Primarygun Posted October 28, 2004 Share Posted October 28, 2004 Two questions. They may be very simple and easy for you, but a brain flash. 1.the square root of 20 consecutive positive intergers have the same intergral number, find that number. 2.(a-b)(c-d)/[(b-c)(d-a)=3 Find (a-c)(b-d)/[(a-b)(c-d)] Link to comment Share on other sites More sharing options...
Primarygun Posted October 29, 2004 Author Share Posted October 29, 2004 Are the questions too vague or else?........... Link to comment Share on other sites More sharing options...
TWJian Posted October 29, 2004 Share Posted October 29, 2004 Forgive me for posting another question,but i don't know how to post a new thread. Ok,can any1 help me to solve this equation:square root of negative 4.BTW,can can any1 teach me how to post my own thread? Link to comment Share on other sites More sharing options...
Primarygun Posted October 29, 2004 Author Share Posted October 29, 2004 Square root of negative 4 is not a real number. First, go to topic area , like Math Problems, then press start a new thread on the left side. Link to comment Share on other sites More sharing options...
TWJian Posted October 30, 2004 Share Posted October 30, 2004 Thx. Link to comment Share on other sites More sharing options...
Severian Posted November 12, 2004 Share Posted November 12, 2004 Two questions.They may be very simple and easy for you' date=' but a brain flash. 1.the square root of 20 consecutive positive intergers have the same intergral number, find that number. 2.(a-b)(c-d)/[(b-c)(d-a)=3 Find (a-c)(b-d)/[(a-b)(c-d)'] I think our problem is that the questions don't have answers. 1. If the number you want is x, then you are inisiting that [math]x=\sqrt{a}=\sqrt{a+1}=\sqrt{a+2}=...[/math] where a is some integer. Clearly this is impossible. Unless I am misunderstanding the question. 2. the eqaution you give gives a relation between a, b, c and d, but doesn't tell you their values. So the object you want to find is not known. Link to comment Share on other sites More sharing options...
matt grime Posted November 13, 2004 Share Posted November 13, 2004 I imagine the first question means "have the same integral part", not "are the same integer", and there are an infinite number of answers, indeed any number greater than 10 is a solution. 2. perhaps can be solved by rearrangement (eg, severian, jsut because i don't know a or b doesn't mean I can't write down a/b given that 3a=4b). Link to comment Share on other sites More sharing options...
Primarygun Posted November 13, 2004 Author Share Posted November 13, 2004 sorry. Indeed , the first question I asked is very vague. Thanks matt grime for clarifying the question again. Ya. The second question has solution by arrangement. The second question is taken from maths olympics. It is designed for students who are aged below 17. The first is for below 12. Link to comment Share on other sites More sharing options...
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