# Math Competition

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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-.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 .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 + ∠.
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