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Number Theory Level 4

a 2 + a 3 + ⋯ + a 99 − a 1 + a 100 \large \sqrt{a_{2}+a_{3}+\cdots+a_{99}}-\sqrt{a_{1}+a_{100}}

Let a 1 , a 2 , a 3 , … , a 100 a_{1},a_{2},a_{3},\ldots,a_{100} is an sequence of consecutive positive integers.

Find the minimum integer value of the expression above. of \ which is an integer.

From Thai High School Maths Exam 2013


The answer is 66.

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Potsawee Manakul by Potsawee Manakul , 25, United Kingdom

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1 solution

Potsawee Manakul
Aug 7, 2015

Let a n = a + n a_{n}=a+n

a 1 + a 100 = 2 a + 101 \sqrt{a_{1}+a_{100}} = \sqrt{2a+101} a 2 + a 3 + . . . + a 99 = 98 a + 4949 = 7 2 a + 101 \sqrt{a_{2}+a_{3}+...+a_{99}} = \sqrt{98a+4949} = 7\sqrt{2a+101} ∴ a 2 + a 3 + . . . + a 99 − a 1 + a 100 = 6 2 a + 101 \therefore \sqrt{a_{2}+a_{3}+...+a_{99}}-\sqrt{a_{1}+a_{100}} = 6\sqrt{2a+101}

It is the minimum integer when 2 a + 101 = 121 2a+101=121 because a ≥ 0 a \geq 0

Thus, the minimum is 6 ⋅ 11 = 66 6\cdot11=66

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