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| − | {{AoPS Wiki:Sandbox/header}} [[Category:Aoum]] [[Category:Creative Projects]] <!-- DO NOT DELETE THIS LINE --> | + | {{Template:Sandbox}} [[Category:Aoum]] [[Category:Creative Projects]] <!-- DO NOT DELETE THIS LINE --> |
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| − | <div style="border: 2px solid #007bff; background-color: #007bff; color: white; padding: 12px 24px; text-align: center; width: fit-content; cursor: pointer; border-radius: 8px; font-size: 16px; font-weight: bold; display: inline-block;">[//{{SERVER}}/wiki/index.php/User:Aoum Aoum]</div>
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| − | <!--
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| − | == Problem 1 ==
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| − | Evaluate <math>5^2 - (10)(6) + 6^2</math>.
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| − | | |
| − | == Solution ==
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| − | It's not hard to expand the expression, but we can also recognize that the given expression is an instance of the formula
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| − | <cmath>(a - b)^2 = a^2 - 2ab + b^2.</cmath>
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| − | | |
| − | We have: <math>5^2 - (10)(6) + 6^2 = 5^2 - (2)(5)(6) + 6^2 = (5-6)^2 = \boxed{1}</math>.
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| − | | |
| − | == Problem 2 ==
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| − | A three digit number <math>``abc"</math> has the property that the product of <math>a</math> and <math>b</math> is equal to <math>c .</math> The digits <math>a, b,</math> and <math>c</math> are not necessarily distinct. What is the greatest possible value of the three digit number?
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| − | | |
| − | == Solution ==
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| − | We want to maximize the hundreds digit above all, so let's first try setting <math>a = 9.</math> We would then like to maximize <math>b,</math> but it must satisfy the constraint <math>9b = c,</math> where <math>b</math> and <math>c</math> are digits. The only digits <math>b</math> and <math>c</math> that satisfy this equation are <math>b = 1</math> and <math>c = 9</math> or <math>b=c=0,</math> so the largest such three digit number is <math>\boxed{919}.</math>
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| − | | |
| − | == Problem 3 ==
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| − | Find the smallest positive integer that is one less than a number that is a multiple of each of <math>3, 5, 7, 9,</math> and <math>11.</math>
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| − | | |
| − | == Solution ==
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| − | Let <math>n</math> be the smallest positive integer that is one less than a multiple of each of <math>3, 5, 7, 9,</math> and <math>11.</math> Then <math>n+1</math> is the smallest positive integer that is a multiple of each of <math>3, 5, 7, 9,</math> and <math>11.</math> In other words, <math>n+1</math> is the least common multiple of <math>3, 5, 7, 9,</math> and <math>11.</math>
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| − | The prime factorizations of these numbers are <math>3, 5, 7, 9 = 3^2,</math> and <math>11,</math> so the lowest common multiple is <math>3^2 \cdot 5 \cdot 7 \cdot 11 = 3465.</math> Therefore, the number we seek is <math>3465 - 1 = \boxed{3464}.</math>
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| − | | |
| − | == Problem 4 ==
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| − | Evaluate: <math>35^4 - 25^4</math>.
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| − | | |
| − | == Solution ==
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| − | We can use difference of squares: <math>35^4-25^4 = (35^2-25^2)(35^2+25^2)</math>. (Note that every fourth power is also a square.) At this point, you can either further factor by using difference of squares on the first factor, or you can simply substitute <math>35^2 = 1225</math> and <math>25^2 = 625</math>, and evaluate: <math>(1225 - 625)(1225 + 625) = 600 \cdot 1850 = \boxed{1110000}</math>.
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| − | | |
| − | == Problem 5 ==
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| − | Four different coins are used in Australia with values of <math>5, 10, 20,</math> and <math>50</math> cents. The Australian coins in Farmer Tim's pocket have a total value of <math>\$3.95,</math> and he has at least one of each coin. What is the smallest number of coins that he could have in his pocket?
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| − | | |
| − | == Solution ==
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| − | Since Farmer Tim has one of each coin, <math>50+20+10+5=85</math> cents are accounted for, leaving <math>395 - 85 = 310</math> cents for the remaining coins. We can try minimizing the number of coins that add up to <math>310</math> cents by using as many large denominations as possible.
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| − | The largest denomination is <math>50</math> cents, so we can take six <math>50</math> cent coins, which add up to <math>6 \cdot 50 = 300</math> cents. We then can take one <math>10</math> cent coin. Thus, we have a combination of seven coins that up to <math>310</math> cents.
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| − | Therefore, the smallest number of coins that Farmer Tim could have is <math>4 + 7 = \boxed{11}</math>.
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| − | | |
| − | == Problem 6 ==
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| − | If the integer <math>152AB1</math> is a perfect square, what is the sum of the digits of its square root?
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| − | | |
| − | == Solution ==
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| − | There are two things that help us here:
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| − | | |
| − | First, <math>400^2 = 160000</math> which is fairly close to <math>152AB1</math>.
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| − | | |
| − | Second, since <math>152AB1</math> ends in a 1, its square root must end in either a 1 or a 9.
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| − | | |
| − | Thus, we can narrow our choices down to square roots slightly lower than 400 that end in a 1 or a 9.
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| − | Now, since <math>152AB1</math> is about 8000 smaller than 160,000, it is reasonable to guess 391 as the square root first (399 is probably a bit too close), and evaluating <math>391^2</math>, we get 152881, which fits the form <math>152AB1</math>. Therefore, the answer is <math>3 + 9 + 1 = \boxed{13}</math>.
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| − | | |
| − | == Problem 7 ==
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| − | What is the largest four-digit number that is equal to the cube of the sum of its digits?
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| − | | |
| − | == Solution ==
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| − | The number we seek, in particular, is a four-digit cube. Thus, we can start with the largest four-digit cube, and work our way down.
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| − | | |
| − | | |
| − | The largest four-digit cube is <math>21^3=9261</math>, but <math>9+2+6+1 \neq 21</math>.
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| − | | |
| − | | |
| − | The next lowest cube is <math>20^3=8000</math>, but <math>8+0+0+0 \neq 20</math>.
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| − | | |
| − | | |
| − | The next lowest cube is <math>19^3=6859</math>, but <math>6+8+5+9 \neq 19</math>.
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| − | | |
| − | The next lowest cube is <math>18^3=5832</math>, and <math>5+8+3+2=18</math>, so the answer is <math>\boxed{5832}</math>.
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| − | | |
| − | == Problem 8 ==
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| − | How many different prime numbers divide evenly into <math>40^4-9^4</math>?
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| − | | |
| − | == Solution ==
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| − | We replace 40 and 9 with <math>a</math> and <math>b</math> for the moment. This gives us: <math>a^4-b^4 = (a^2)^2-(b^2)^2</math>.
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| − | | |
| − | Invoking the difference of squares identity two separate times, we get: <math>a^4-b^4= (a^2+b^2)(a^2-b^2)=(a^2+b^2)(a+b)(a-b)</math>.
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| − | | |
| − | | |
| − | Substituting 40 and 9 back in, we have: <math>(40^2+9^2)(40+9)(40-9)</math>. The first term of this is easy to simplify if we remember the Pythagorean triple (9,40,41). We get: <math>(41^2)(49)(31)=7^2\cdot 31 \cdot 41^2</math>.
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| − | | |
| − | Now that we have the prime factorization of the given expression, we see it is divisible by the primes 7, 31, and 41, so the answer is <math>\boxed{3}</math>.
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| − | | |
| − | == Problem 9 ==
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| − | Twenty-five blue and twenty-five yellow marbles are placed in a jar. Thirty-four marbles are removed from the jar and put in a second jar. What is the positive difference between the number of blue marbles in the second jar and the number of yellow marbles remaining in the first jar?
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| − | | |
| − | == Solution ==
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| − | The first jar initially contains <math>25 + 25 = 50</math> marbles. After 34 marbles are moved to the second jar, the first jar has <math>50 - 34 = 16</math> marbles remaining.
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| − | | |
| − | Let <math>b</math> be the number of blue marbles in the first jar after the marble transfer. Then the number of yellow marbles in the first jar is <math>16 - b</math>.
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| − | | |
| − | There are 25 blue marbles combined between the jars, so the second jar contains <math>25 - b</math> blue marbles. Similarly, there are a total of 25 yellow marbles, so the second jar contains <math>25 - (16 - b) = b + 9</math> yellow marbles. Note that <math>(25 - b) + (b + 9) = 34</math>.
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| − | Therefore, the positive difference between the number of blue marbles in the second jar and the number of yellow marbles in the first jar is <math>(25 - b) - (16 - b) = \boxed{9}</math>.
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| − | | |
| − | == Problem 10 ==
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| − | Among all fractions <math>x</math> that have a positive integer numerator and denominator and satisfy
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| − | <cmath>\frac{9}{11} \le x \le \frac{11}{13},</cmath>
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| − | which fraction has the smallest denominator?
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| − | | |
| − | == Solution ==
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| − | Suppose the fraction <math>a/b</math> satisfies
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| − | <cmath>\frac{9}{11} \le \frac{a}{b} \le \frac{11}{13}.</cmath>
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| − | Then
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| − | <cmath>\frac{9b}{11} \le a \le \frac{11b}{13}.</cmath>
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| − | This inequality tells us that the positive integer <math>a</math> must lie in a certain interval.
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| − | | |
| − | We want to minimize <math>b</math>, so we test values of <math>b</math> starting with 1. For <math>b = 1</math>, the inequality above becomes
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| − | <cmath>\frac{9}{11} \le a \le \frac{11}{13}.</cmath>
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| − | No positive integer <math>a</math> satisfies this inequality since <math>\dfrac{11}{13}<1</math>.
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| − | | |
| − | For <math>b = 2</math>, the inequality above becomes
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| − | <cmath>\frac{18}{11} \le a \le \frac{22}{13}.</cmath>
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| − | No positive integer <math>a</math> satisfies this inequality because the inequality forces <math>1<a<2</math>.
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| − | | |
| − | We keep going, until we hit <math>b = 6</math>:
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| − | <cmath>\frac{54}{11} \le a \le \frac{66}{13}.</cmath>
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| − | The positive integer <math>a = 5</math> satisfies this inequality. Therefore, the fraction <math>x</math> that has the smallest denominator is <math>a/b = \boxed{5/6}</math>.
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| − | | |
| − | == Problem 11 ==
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| − | Evaluate: <math>\dfrac{(x+y)^2 - (x-y)^2}{y}</math> for <math>x=6</math>, <math>y \not= 0</math>.
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| − | | |
| − | == Solution ==
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| − | To evaluate the expression <math>(x + y)^2 - (x - y)^2</math>, we can expand it. We can also recognize that it is a difference of squares:
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| − | <cmath>(x + y)^2 - (x - y)^2 = [(x + y) + (x - y)][(x + y) - (x - y)] = (2x)(2y) = 4xy.</cmath>
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| − | | |
| − | | |
| − | Thus,
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| − | <cmath>\frac{(x + y)^2 - (x - y)^2}{y} = \frac{4xy}{y} = 4x = \boxed{24}.</cmath>
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| − | | |
| − | == Problem 12 ==
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| − | Some perfect squares (such as 121) have a digit sum <math>(1 + 2 + 1 = 4)</math> that is equal to the square of the digit sum of their square root <math>(\sqrt{121}=11</math>, and <math>(1 + 1)^2 = 4)</math>.
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| − | | |
| − | What is the smallest perfect square greater than 100 that does not have this property?
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| − | | |
| − | == Solution ==
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| − | We can start with <math>12^2</math> and work our way up.
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| − | | |
| − | We have that <math>12^2 = 144</math>, and <math>1 + 4 + 4 = 9 = (1+2)^2</math>.
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| − | | |
| − | Similarly, for <math>13^2 = 169</math>, we have <math>1 + 6 + 9 = 16 = (1 + 3)^2</math>.
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| − | | |
| − | However, for <math>14^2 = 196</math>, we have <math>1 + 9 + 6 = 16</math>, which is not equal to <math>(1+4)^2 = 25</math>. Thus, <math>\boxed{196}</math> is the smallest such square greater than 100.
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| − | | |
| − | (It is not a coincidence that the long multiplications for <math>11\times 11</math>, <math>12\times 12</math>, and <math>13\times 13</math> don't require carrying, but <math>14\times 14</math> does.)
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| − | | |
| − | == Problem 13 ==
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| − | In the arrangement below, each number is the non-negative difference of the two numbers above it. What is the sum of the four greatest distinct possible values for <math>z</math>?
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| − | | |
| − | <asy>
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| − | unitsize(1 cm);
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| − | | |
| − | label("$40$", (0,0));
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| − | label("$4$", (1,0));
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| − | label("$18$", (2,0));
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| − | label("$48$", (3,0));
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| − | label("$40$", (4,0));
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| − | label("$76$", (5,0));
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| − | label("$z$", (6,0));
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| − | label("$36$", (0.5,-0.5));
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| − | label("$14$", (1.5,-0.5));
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| − | label("$30$", (2.5,-0.5));
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| − | label("$8$", (3.5,-0.5));
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| − | label("$36$", (4.5,-0.5));
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| − | label("$y$", (5.5,-0.5));
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| − | label("$22$", (1,-1));
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| − | label("$16$", (2,-1));
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| − | label("$22$", (3,-1));
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| − | label("$28$", (4,-1));
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| − | label("$x$", (5,-1));
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| − | label("$6$", (1.5,-1.5));
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| − | label("$6$", (2.5,-1.5));
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| − | label("$6$", (3.5,-1.5));
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| − | label("$w$", (4.5,-1.5));
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| − | label("$0$", (2,-2));
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| − | label("$0$", (3,-2));
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| − | label("$0$", (4,-2));
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| − | </asy>
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| − | | |
| − | == Solution ==
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| − | We start with the bottom of the table, and work our way up. The difference between <math>w</math> and 6 is 0, so <math>w</math> must be equal to 6.
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| − | | |
| − | Then the difference between <math>x</math> and 28 is 6, so <math>x</math> could be either <math>28 + 6 = 34</math> or <math>28 - 6 = 22</math>.
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| − | | |
| − | The difference between <math>y</math> and 36 is <math>x</math>. If <math>x = 34</math>, then <math>y</math> could be either <math>36 + 34 = 70</math> or <math>36 - 34 = 2</math>. Similarly, if <math>x = 22</math>, then <math>y</math> could be either <math>36 + 22 = 58</math> or <math>36 - 22 = 14</math>. This gives us four possible values of <math>y</math>.
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| − | The difference between <math>z</math> and 76 is <math>y</math>. For each possible value of <math>y</math>, <math>z</math> could be either <math>76 + y</math> or <math>76 - y</math>. Going through the possible values of <math>y</math>, we see that the possible values of <math>z</math> are 146, 6, 78, 74, 134, 18, 90, and 62. The largest four possible values are 146, 134, 90, and 78, and their sum is <math>146 + 134 + 90 + 78 = \boxed{448}</math>.
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| − | Alternative solution:
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| − | We have <math>w=6\pm 0,~x=28\pm w,~y=36\pm x,</math> and <math>z=76\pm y</math>. Through a series of substitutions, we obtain
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| − | <cmath>z = 76\pm 36\pm 28\pm 6,</cmath>which gives eight possible values for <math>z</math> (since each sign "<math>\pm</math>" can independently be chosen in two different ways). Since <math>36</math> by itself is larger than <math>28+6</math>, the four largest values of <math>z</math> are those with <math>+36</math>, regardless of the signs of <math>28</math> and <math>6</math>. That is, the four largest values of <math>z</math> are
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| − | \begin{align*}
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| − | &76+36+28+6, \\
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| − | &76+36+28-6, \\
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| − | &76+36-28+6, \\
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| − | &76+36-28-6.
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| − | \end{align*}Summing, we get <math>4(76)+4(36) = \boxed{448}</math> (note that the <math>\pm28\pm6</math> terms just cancel out).
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| − | == Problem 14 ==
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| − | A collection of nickels, dimes and pennies has an average value of 7 cents per coin. If a nickel were replaced by five pennies, the average would drop to 6 cents per coin. What is the number of dimes in the collection?
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| − | == Solution ==
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| − | Let <math>c</math> be the number of coins and <math>v</math> the total value of the coins in cents. Our first average fact states that <math>\dfrac{v}{c}=7</math>, or <math>v=7c</math>. Once we do the exchange, we have <math>4</math> more coins and the same total value, so <math>\dfrac{v}{c+4}=6</math>, or <math>v=6(c+4)</math>.
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| − | Setting these equations equal, we have <math>7c=6(c+4)</math>, so <math>c=24</math>. Plugging this into either equation above shows <math>v=168</math> cents.
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| − | Now, let <math>p</math> be the number of pennies, <math>n</math> the number of nickels, and <math>d</math> the number of dimes. We can write the equations
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| − | \begin{align*}
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| − | d + n + p &= 24, \\
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| − | 10d + 5n + p &= 168.
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| − | \end{align*}If we multiply the first equation by 10, and then subtract the second equation, we get <math>5n + 9p = 72</math>. Solving for <math>n</math>, we find <math>n = \dfrac{72 - 9p}{5}=\dfrac{9(8-p)}{5}</math>.
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| − | From the factorization, we see that the only <math>p</math> value which makes <math>n</math> a positive integer is <math>p=3</math>. This <math>p</math> value forces <math>n=9</math>, and thus <math>d = 24 - n - p = 24 - 3 - 9 = \boxed{12}</math>.
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