New Year Problem (6)

Algebra Level 2

What is the value of the following expression? ∑ n = 0 2020 ⌊ n 10 ⌋ \sum_{n=0}^{2020} \left\lfloor{\frac{\sqrt{n}}{10}}\right\rfloor


The answer is 5084.

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Chew-Seong Cheong
Dec 27, 2019

∑ n = 0 2020 ⌊ n 10 ⌋ = ∑ n = 0 99 0 + ∑ n = 100 399 1 + ∑ n = 400 899 2 + ∑ n = 900 1599 3 + ∑ n = 1600 2020 4 = 0 + ( 400 − 100 ) + 2 ( 900 − 400 ) + 3 ( 1600 − 900 ) + 4 ( 2021 − 1600 ) = 0 + 300 + 1000 + 2100 + 1684 = 5084 \begin{aligned} \sum_{n=0}^{2020} \left \lfloor \frac {\sqrt n}{10} \right \rfloor & = \sum_{n=0}^{99} 0 + \sum_{n=100}^{399} 1 + \sum_{n=400}^{899} 2 + \sum_{n=900}^{1599} 3 + \sum_{n=1600}^{2020} 4 \\ & = 0 + (400-100) + 2(900-400) + 3(1600-900) + 4(2021-1600) \\ &= 0 + 300 + 1000 + 2100 + 1684 \\ & = \boxed{5084} \end{aligned}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

Thank you for sharing a nice solution. This problem is original and l was a bit concerned about the solution.

Hana Wehbi - 1 year, 5 months ago

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A day before yesterday when i solved this question i got the answer to be 5084. But your's solution was 5048. And i was totally confused at that moment.

Syed Shahabudeen - 1 year, 5 months ago

Hana, since you like to set problems, you should learn up Wolfram Alpha app, which is available online free of charge. It allow you to check the solution. The solution for this problem is here .

Chew-Seong Cheong - 1 year, 5 months ago

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Thank you, l will.

Hana Wehbi - 1 year, 5 months ago

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