There-exists-a-unique-positive-integer-a-for-which-The-sum-u-n-1-2023-n-2-na-5-is-an-integer-trictly-between-1000-amp-1000-find-a-u-

Question Number 193585 by York12 last updated on 16/Jun/23

T h e r e e x i s t s a u n i q u e p o s i t i v e i n t e g e r a f o r

w h i c h T h e s u m u = \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - n a}{5} ⌋ i s a n i n t e g e r

t r i c t l y b e t w e e n - 1000 & 1000 f i n d a + u .

Answered by York12 last updated on 16/Jun/23

u = \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - n a}{5} ⌋, ⌊ \frac{n^{2} - n a}{5} ⌋ = \frac{n^{2} - n a}{5} - {\frac{n^{2} - n a}{5}}

w h e r e {⊛} i n d i c a t e s t h e f r a c t i o n a l

p a r t f u n c t i o n

0 ⩽ {\frac{n^{2} - n a}{5}} < 1 \Rightarrow 0 ⩽ \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - n a}{5}} < 2023

- 1000 < \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - n a}{5} ⌋ < 1000 \Rightarrow 0 < \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - n a}{5}) - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - n a}{5}} < 1000

\Rightarrow - 3023 < \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - n a}{5}) < 1000

\Rightarrow - 3023 < \frac{\underset{n = 1}{\sum^{2023}} (n^{2})}{5} - \frac{a \underset{n = 1}{\sum^{2023}} (n)}{5} < 1000

\Rightarrow 5 \times (3023 + \frac{\underset{n = 1}{\sum^{2023}} (n^{2})}{5}) > a > 5 \times (\frac{\underset{n = 1}{\sum^{2023}} (n)}{5} - 1000)

\Rightarrow 1349.007 > a > 1348.998 \Rightarrow a = 1349

∴ u = \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - 1349 n}{5} ⌋ = \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - 1349 n}{5}) - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}}

\underset{n = 1}{\sum^{2023}} (\frac{n^{2} - 1349 n}{5}) = 0 \Rightarrow u = - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}}

s o n o w w e a r e o n l y i n t r e s t e d i n t h e v a l u e

o f t h e R e m a i n d e r o f t h e e x p r e s s i o n

\frac{n^{2} - 1349 n}{5}, o n e o f t h e s p e c i a l p r o p e r t i e s

o f m o d (5) & m o d (10) t h a t

a_{n} a_{n - 1} a_{n - 2} a_{n - 2} \dots a_{1} a_{0} \equiv a_{0} m o d (5)

a_{n} a_{n - 1} a_{n - 2} a_{n - 2} \dots a_{1} a_{0} \equiv a_{0} m o d (10)

\underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}} = ⌊ \frac{2023}{5} ⌋ \underset{n = 1}{\sum^{5}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) + \underset{n = 1}{\sum^{2023 m o d (5)}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5})

\underset{n = 1}{\sum^{5}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) = 1, \underset{n = 1}{\sum^{2023 m o d (5)}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) = 1

\Rightarrow u = - 1 \times ((404 \times 1) + 1) = - 405

\Rightarrow u + a = 944

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