Is it average?

Algebra Level 5

1 2 + 1 2 + 1 100 + 1 2 + 2 100 + 1 2 + 3 100 + + 1 2 + 299 100 = ? \large{\left \lfloor \dfrac{1}{2} \right \rfloor + \left \lceil \dfrac{1}{2} + \dfrac {1}{100} \right \rceil + \left \lfloor \dfrac{1}{2} + \dfrac {2}{100} \right \rfloor + \left \lceil \dfrac{1}{2} + \dfrac {3}{100} \right \rceil + \ldots + \left \lceil \dfrac {1}{2} + \dfrac {299}{100} \right \rceil = \ ?}


Inspired from this and this .


The answer is 600.

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1 solution

Chew-Seong Cheong
Sep 28, 2015

Let S S be the sum of the expression, then we have:

S = 1 2 + 1 2 + 1 100 + 1 2 + 2 100 + 1 2 + 3 100 + . . . + 1 2 + 298 100 + 1 2 + 299 100 = k = 0 149 1 2 + 2 k 100 + k = 0 149 1 2 + 2 k + 1 100 = k = 0 24 1 2 + 2 k 100 + k = 25 74 1 2 + 2 k 100 + k = 75 124 1 2 + 2 k 100 + k = 125 149 1 2 + 2 k 100 + k = 0 24 1 2 + 2 k + 1 100 + k = 25 74 1 2 + 2 k + 1 100 + k = 75 124 1 2 + 2 k + 1 100 + k = 125 149 1 2 + 2 k + 1 100 = 25 ( 0 ) + 50 ( 1 ) + 50 ( 2 ) + 25 ( 3 ) + 25 ( 1 ) + 50 ( 2 ) + 50 ( 3 ) + 25 ( 4 ) = 0 + 50 + 100 + 75 + 25 + 100 + 150 + 100 = 600 \begin{aligned} S & = \left \lfloor \frac{1}{2} \right \rfloor + \left \lceil \frac{1}{2} + \frac{1}{100} \right \rceil + \left \lfloor \frac{1}{2} + \frac{2}{100} \right \rfloor + \left \lceil \frac{1}{2} + \frac{3}{100} \right \rceil + ... + \left \lfloor \frac{1}{2} + \frac{298}{100} \right \rfloor + \left \lceil \frac{1}{2} + \frac{299}{100} \right \rceil \\ & = \sum_{k=0}^{149} \left \lfloor \frac{1}{2} + \frac{2k}{100} \right \rfloor + \sum_{k=0}^{149} \left \lceil \frac{1}{2} + \frac{2k+1}{100} \right \rceil \\ & = \sum_{k=0}^{24} \left \lfloor \frac{1}{2} + \frac{2k}{100} \right \rfloor + \sum_{k=25}^{74} \left \lfloor \frac{1}{2} + \frac{2k}{100} \right \rfloor + \sum_{k=75}^{124} \left \lfloor \frac{1}{2} + \frac{2k}{100} \right \rfloor + \sum_{k=125}^{149} \left \lfloor \frac{1}{2} + \frac{2k}{100} \right \rfloor \\ & \quad \quad + \sum_{k=0}^{24} \left \lceil \frac{1}{2} + \frac{2k+1}{100} \right \rceil + \sum_{k=25}^{74} \left \lceil \frac{1}{2} + \frac{2k+1}{100} \right \rceil + \sum_{k=75}^{124} \left \lceil \frac{1}{2} + \frac{2k+1}{100} \right \rceil + \sum_{k=125}^{149} \left \lceil \frac{1}{2} + \frac{2k+1}{100} \right \rceil \\ & = 25(0) + 50(1) + 50(2) + 25(3) + 25(1) + 50(2) + 50(3) + 25(4) \\ & = 0 + 50 + 100 + 75 + 25 + 100 + 150 + 100 \\ & = \boxed{600} \end{aligned}

Same Way Sir, and superb question @Sharky Kesa , nice mixture of both questions. Interesting thing that all 3 questions became level 4.

Kushagra Sahni - 5 years, 8 months ago

I factorized into 1/2(300) + 1/100(1+2+3...+299)

Why is my factorization wrong ?

Kevin Hernández - 5 years, 8 months ago

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Each one of the terms in the summation is an integer not a fraction because of the greatest integer function or floor (hooks at the bottom) function x \lfloor x \rfloor and the least integer function or ceiling (hooks on top) function x \lceil x \rceil . They are not brackets. 1.7 = 1 \lfloor 1.7 \rfloor = 1 , the greatest integer smaller than 1.7 1.7 or the integer part of 1.7 1.7 and 1.2 = 2 \lceil 1.2 \rceil = 2 , the least integer larger than 1.2 1.2 .

Chew-Seong Cheong - 5 years, 8 months ago

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