let p be a prime number & let a_1 ,a_2 ,a_3 ,...,a_(p ) be integers show that , there exists an integer k such that the numbers a_1 +k, a_2 +k,a_3 +k,....,a_p +k produce at least (1/2)p distinct remainders when divided by p.


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Question Number 194710 by York12 last updated on 14/Jul/23

$l e t p b e a p r i m e n u m b e r$ $& l e t a_{1}, a_{2}, a_{3}, . . ., a_{p} b e i n t e g e r s$ $s h o w t h a t, t h e r e e x i s t s a n i n t e g e r k s u c h t h a t t h e n u m b e r s$ $a_{1} + k, a_{2} + k, a_{3} + k, . . . ., a_{p} + k$ $p r o d u c e a t l e a s t \frac{1}{2} p d i s t i n c t r e m a i n d e r s$ $w h e n d i v i d e d b y p .$
Commented by Tinku Tara last updated on 14/Jul/23

$Question isnt clear$ $a_{1} = p, a_{2} = 2 p, a_{3} = 3 p, . . ., a_{p} = p^{2}$ $They all produce (k \mod p) as$ $remainder when k is added and divided by p$
Commented by York12 last updated on 14/Jul/23

$s i r a r e y o u t h e o w n e r o f t h e a p p$
Commented by York12 last updated on 14/Jul/23

$i f s o p l e a s e w o r k o n c o n v e r t i n g i n t o$ $l a t e x t o o l$ $i t p r o d u c e s w r o n g l a t a x c o d e s$
Commented by Tinku Tara last updated on 14/Jul/23

$Type the equation that generated$ $wrong latex as comment here and$ $and we will check .$
Commented by York12 last updated on 14/Jul/23

$o k a y s i r t h a n k s$
Commented by Tinku Tara last updated on 14/Jul/23
You can visit Https://www.tinkutara.com. The web site has the same forum displayed as latex.
Commented by York12 last updated on 14/Jul/23

$o k a y s i r I w i l l c h e c k$
Commented by York12 last updated on 14/Jul/23

$a^{n} - 1 = (a - 1) (a^{n - 1} + a^{n - 2} + a^{n - 3} + . . . + 1)$ $n (a^{\frac{n + 1}{2}} - a^{\frac{n - 1}{2}}) = {n a}^{\frac{n - 1}{2}} (a - 1)$ $i t i s s u f f e c i e n t t o p r o v e t h a t$ ${n a}^{\frac{n - 1}{2}} < (a^{n - 1} + a^{n - 2} + a^{n - 3} + . . . + 1)$ $n^{n} \sqrt{a^{\frac{n (n - 1)}{2}}} = {n a}^{\frac{n - 1}{2}} < (a^{n - 1} + a^{n - 2} + a^{n - 3} + . . . + 1)$ $a n d t h e e q u a l i t y d o e s n o t h o l d$ $s i n c e$ $a > 1$ $a n d a^{n - 1}, a^{n - 2}, . . . ., 1 c a n b e e q u a l I F F a = 1$
Commented by York12 last updated on 14/Jul/23
$u=Σ_(n=1) ^(2023) ⌊((n^2 −na)/5)⌋ , ⌊((n^2 −na)/5)⌋=((n^2 −na)/5)−{((n^2 −na)/5)} where {⊛} indicates the fractional part function 0≤{((n^2 −na)/5)}<1 ⇒ 0≤Σ_(n=1) ^(2023) {((n^2 −na)/5)}<2023 −1000<Σ_(n=1) ^(2023) ⌊((n^2 −na)/5)⌋<1000 ⇒ 0<Σ_(n=1) ^(2023) (((n^2 −na)/5))−Σ_(n=1) ^(2023) {((n^2 −na)/5)}<1000 ⇒−3023<Σ_(n=1) ^(2023) (((n^2 −na)/5))<1000 ⇒−3023<((Σ_(n=1) ^(2023) (n^2 ))/5)−((aΣ_(n=1) ^(2023) (n))/5)<1000 ⇒5×(3023+((Σ_(n=1) ^(2023) (n^2 ))/5))>a>5×(((Σ_(n=1) ^(2023) (n))/5)−1000) ⇒ 1349.007>a>1348.998 ⇒ a=1349 ∴u=Σ_(n=1) ^(2023) ⌊((n^2 −1349n)/5)⌋=Σ_(n=1) ^(2023) (((n^2 −1349n)/5))−Σ_(n=1) ^(2023) {((n^2 −1349n)/5)} Σ_(n=1) ^(2023) (((n^2 −1349n)/5))=0 ⇒ u=−Σ_(n=1) ^(2023) {((n^2 −1349n)/5)} so now we are only intrested in the value of the Remainder of the expression ((n^2 −1349n)/5) , one of the special properties of mod (5) & mod (10) that a_n a_(n−1) a_(n−2) a_(n−2) ...a_1 a_0 ≡ a_0 mod (5) a_n a_(n−1) a_(n−2) a_(n−2) ...a_1 a_0 ≡ a_0 mod(10) Σ_(n=1) ^(2023) {((n^2 −1349n)/5)}=⌊((2023)/5)⌋Σ_(n=1) ^5 ((((((n^2 −1349n)/5))mod(5))/5))+Σ_(n=1) ^(2023 mod(5)) ((((((n^2 −1349n)/5))mod(5))/5)) Σ_(n=1) ^5 ((((((n^2 −1349n)/5))mod(5))/5))=1 , Σ_(n=1) ^(2023 mod(5)) ((((((n^2 −1349n)/5))mod(5))/5))=1 ⇒u= −1×((404×1)+1)=−405 ⇒ u +a =944$
$u = \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - n a}{5} ⌋, ⌊ \frac{n^{2} - n a}{5} ⌋ = \frac{n^{2} - n a}{5} - {\frac{n^{2} - n a}{5}}$ $w h e r e {⊛} i n d i c a t e s t h e f r a c t i o n a l$ $p a r t f u n c t i o n$ $0 ⩽ {\frac{n^{2} - n a}{5}} < 1 \Rightarrow 0 ⩽ \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - n a}{5}} < 2023$ $- 1000 < \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - n a}{5} ⌋ < 1000 \Rightarrow 0 < \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - n a}{5}) - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - n a}{5}} < 1000$ $\Rightarrow - 3023 < \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - n a}{5}) < 1000$ $\Rightarrow - 3023 < \frac{\underset{n = 1}{\sum^{2023}} (n^{2})}{5} - \frac{a \underset{n = 1}{\sum^{2023}} (n)}{5} < 1000$ $\Rightarrow 5 \times (3023 + \frac{\underset{n = 1}{\sum^{2023}} (n^{2})}{5}) > a > 5 \times (\frac{\underset{n = 1}{\sum^{2023}} (n)}{5} - 1000)$ $\Rightarrow 1349.007 > a > 1348.998 \Rightarrow a = 1349$ $∴ u = \underset{n = 1}{\sum^{2023}} ⌊ \frac{n^{2} - 1349 n}{5} ⌋ = \underset{n = 1}{\sum^{2023}} (\frac{n^{2} - 1349 n}{5}) - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}}$ $\underset{n = 1}{\sum^{2023}} (\frac{n^{2} - 1349 n}{5}) = 0 \Rightarrow u = - \underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}}$ $s o n o w w e a r e o n l y i n t r e s t e d i n t h e v a l u e$ $o f t h e R e m a i n d e r o f t h e e x p r e s s i o n$ $\frac{n^{2} - 1349 n}{5}, o n e o f t h e s p e c i a l p r o p e r t i e s$ $o f m o d (5) & m o d (10) t h a t$ $a_{n} a_{n - 1} a_{n - 2} a_{n - 2} . . . a_{1} a_{0} \equiv a_{0} m o d (5)$ $a_{n} a_{n - 1} a_{n - 2} a_{n - 2} . . . a_{1} a_{0} \equiv a_{0} m o d (10)$ $\underset{n = 1}{\sum^{2023}} {\frac{n^{2} - 1349 n}{5}} = ⌊ \frac{2023}{5} ⌋ \underset{n = 1}{\sum^{5}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) + \underset{n = 1}{\sum^{2023 m o d (5)}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5})$ $\underset{n = 1}{\sum^{5}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) = 1, \underset{n = 1}{\sum^{2023 m o d (5)}} (\frac{(\frac{n^{2} - 1349 n}{5}) m o d (5)}{5}) = 1$ $\Rightarrow u = - 1 \times ((404 \times 1) + 1) = - 405$ $\Rightarrow u + a = 944$
Commented by York12 last updated on 14/Jul/23

$a n d I h o p e i f y o u c a n a d d s o m e c o d e s f o r$ $t e x t c o l o u r$
Commented by Tinku Tara last updated on 14/Jul/23

$for the equation with a^{n} latex is correct$ $tap on 3 dots \to g e t l a t e x$ $p a s t e l a t e x i n m a t h j a x t o v a l i d a t e$
Commented by York12 last updated on 14/Jul/23

$I w a s u s i n g o v e r l e a f$ $o k a y s i r I w i l l t r y$
Commented by Tinku Tara last updated on 14/Jul/23
https://www.mathjax.org/#demo
Commented by Tinku Tara last updated on 14/Jul/23

$text color is additional latex module not$ $supported in many places so colors$ $are not coded$
Commented by York12 last updated on 14/Jul/23

$o k a y s i r t h a n k s$
Commented by York12 last updated on 18/Aug/23

$s i r c a n n o t T h e g e n e r a t e d L a t e x c o d e b e s h o r t e r$
Commented by York12 last updated on 18/Aug/23

$c a n o t y o u m a k e i t g e n d r a t e$ $l a t e x u s i n g a s m a t h p a c k a g e$
Commented by Tinku Tara last updated on 18/Aug/23

$The generate could probably be shorter$ $but you really need to make minor modification$ $only$ $Microsoft Equation Editor and MathJax$ $are able to show this code without$ $modification$
Commented by York12 last updated on 18/Aug/23

$t h a n k s$
Commented by Tinku Tara last updated on 18/Aug/23

$what is the error that you get in$ $asmath ?$
Commented by York12 last updated on 18/Aug/23

$O v e a r l e a f w a s g i v i n g a l e r t s f o r e r r o r s$ $b u t i t i s o k a y n o w$
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