Algebra

Number Theory Level 4

Let a + b + c + d + e = 20 a+b+c+d+e = 20 , where a a , b b , c c , d d and e e are positive integers. Find the maximum value of a b + a d + b c + b e + c d + d e ab+ad+bc+be+cd+de .


The answer is 100.

Ferdinand Halim Santoso by Ferdinand Halim Santoso , 18, Indonesia

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1 solution

Otto Bretscher
May 31, 2016

Let x = a + c + e x=a+c+e and y = b + d y=b+d . Given that a + b + c + d + e = x + y = 20 a+b+c+d+e=x+y=20 , we have a b + a d + b c + b e + c d + d e ab+ad+bc+be+cd+de = x y ≤ ( x + y ) 2 4 = 100 =xy\leq \frac{(x+y)^2}{4}=\boxed{100} . The maximum is attained when b = d = 5 , a = c = 4 , e = 2 b=d=5,a=c=4,e=2 , for example.

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