There exists a unique positive integer a for which The sum u = Σ_(n=1) ^(2023) ⌊((n^2 −na)/5)⌋ is an integer strictly between −1000 & 1000 find a+u.


Question and Answers Forum

All Questions Topic List
Logarithms Questions
Previous in All Question Next in All Question
Previous in Logarithms Next in Logarithms
Question Number 193183 by York12 last updated on 06/Jun/23

$T h e r e e x i s t s a u n i q u e p o s i t i v e i n t e g e r a f o r$ $w h i c h T h e s u m u = \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - n a}{5} ⌋ i s a n i n t e g e r$ $s t r i c t l y b e t w e e n - 1000 & 1000 f i n d a + u .$
	Answered by York12 last updated on 12/Jun/23
	$u=Σ_(n=1) ^(2023) ⌊((n^2 −na)/5)⌋ , ⌊((n^2 −na)/5)⌋=((n^2 −na)/5)−{((n^2 −na)/5)} where {⊛} indicates the fractional part function 0≤{((n^2 −na)/5)}<1 ⇒ 0≤Σ_(n=1) ^(2023) {((n^2 −na)/5)}<2023 −1000<Σ_(n=1) ^(2023) ⌊((n^2 −na)/5)⌋<1000 ⇒ 0<Σ_(n=1) ^(2023) (((n^2 −na)/5))−Σ_(n=1) ^(2023) {((n^2 −na)/5)}<1000 ⇒−3023<Σ_(n=1) ^(2023) (((n^2 −na)/5))<1000 ⇒−3023<((Σ_(n=1) ^(2023) (n^2 ))/5)−((aΣ_(n=1) ^(2023) (n))/5)<1000 ⇒5×(3023+((Σ_(n=1) ^(2023) (n^2 ))/5))>a>5×(((Σ_(n=1) ^(2023) (n))/5)−1000) ⇒ 1349.007>a>1348.998 ⇒ a=1349 ∴u=Σ_(n=1) ^(2023) ⌊((n^2 −1349n)/5)⌋=Σ_(n=1) ^(2023) (((n^2 −1349n)/5))−Σ_(n=1) ^(2023) {((n^2 −1349n)/5)} Σ_(n=1) ^(2023) (((n^2 −1349n)/5))=0 ⇒ u=−Σ_(n=1) ^(2023) {((n^2 −1349n)/5)} so now we are only intrested in the value of the Remainder of the expression ((n^2 −1349n)/5) , one of the special properties of mod (5) & mod (10) that a_n a_(n−1) a_(n−2) a_(n−2) ...a_1 a_0 ≡ a_0 mod (5) a_n a_(n−1) a_(n−2) a_(n−2) ...a_1 a_0 ≡ a_0 mod(10) Σ_(n=1) ^(2023) {((n^2 −1349n)/5)}=⌊((2023)/5)⌋Σ_(n=1) ^5 ((((((n^2 −1349n)/5))mod(5))/5))+Σ_(n=1) ^(2023 mod(5)) ((((((n^2 −1349n)/5))mod(5))/5)) Σ_(n=1) ^5 ((((((n^2 −1349n)/5))mod(5))/5))=1 , Σ_(n=1) ^(2023 mod(5)) ((((((n^2 −1349n)/5))mod(5))/5))=1 ⇒u= −1×((404×1)+1)=−405 ⇒ u +a =944$
	$u = \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - n a}{5} ⌋, ⌊ \frac{n^{2} - n a}{5} ⌋ = \frac{n^{2} - n a}{5} - {\frac{n^{2} - n a}{5}}$ $w h e r e {⊛} i n d i c a t e s t h e f r a c t i o n a l$ $p a r t f u n c t i o n$ $0 ⩽ {\frac{n^{2} - n a}{5}} < 1 \Rightarrow 0 ⩽ \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - n a}{5}} < 2023$ $- 1000 < \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - n a}{5} ⌋ < 1000 \Rightarrow 0 < \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - n a}{5}) - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - n a}{5}} < 1000$ $\Rightarrow - 3023 < \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - n a}{5}) < 1000$ $\Rightarrow - 3023 < \frac{\underset{n = 1}{\sum^{2023}} (n^{2})}{5} - \frac{a \underset{n = 1}{\sum^{2023}} (n)}{5} < 1000$ $\Rightarrow 5 \times (3023 + \frac{\underset{n = 1}{\sum^{2023}} (n^{2})}{5}) > a > 5 \times (\frac{\underset{n = 1}{\sum^{2023}} (n)}{5} - 1000)$ $\Rightarrow 1349.007 > a > 1348.998 \Rightarrow a = 1349$ $∴ u = \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - 1349 n}{5} ⌋ = \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - 1349 n}{5}) - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}}$ $\underset{n = 1}{\sum^{2023}} (\frac{n^{2} - 1349 n}{5}) = 0 \Rightarrow u = - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}}$ $s o n o w w e a r e o n l y i n t r e s t e d i n t h e v a l u e$ $o f t h e R e m a i n d e r o f t h e e x p r e s s i o n$ $\frac{n^{2} - 1349 n}{5}, o n e o f t h e s p e c i a l p r o p e r t i e s$ $o f m o d (5) & m o d (10) t h a t$ $a_{n} a_{n - 1} a_{n - 2} a_{n - 2} . . . a_{1} a_{0} \equiv a_{0} m o d (5)$ $a_{n} a_{n - 1} a_{n - 2} a_{n - 2} . . . a_{1} a_{0} \equiv a_{0} m o d (10)$ $\underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}} = ⌊ \frac{2023}{5} ⌋ \underset{n = 1}{\sum^{5}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) + \underset{n = 1}{\sum^{2023 m o d (5)}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5})$ $\underset{n = 1}{\sum^{5}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) = 1, \underset{n = 1}{\sum^{2023 m o d (5)}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) = 1$ $\Rightarrow u = - 1 \times ((404 \times 1) + 1) = - 405$ $\Rightarrow u + a = 944$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum