Math Competition

Sponsored link: Download Math Competition
On this page you can read or download Math Competition in PDF format. We also recommend you to learn related results, that can be interesting for you. If you didn't find any matches, try to search the book, using another keywords.
Solutions: Math League Contest #5, 2/24/2009 Achs Math Competition
solutions: math league contest #5, 2/24/2009 achs math competition
What is the least possible positive difference between two three-digit numbers which together use the digits 4, 5, 6, 7, 8, and 9? To minimize the difference d, the hundred’s digits of the two numbers should be consecutive integers. Consider the two-digit numbers obtained from the remaining (least significant two) digits of our three-digit numbers. d will be minimized if the two-digit number obtained from the smaller three-digit number is maximized, and if the two-digit number obtained from the larger three-.

Language: english
PDF pages: 32, PDF size: 0.2 MB
Report
Estonian Math Competitions 2010/2011
estonian math competitions 2010/2011
Answer: a) no; b) yes. Solutiona) Both the step that involves increasing two of the numbers by 3 and the step that involves decreasing one of the numbers by 6 result in the sum of all three numbers being changed byThus the remainder when the sum of the three numbers is divided by 6 will always be the same regardless of the number of steps taken. But as the sums 0 + 1 + 2 and 1 + 2 + 3 give different remainders when divided by 6, it is impossible to reach the required end situation from the given initial .

Language: english
PDF pages: 30, PDF size: 0.23 MB
Report
Estonian Math Competitions 2007/2008
estonian math competitions 2007/2008
Language: english
PDF pages: 26, PDF size: 0.25 MB
Report
Estonian Math Competitions 2008/2009
estonian math competitions 2008/2009
Language: english
PDF pages: 26, PDF size: 0.24 MB
Report
Estonian Math Competitions 2009/2010
estonian math competitions 2009/2010
SolutionWe use the same notation as in the SolutionTriangle BCD is isosceles, hence ∠ DCB = ∠ DBC =As D is the circumcenter of ABC, we have ∠ BDC = 2∠ BAC = 2β. The sizes of the angles of triangle BCD are therefore β, β, and 2β; thus β + β + 2β = 180 , whence β = 45 . As ∠ BCA = ∠BDA , we have ∠ BCD = α + α = β, whence α = 2 β = 30 . Consequently, 2 2 3 the sizes of the angles of quadrangle ABCD are ∠ DAB = ∠ ABD = 1802 −α = 75 , ∠ ABC = ∠ ABD + ∠CBD = 75 + 45 = 120 , ∠ BCD = β = 45 , and ∠CDA = ∠CDB + ∠.

Language: english
PDF pages: 24, PDF size: 0.19 MB
Report
1   2   3   4   5   6   7   8   9   10   Next page →
English ▼
Home Copyright Information Privacy Policy Contact us

PDFSB.NET | All Rights Reserved
This project is a PDF search engine and do not store, hold or retain any files.