A trip along Route 66

Algebra Level 5

{ ∑ k = 1 66 a k = 2015 ∑ k = 1 66 a k 2 = 62465 \large \begin{cases} {\displaystyle\sum_{k=1}^{66} a_{k} = 2015} \\ {\displaystyle\sum_{k=1}^{66} a_{k}^{2} = 62465} \end{cases}

Suppose real numbers a 1 , a 2 , a 3 , … , a 66 a_1, a_2, a_3, \ldots,a_{66} satisfy the two equations above with S = max ⁡ ( a 1 , a 2 , a 3 , … , a 66 ) S = \max(a_1, a_2, a_3, \ldots, a_{66}) .

If S = m n , S = \dfrac{m}{n}, where m m and n n are positive coprime integers, then find m + n . m + n.


The answer is 2048.

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Brian Charlesworth by Brian Charlesworth , 55, Canada

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1 solution

Without loss of generality we will isolate a 1 a_{1} and determine its maximum possible value. Then by Cauchy's Inequality , with the b k b_{k} 's all equalling 1 , 1, we have that

( ∑ k = 2 66 a k 2 ) ( ∑ k = 2 66 1 ) ≥ ( ∑ k = 2 66 a k ) 2 \displaystyle\left(\sum_{k=2}^{66} a_{k}^{2} \right) \left(\sum_{k=2}^{66} 1 \right) \ge \left( \sum_{k=2}^{66} a_{k} \right)^{2}

⟹ 65 × ( 62465 − a 1 2 ) ≥ ( 2015 − a 1 ) 2 \Longrightarrow 65 \times (62465 - a_{1}^{2}) \ge (2015 - a_{1})^{2}

⟹ 65 × 62465 − 201 5 2 ≥ 66 a 1 2 − 4030 a 1 \Longrightarrow 65 \times 62465 - 2015^{2} \ge 66a_{1}^{2} - 4030a_{1}

⟹ 0 ≥ 2 a 1 ( 33 a 1 − 2015 ) ⟹ 0 ≤ a 1 ≤ 2015 33 . \Longrightarrow 0 \ge 2a_{1}(33a_{1} - 2015) \Longrightarrow 0 \le a_{1} \le \dfrac{2015}{33}.

The maximum value of a 1 = 2015 33 a_{1} = \dfrac{2015}{33} can be realized by setting all other a k a_{k} 's to 992 33 . \dfrac{992}{33}.

Thus S = 2015 33 , S = \dfrac{2015}{33}, and so m + n = 2015 + 33 = 2048 . m + n = 2015 + 33 = \boxed{2048}.

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Moderator note:

Right, one should be able to identify that Cauchy Schwarz Inequality is applicable when we are all given the sum of squares of all terms.

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