Question-159152

Question Number 159152 by HongKing last updated on 13/Nov/21

Answered by mr W last updated on 13/Nov/21

l e t x = n + f w i t h n \in Z, 0 ⩽ f < 1

R H S = n + 2022 = L H S ⩾ 2021 n

\Rightarrow n ⩽ \frac{2022}{2020} \Rightarrow n ⩽ 1 \dots (i)

R H S = n + 2022 = L H S < 2021 (n + 1)

\Rightarrow n > \frac{1}{2020} \Rightarrow n ⩾ 1 \dots (i i)

\Rightarrow n = 1

\Rightarrow x = 1 + f

1 + [f] + 1 + [f + \frac{1}{2021}] + 1 + [f + \frac{2}{2021}] + \dots + 1 + [f + \frac{2020}{2021}] = 1 + 2022

2021 + [f] + [f + \frac{1}{2021}] + [f + \frac{2}{2021}] + \dots + [f + \frac{2020}{2021}] = 1 + 2022

[f] + [f + \frac{1}{2021}] + [f + \frac{2}{2021}] + \dots + [f + \frac{2018}{2021}] + [f + \frac{2019}{2021}] + [f + \frac{2020}{2021}] = 2

e a c h o f t h e l a s t t w o t e r m s s h o u l d b e 1

a n d a l l o t h e r t e r m s b e f o r e t h e m

s h o u l d b e z e r o .

1 ⩽ f + \frac{2019}{2021} \land f + \frac{2018}{2021} < 1 \Rightarrow \frac{2}{2021} ⩽ f < \frac{3}{2021}

\Rightarrow s o l u t i o n i s 1 + \frac{2}{2021} ⩽ x < 1 + \frac{3}{2021}

Commented by HongKing last updated on 13/Nov/21

Very nice solution thank you so much

my dear Ser

Commented by Tawa11 last updated on 14/Nov/21

Great sir .

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