard’s treasure chest holds 100 grams of a mix of gold, silver and bronze coins. The gold, silver and bronze coins each weigh 6 grams, 2 grams and 5 grams, respectively, and are worth 30 droubles, 8 droubles and 15 droubles, respectively. Greenbeard has at least one of each kind of coin. What is the greatest possible total value, in droubles, of the coins in the chest?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Since we are told that Greenbeard has at least one of each kind of coin, that accounts for 6 + 2 + 5 = 13 grams of the total 100 grams held in the treasure chest. Let’s determine the maximum possible value of the remaining 100 – 13 = 87 grams. The gold coins are the most valuable, so we want as many of those as possible. At a weight of 6 grams each, there could be at most another 14 gold coins, for an additional 14 × 6 = 84 grams. But there is no way to reach a total weight of exactly 100 grams with additional silver and bronze coins. If, however, there were another 13 gold coins, for an additional 13 × 6 = 78 grams, that would leave 87 – 78 = 9 grams to account for. Since the bronze coins are more valuable than the silver coins, let’s determine the maximum number of additional bronze coins next. The remaining 9 grams of weight could come from at most one additional bronze coin weighing 5 grams. That leaves 9 – 5 = 4 grams to account for. Two silver coins weighing 2 grams each could account for the remaining 4 grams. That’s 14 gold coins, 2 bronze coins and 3 silver coins, with a total weight of 14 × 6 + 2 × 5 + 3 × 2 = 84 + 10 + 6 = 100 grams. This yields the largest possible value of the coins in the treasure chest, which is 14 × 30 + 2 × 15 + 3 × 8 = 420 + 24 + 30 = </em></span><span style="color:#272561;"><em><strong><em>474</em></strong></em></span><span style="color:#19a2bf;"><em> droubles.</em></span></p>
<p style="text-align: center;"><strong>♦ <span style="color:#f47321;"> Page 1 </span> of the linked PDF contains PROBLEMS & SOLUTIONS.<br>
<span style="color:#f47321;">Page 2 </span> contains ONLY PROBLEMS. ♦ </strong></p>
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Mon, 06 May 2019 14:15:11 +0000kera.johnson242169 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/national-competition-8#commentsMini #90 - Even More Analytic Geometry
https://www.mathcounts.org/resources/video-library/mathcounts-minis/mini-90-even-more-analytic-geometry
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<h3>May 2019</h3>
<p><em>This video shows multiple ways to solve problems involving analytic geometry</em><em>.</em></p>
<ul><li><a href="/sites/default/files/May2019MinisHandout.pdf">Download the Activity Sheet here</a>.</li>
<li>Download the Solutions here.</li>
</ul><p><em>Video by Art of Problem Solving's Richard Rusczyk, a MATHCOUNTS alum. Visit <a href="http://www.artofproblemsolving.com">Art of Problem Solving</a> for many more educational resources.</em></p>
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<div class="field field-name-field-video-category field-type-taxonomy-term-reference field-label-inline clearfix field--video-category"><div class="field-label">Video Category: </div><div class="field-items"><div class="field-item even"><a href="/resources/video-library/mathcounts-minis">MATHCOUNTS Minis</a></div></div></div>Wed, 01 May 2019 18:49:57 +0000kera.johnson242163 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/video-library/mathcounts-minis/mini-90-even-more-analytic-geometry#commentsCounting Down to Nationals
https://www.mathcounts.org/resources/problem-of-the-week/counting-down-nationals-0
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-04-29T06:00:00-07:00">April 29, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/algebraic-expressions-equations">Algebraic Expressions & Equations</a>, <a href="/problem-topic-content-area/number-theory">Number Theory</a>, <a href="/problem-topic-content-area/statistics-data">Statistics & Data</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>The countdown round is often the most anticipated and exciting part of competition for Mathletes<sup>®</sup> and spectators alike. Here are some of our favorite countdown round problems from the 2019 competitions</p>
<p><strong><u>School #16</u></strong><br>
On a certain farm, each chicken has two feet and each rabbit has four feet. If the combined number of chickens and rabbits on the farm is 100 and there are a total of 260 feet on these animals, how many chickens are there?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Let c and r represent the numbers of chickens and rabbits, respectively, on this farm. We can write the following equations: 2c + 4r = 260 and c + r = 100. Dividing both sides of the first equation by 2 yields c + 2r = 130. If we subtract the two equations c + 2r = 130 and c + r = 100, we get (c + 2r) − (c + r) = 130 − 100 ® r = 30. So, there are 30 rabbits on this farm and 100 − 30 = </em></span><span style="color:#272561;"><em><strong>70</strong> </em></span><span style="color:#19a2bf;"><em>chickens. </em></span></p>
<p><strong><u>Chapter #59</u></strong><br>
Seven consecutive even integers have a sum of 406. What is the sum of the least and the greatest of the seven integers?</p>
<div>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>For a list of consecutive integers, the mean is equal to the median of the integers. So, for these seven consecutive integers, the mean and median are both equal to 406/7 = 58. Since there are seven integers in this list, the median is middle integer, which is also the fourth integer, when the list is arranged ordered. If the fourth integer is 58, then the least of the seven integers is three less than 58, or 55; the greatest of these integers is three more than 58, or 61. The sum of the least and greatest of the seven integers, then, is 55 + 61 = </em></span><span style="color:#272561;"><strong><em>116</em></strong></span><span style="color:#19a2bf;"><em>.</em></span></p>
</div>
<p><strong><u>State #43</u></strong><br>
Scott is thinking of a positive integer that is one more than, or one less than a multiple of 4. Quincy is thinking of an integer that is one more than, or one less than a multiple of 10. Stella is thinking of an integer with an even number of positive integer divisors. If all three are thinking of the same integer <em>n</em>, what is the least possible value of <em>n</em>?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>If Stella’s number has an even number of positive divisors, it must not be a perfect square. Scott’s number is positive and is one more or one less than a multiple of 4. Listing some multiples of 4, we have 0, 4, 8, 12, 16, 20, 24, …. So, excluding perfect squares, Scott’s number could be, 3, 5, 7, 11, 13, 15, 17, 19, 21, 23, …. Quincy’s number, which must also be positive, is one more or one less than a multiple of 10. Let’s see if any numbers on the list we've started of possible numbers Scott is thinking of is one more or one less than a multiple of 10. We see that the list we've started includes 11 and 21, both of which are one more than a multiple of 10, and 19, which is one less than a multiple of 10. So, the least possible value of the integer that these three are thinking of is n = </em></span><span style="color:#272561;"><em><strong>11</strong></em></span><span style="color:#19a2bf;"><em>.</em></span></p>
<p style="text-align: center;">
<strong>
♦
<span style="color:#f47321;">
Page 1
</span>
of the linked PDF contains PROBLEMS & SOLUTIONS.<br>
<span style="color:#f47321;">
Page 2
</span>
contains ONLY PROBLEMS. ♦
</strong>
</p>
</div>
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Mon, 29 Apr 2019 04:47:03 +0000kera.johnson242162 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/counting-down-nationals-0#commentsCounting Down to Nationals
https://www.mathcounts.org/resources/problem-of-the-week/counting-down-nationals
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2019-04-29T06:00:00-07:00">April 29, 2019</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/algebraic-expressions-equations">Algebraic Expressions & Equations</a>, <a href="/problem-topic-content-area/number-theory">Number Theory</a>, <a href="/problem-topic-content-area/statistics-data">Statistics & Data</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>The countdown round is often the most anticipated and exciting part of competition for Mathletes<sup>®</sup> and spectators alike. Here are some of our favorite countdown round problems from the 2019 competitions</p>
<p><strong><u>School #16</u></strong></br>
On a certain farm, each chicken has two feet and each rabbit has four feet. If the combined number of chickens and rabbits on the farm is 100 and there are a total of 260 feet on these animals, how many chickens are there?</p>
<p><strong><u>Chapter #59</u></strong></br>
Seven consecutive even integers have a sum of 406. What is the sum of the least and the greatest of the seven integers?</p>
<p><strong><u>State #43</u></strong></br>
Scott is thinking of a positive integer that is one more than, or one less than a multiple of 4. Quincy is thinking of an integer that is one more than, or one less than a multiple of 10. Stella is thinking of an integer with an even number of positive integer divisors. If all three are thinking of the same integer <em>n</em>, what is the least possible value of <em>n</em>?</p>
<p style="text-align: center;"><strong>CHECK THE <a href="https://www.mathcounts.org/resources/problem-archive"> PROBLEM OF THE WEEK ARCHIVE</a><br>
FOR SOLUTIONS TO PREVIOUS PROBLEMS </strong></p>
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Mon, 29 Apr 2019 04:44:04 +0000kera.johnson242161 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/counting-down-nationals#comments