Difference between revisions of "User:Aoum/Sandbox"
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<span style="font-family:aops">…</span> | <span style="font-family:aops">…</span> | ||
<!-- | <!-- | ||
| + | = Week 1 = | ||
| + | |||
== Problem 1 == | == Problem 1 == | ||
Evaluate <math>5^2 - (10)(6) + 6^2</math>. | Evaluate <math>5^2 - (10)(6) + 6^2</math>. | ||
| − | == Solution == | + | === Solution === |
It's not hard to expand the expression, but we can also recognize that the given expression is an instance of the formula | It's not hard to expand the expression, but we can also recognize that the given expression is an instance of the formula | ||
<cmath>(a - b)^2 = a^2 - 2ab + b^2.</cmath> | <cmath>(a - b)^2 = a^2 - 2ab + b^2.</cmath> | ||
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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? | 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? | ||
| − | == Solution == | + | === Solution === |
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> | 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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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> | 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> | ||
| − | == Solution == | + | === Solution === |
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> | 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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Evaluate: <math>35^4 - 25^4</math>. | Evaluate: <math>35^4 - 25^4</math>. | ||
| − | == Solution == | + | === Solution === |
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>. | 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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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? | 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? | ||
| − | == Solution == | + | === Solution === |
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. | 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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If the integer <math>152AB1</math> is a perfect square, what is the sum of the digits of its square root? | If the integer <math>152AB1</math> is a perfect square, what is the sum of the digits of its square root? | ||
| − | == Solution == | + | === Solution === |
There are two things that help us here: | There are two things that help us here: | ||
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What is the largest four-digit number that is equal to the cube of the sum of its digits? | What is the largest four-digit number that is equal to the cube of the sum of its digits? | ||
| − | == Solution == | + | === Solution === |
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. | 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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How many different prime numbers divide evenly into <math>40^4-9^4</math>? | How many different prime numbers divide evenly into <math>40^4-9^4</math>? | ||
| − | == Solution == | + | === Solution === |
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>. | 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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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? | 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? | ||
| − | == Solution == | + | === Solution === |
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. | 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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which fraction has the smallest denominator? | which fraction has the smallest denominator? | ||
| − | == Solution == | + | === Solution === |
Suppose the fraction <math>a/b</math> satisfies | Suppose the fraction <math>a/b</math> satisfies | ||
<cmath>\frac{9}{11} \le \frac{a}{b} \le \frac{11}{13}.</cmath> | <cmath>\frac{9}{11} \le \frac{a}{b} \le \frac{11}{13}.</cmath> | ||
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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>. | Evaluate: <math>\dfrac{(x+y)^2 - (x-y)^2}{y}</math> for <math>x=6</math>, <math>y \not= 0</math>. | ||
| − | == Solution == | + | === Solution === |
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: | 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: | ||
<cmath>(x + y)^2 - (x - y)^2 = [(x + y) + (x - y)][(x + y) - (x - y)] = (2x)(2y) = 4xy.</cmath> | <cmath>(x + y)^2 - (x - y)^2 = [(x + y) + (x - y)][(x + y) - (x - y)] = (2x)(2y) = 4xy.</cmath> | ||
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What is the smallest perfect square greater than 100 that does not have this property? | What is the smallest perfect square greater than 100 that does not have this property? | ||
| − | == Solution == | + | === Solution === |
We can start with <math>12^2</math> and work our way up. | We can start with <math>12^2</math> and work our way up. | ||
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unitsize(1 cm); | unitsize(1 cm); | ||
| − | label("$40$", (0,0)); | + | // Labels for the values on the x-axis |
| − | label("$4$", (1,0)); | + | label("$40$", (0,0), S); |
| − | label("$18$", (2,0)); | + | label("$4$", (1,0), S); |
| − | label("$48$", (3,0)); | + | label("$18$", (2,0), S); |
| − | label("$40$", (4,0)); | + | label("$48$", (3,0), S); |
| − | label("$76$", (5,0)); | + | label("$40$", (4,0), S); |
| − | label("$z$", (6,0)); | + | label("$76$", (5,0), S); |
| − | label("$36$", (0.5,-0.5)); | + | label("$z$", (6,0), S); |
| − | label("$14$", (1.5,-0.5)); | + | |
| − | label("$30$", (2.5,-0.5)); | + | // Horizontal distances from the axis |
| − | label("$8$", (3.5,-0.5)); | + | label("$36$", (0.5,-0.5), S); |
| − | label("$36$", (4.5,-0.5)); | + | label("$14$", (1.5,-0.5), S); |
| − | label("$y$", (5.5,-0.5)); | + | label("$30$", (2.5,-0.5), S); |
| − | label("$22$", (1,-1)); | + | label("$8$", (3.5,-0.5), S); |
| − | label("$16$", (2,-1)); | + | label("$36$", (4.5,-0.5), S); |
| − | label("$22$", (3,-1)); | + | label("$y$", (5.5,-0.5), S); |
| − | label("$28$", (4,-1)); | + | |
| − | label("$x$", (5,-1)); | + | // Labels for vertical distances |
| − | label("$6$", (1.5,-1.5)); | + | label("$22$", (1,-1), S); |
| − | label("$6$", (2.5,-1.5)); | + | label("$16$", (2,-1), S); |
| − | label("$6$", (3.5,-1.5)); | + | label("$22$", (3,-1), S); |
| − | label("$w$", (4.5,-1.5)); | + | label("$28$", (4,-1), S); |
| − | label("$0$", (2,-2)); | + | label("$x$", (5,-1), S); |
| − | label("$0$", (3,-2)); | + | |
| − | label("$0$", (4,-2)); | + | // Additional horizontal distances below the x-axis |
| + | label("$6$", (1.5,-1.5), S); | ||
| + | label("$6$", (2.5,-1.5), S); | ||
| + | label("$6$", (3.5,-1.5), S); | ||
| + | label("$w$", (4.5,-1.5), S); | ||
| + | |||
| + | // Zero markers (below the axis) | ||
| + | label("$0$", (2,-2), S); | ||
| + | label("$0$", (3,-2), S); | ||
| + | label("$0$", (4,-2), S); | ||
</asy> | </asy> | ||
| − | == Solution == | + | === Solution === |
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. | 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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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? | 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? | ||
| − | == Solution == | + | === Solution === |
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>. | 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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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>. | 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>. | ||
| + | |||
| + | = Week 2 = | ||
| + | |||
| + | == Problem 1 == | ||
| + | What is the largest prime factor of 360? | ||
| + | |||
| + | === Solution === | ||
| + | The prime factorization of 360 is <math>2^3\cdot 3^2\cdot 5</math> , so the largest prime factor is <math>\boxed{5}</math>. | ||
| + | |||
| + | == Problem 2 == | ||
| + | What is the smallest prime whose digits sum to 19? | ||
| + | |||
| + | === Solution === | ||
| + | If the prime has at most two digits, then the sum of the digits is at most <math>9+9= 18</math>, so the prime must have at least three digits. The smallest three-digit number where the sum of the digits is 19 is <math>\boxed{199}</math>, which is prime, so this is the answer. | ||
| + | |||
| + | == Problem 3 == | ||
| + | What is the least common multiple (LCM) of 45, 60, and 75? | ||
| + | |||
| + | === Solution === | ||
| + | The prime factorizations of 45, 60, and 75 are | ||
| + | \begin{align*} | ||
| + | 45 &= 3^2\cdot 5,\\ | ||
| + | 60 &= 2^2\cdot 3\cdot 5,\\ | ||
| + | 75 &= 3\cdot 5^2. | ||
| + | \end{align*} | ||
| + | Taking the largest power of each prime factor, we find that the LCM is <math>2^2\cdot 3^2\cdot 5^2= \boxed{900}</math>. | ||
| + | |||
| + | === Problem 4 == | ||
| + | What is the smallest three-digit number divisible by the first three prime numbers and the first three composite numbers? | ||
| + | |||
| + | === Solution === | ||
| + | The first three primes are 2, 3, and 5, and the first three composite numbers are 4, 6, and 8. The LCM of these numbers is <math>2^3 \cdot 3 \cdot 5 = 120</math>, which is a three-digit number, so the answer is <math>\boxed{120}</math>. | ||
| + | |||
| + | == Problem 5 == | ||
| + | What is the greatest common divisor of 238 and 374? | ||
| + | |||
| + | === Solution === | ||
| + | Since these numbers are small, it is relatively easy to use prime factorization. The prime factorizations of 238 and 374 are <math>238=2\cdot 7\cdot 17</math> and <math>374=2\cdot 11\cdot 17</math>, so their GCD is <math>2\cdot 17= \boxed{34}</math>. | ||
| + | |||
--> | --> | ||
Revision as of 13:34, 28 June 2025
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