p1. Given the set $H = \{(x, y)|(x -y)^2 + x^2 - 15x + 50 = 0$ where x and y are natural numbers $\}$. Find the number of subsets of $H$. p2. A magician claims to be an expert at guessing minds with following show. One of the viewers was initially asked to hidden write a five-digit number, then subtract it with the sum of the digits that make up the number, then name four of the five digits that make up the resulting number (in order of any). Then the magician can guess the numbers hidden. For example, if the audience mentions four numbers result: $0, 1, 2, 3$, then the magician will know that the hidden number is $3$. a. Give an example of your own from the above process. b. Explain mathematically the general form of the process. p3. In a fruit basket there are $20$ apples, $18$ oranges, $16$ mangoes, $10$ pineapples and $6$ papayas. If someone wants to take $10$ pieces from the basket. After that, how many possible compositions of fruit are drawn? p4. Inside the Equator Park, a pyramid-shaped building will be made with base of an equilateral triangle made of translucent material with a side length of the base $8\sqrt3$ m long and $8$ m high. A globe will be placed in a pyramid the. Ignoring the thickness of the pyramidal material, determine the greatest possible length of the radius of the globe that can be made. p5. What is the remainder of $2012^{2012} + 2014^{2012}$ divided by $2013^2$?
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Tags: algebra, number theory, combinatorics, geometry, indonesia juniors
04.11.2021 05:09
parmenides51 wrote: p5. What is the remainder of $2012^{2012} + 2014^{2012}$ divided by $2013^2$?
04.11.2021 05:11
I don't understand P1. What is $f$?
04.11.2021 05:24
parmenides51 wrote: p1. Given the set $H = \{f (x, y)|(x -y)^2 + x^2 - 15x + 50 = 0$ where x and y are natural numbers $\}$. Find the number of subsets of $H$.?
04.11.2021 15:40
parmenides51 wrote: p3. In a fruit basket there are $20$ apples, $18$ oranges, $16$ mangoes, $10$ pineapples and $6$ papayas. If someone wants to take $10$ pieces from the basket. After that, how many possible compositions of fruit are drawn?
04.11.2021 16:02
parmenides51 wrote: p3. In a fruit basket there are $20$ apples, $18$ oranges, $16$ mangoes, $10$ pineapples and $6$ papayas. If someone wants to take $10$ pieces from the basket. After that, how many possible compositions of fruit are drawn?
04.11.2021 16:12
BackToSchool wrote: parmenides51 wrote: p3. In a fruit basket there are $20$ apples, $18$ oranges, $16$ mangoes, $10$ pineapples and $6$ papayas. If someone wants to take $10$ pieces from the basket. After that, how many possible compositions of fruit are drawn?
04.11.2021 18:37
$P1$, $$y = x -\sqrt{-x^2 + 15 x - 50} $$or $$y = x + \sqrt{-x^2 + 15 x - 50} $$so, $-x^2 + 15 x - 50>=0$ must be... $5<=x<=10$ x={ 5,6,7,8,9,10} by quickly examining the elements of his set; We see that there are $6$ $(x,y)$ ordered pairs. Then, the number of subsets of $H$ is $2^6$=$64$.
23.07.2022 13:24
BackToSchool wrote: BackToSchool wrote: parmenides51 wrote: p3. In a fruit basket there are $20$ apples, $18$ oranges, $16$ mangoes, $10$ pineapples and $6$ papayas. If someone wants to take $10$ pieces from the basket. After that, how many possible compositions of fruit are drawn?
your bars and stars theorem is wrong, check it again. the correct answer is 966
23.07.2022 22:44
amogususususus wrote: BackToSchool wrote:
your bars and stars theorem is wrong, check it again. the correct answer is 966 Your solutions are welcome. Otherwise how you know the "correct answer" is correct.
25.07.2022 04:45
Because the number of apple, orange, mango, pineapple is greater than $10$, we just have to look at the possible number of papaya on the configuration. Let the number of apple, orange, mango, pineapple, papaya taken is non-negative integer $a,b,c,d,e$. Therefore, the number of possible configurations is the same as the number of solutions $(a,b,c,d,e)$ to $a+b+c+d+e=10$ It is known that $e\le6$, therefore $4\le(a+b+c+d)\le10$. Then, with bars and stars theorem. The number of possible solutions $(a,b,c,d) = {7 \choose 3} + {8 \choose 3} + {9 \choose 3} + {10 \choose 3} + {11 \choose 3} + {12 \choose 3} + {13\choose 3} = 966$ Because $a,b,c,d$ is bijective to $e$, the number of solution $(a,b,c,d,e)$ is $966$.
25.07.2022 04:49
BackToSchool wrote: amogususususus wrote: BackToSchool wrote:
your bars and stars theorem is wrong, check it again. the correct answer is 966 Your solutions are welcome. Otherwise how you know the "correct answer" is correct. Yes, indeed my answer is correct because i already checked the official solution.