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CEMC Cayley

Algebra Difficulty 2.4 Junior Find the answer

In the star shown, the sum of the four integers along each straight line is to be the same. Five numbers have been entered. The five missing numbers are 19, 21, 23, 25, and 27. Which number is represented by q q ?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Suppose that the sum of the four integers along each straight line equals S S . Then S=9+p+q+7=3+p+u+15=3+q+r+11=9+u+s+11=15+s+r+7 S=9+p+q+7=3+p+u+15=3+q+r+11=9+u+s+11=15+s+r+7 . Thus, 5S=(9+p+q+7)+(3+p+u+15)+(3+q+r+11)+(9+u+s+11)+(15+s+r+7)=2p+2q+2r+2s+2u+90 5S = (9+p+q+7)+(3+p+u+15)+(3+q+r+11)+(9+u+s+11)+(15+s+r+7) = 2p+2q+2r+2s+2u+90 . Since p,q,r,s, p, q, r, s, and u u are the numbers 19, 21, 23, 25, and 27 in some order, then p+q+r+s+u=19+21+23+25+27=115 p+q+r+s+u=19+21+23+25+27=115 and so 5S=2(115)+90=320 5S=2(115)+90=320 or S=64 S=64 . Since S=64 S=64 , then 3+p+u+15=64 3+p+u+15=64 or p+u=46 p+u=46 . Since S=64 S=64 , then 15+s+r+7=64 15+s+r+7=64 or s+r=42 s+r=42 . Therefore, q=(p+q+r+s+u)−(p+u)−(s+r)=115−46−42=27 q=(p+q+r+s+u)-(p+u)-(s+r)=115-46-42=27 .

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