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

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.

$${let}\:{p}\:{be}\:{a}\:{prime}\:{number} \\ $$$$\&\:{let}\:{a}_{\mathrm{1}} \:,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,...,{a}_{{p}\:} {be}\:{integers} \\ $$$${show}\:{that}\:,\:{there}\:{exists}\:{an}\:{integer}\:{k}\:{such}\:{that}\:{the}\:{numbers} \\ $$$${a}_{\mathrm{1}} +{k},\:{a}_{\mathrm{2}} +{k},{a}_{\mathrm{3}} +{k},....,{a}_{{p}} +{k} \\ $$$${produce}\:{at}\:{least}\:\frac{\mathrm{1}}{\mathrm{2}}{p}\:{distinct}\:{remainders} \\ $$$${when}\:{divided}\:{by}\:{p}. \\ $$

Commented by Tinku Tara last updated on 14/Jul/23

Question isnt clear a_1 =p,a_2 =2p,a_3 =3p,...,a_p =p^2 They all produce (k mod p) as remainder when k is added and divided by p

$$\mathrm{Question}\:\mathrm{isnt}\:\mathrm{clear} \\ $$$$\mathrm{a}_{\mathrm{1}} ={p},{a}_{\mathrm{2}} =\mathrm{2}{p},{a}_{\mathrm{3}} =\mathrm{3}{p},...,{a}_{{p}} ={p}^{\mathrm{2}} \\ $$$$\mathrm{They}\:\mathrm{all}\:\mathrm{produce}\:\left(\mathrm{k}\:\mathrm{mod}\:\mathrm{p}\right)\:\mathrm{as} \\ $$$$\mathrm{remainder}\:\mathrm{when}\:\mathrm{k}\:\mathrm{is}\:\mathrm{added}\:\mathrm{and}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{p} \\ $$

Commented by York12 last updated on 14/Jul/23

sir are you the owner of the app

$${sir}\:{are}\:{you}\:{the}\:{owner}\:{of}\:{the}\:{app} \\ $$

Commented by York12 last updated on 14/Jul/23

if so please work on converting into latex tool it produces wrong latax codes

$${if}\:{so}\:{please}\:{work}\:{on}\:{converting}\:{into}\: \\ $$$${latex}\:{tool} \\ $$$${it}\:{produces}\:{wrong}\:{latax}\:{codes} \\ $$$$ \\ $$

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.

$$\mathrm{Type}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{that}\:\mathrm{generated} \\ $$$$\mathrm{wrong}\:\mathrm{latex}\:\mathrm{as}\:\mathrm{comment}\:\mathrm{here}\:\mathrm{and} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{will}\:\mathrm{check}. \\ $$

Commented by York12 last updated on 14/Jul/23

okay sir thanks

$${okay}\:{sir}\:{thanks} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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

okay sir I will check

$${okay}\:{sir}\:{I}\:{will}\:{check} \\ $$

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^((n+1)/2) −a^((n−1)/2) )=na^((n−1)/2) (a−1) it is suffecient to prove that na^((n−1)/2) <(a^(n−1) +a^(n−2) +a^(n−3) +...+1) n^n (√a^((n(n−1))/2) )=na^((n−1)/2) <(a^(n−1) +a^(n−2) +a^(n−3) +...+1) and the equality doesnot hold since a>1 and a^(n−1) ,a^(n−2) ,....,1 can be equal IFF a=1

$$ \\ $$$${a}^{{n}} −\mathrm{1}=\left({a}−\mathrm{1}\right)\left({a}^{{n}−\mathrm{1}} +{a}^{{n}−\mathrm{2}} +{a}^{{n}−\mathrm{3}} +...+\mathrm{1}\right) \\ $$$${n}\left({a}^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} −{a}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} \right)={na}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} \left({a}−\mathrm{1}\right) \\ $$$${it}\:{is}\:{suffecient}\:{to}\:{prove}\:{that} \\ $$$${na}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} <\left({a}^{{n}−\mathrm{1}} +{a}^{{n}−\mathrm{2}} +{a}^{{n}−\mathrm{3}} +...+\mathrm{1}\right) \\ $$$${n}\:^{{n}} \sqrt{{a}^{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} }={na}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} <\left({a}^{{n}−\mathrm{1}} +{a}^{{n}−\mathrm{2}} +{a}^{{n}−\mathrm{3}} +...+\mathrm{1}\right) \\ $$$${and}\:{the}\:{equality}\:{doesnot}\:{hold} \\ $$$${since} \\ $$$${a}>\mathrm{1} \\ $$$${and}\:{a}^{{n}−\mathrm{1}} \:,{a}^{{n}−\mathrm{2}} ,....,\mathrm{1}\:\:{can}\:{be}\:{equal}\:{IFF}\:{a}=\mathrm{1} \\ $$

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

$$ \\ $$$$ \\ $$$$ \\ $$$$\boldsymbol{{u}}=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\lfloor\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\rfloor\:,\:\lfloor\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\rfloor=\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}−\left\{\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\right\} \\ $$$$\boldsymbol{{where}}\:\left\{\circledast\right\}\:\boldsymbol{{indicates}}\:\boldsymbol{{the}}\:\boldsymbol{{fractional}}\: \\ $$$$\boldsymbol{{part}}\:\boldsymbol{{function}}\: \\ $$$$\mathrm{0}\leqslant\left\{\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\right\}<\mathrm{1}\:\Rightarrow\:\mathrm{0}\leqslant\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left\{\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\right\}<\mathrm{2023} \\ $$$$−\mathrm{1000}<\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\lfloor\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\rfloor<\mathrm{1000}\:\Rightarrow\:\mathrm{0}<\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left(\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\right)−\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left\{\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\right\}<\mathrm{1000} \\ $$$$\Rightarrow−\mathrm{3023}<\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left(\frac{\boldsymbol{{n}}^{\mathrm{2}} −\boldsymbol{{na}}}{\mathrm{5}}\right)<\mathrm{1000} \\ $$$$\Rightarrow−\mathrm{3023}<\frac{\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left(\boldsymbol{{n}}^{\mathrm{2}} \right)}{\mathrm{5}}−\frac{\boldsymbol{{a}}\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left(\boldsymbol{{n}}\right)}{\mathrm{5}}<\mathrm{1000} \\ $$$$\Rightarrow\mathrm{5}×\left(\mathrm{3023}+\frac{\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left(\boldsymbol{{n}}^{\mathrm{2}} \right)}{\mathrm{5}}\right)>\boldsymbol{{a}}>\mathrm{5}×\left(\frac{\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left(\boldsymbol{{n}}\right)}{\mathrm{5}}−\mathrm{1000}\right) \\ $$$$\Rightarrow\:\mathrm{1349}.\mathrm{007}>{a}>\mathrm{1348}.\mathrm{998}\:\Rightarrow\:{a}=\mathrm{1349} \\ $$$$\therefore\boldsymbol{{u}}=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\lfloor\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\rfloor=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left(\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right)−\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left\{\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right\} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left(\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right)=\mathrm{0}\:\Rightarrow\:\boldsymbol{{u}}=−\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left\{\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right\} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{now}}\:\boldsymbol{{we}}\:\boldsymbol{{are}}\:\boldsymbol{{only}}\:\boldsymbol{{intrested}}\:\boldsymbol{{in}}\:\boldsymbol{{the}}\:\boldsymbol{{value}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{Remainder}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{expression}} \\ $$$$\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\:\:,\:\boldsymbol{{one}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{special}}\:\boldsymbol{{properties}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{mod}}\:\left(\mathrm{5}\right)\:\&\:\boldsymbol{{mod}}\:\left(\mathrm{10}\right)\:\boldsymbol{{that}}\: \\ $$$$\boldsymbol{{a}}_{\boldsymbol{{n}}} \boldsymbol{{a}}_{\boldsymbol{{n}}−\mathrm{1}} \boldsymbol{{a}}_{\boldsymbol{{n}}−\mathrm{2}} \boldsymbol{{a}}_{\boldsymbol{{n}}−\mathrm{2}} ...\boldsymbol{{a}}_{\mathrm{1}} \boldsymbol{{a}}_{\mathrm{0}} \:\equiv\:\boldsymbol{{a}}_{\mathrm{0}} \:\boldsymbol{{mod}}\:\left(\mathrm{5}\right) \\ $$$$\boldsymbol{{a}}_{\boldsymbol{{n}}} \boldsymbol{{a}}_{\boldsymbol{{n}}−\mathrm{1}} \boldsymbol{{a}}_{\boldsymbol{{n}}−\mathrm{2}} \boldsymbol{{a}}_{\boldsymbol{{n}}−\mathrm{2}} ...\boldsymbol{{a}}_{\mathrm{1}} \boldsymbol{{a}}_{\mathrm{0}} \:\equiv\:\boldsymbol{{a}}_{\mathrm{0}} \:\boldsymbol{{mod}}\left(\mathrm{10}\right) \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left\{\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right\}=\lfloor\frac{\mathrm{2023}}{\mathrm{5}}\rfloor\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}\left(\frac{\left(\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right)\boldsymbol{{mod}}\left(\mathrm{5}\right)}{\mathrm{5}}\right)+\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}\:\boldsymbol{{mod}}\left(\mathrm{5}\right)} {\sum}}\left(\frac{\left(\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right)\boldsymbol{{mod}}\left(\mathrm{5}\right)}{\mathrm{5}}\right) \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}\left(\frac{\left(\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right)\boldsymbol{{mod}}\left(\mathrm{5}\right)}{\mathrm{5}}\right)=\mathrm{1}\:,\:\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{2023}\:\boldsymbol{{mod}}\left(\mathrm{5}\right)} {\sum}}\left(\frac{\left(\frac{\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1349}\boldsymbol{{n}}}{\mathrm{5}}\right)\boldsymbol{{mod}}\left(\mathrm{5}\right)}{\mathrm{5}}\right)=\mathrm{1} \\ $$$$\Rightarrow\boldsymbol{{u}}=\:−\mathrm{1}×\left(\left(\mathrm{404}×\mathrm{1}\right)+\mathrm{1}\right)=−\mathrm{405} \\ $$$$\Rightarrow\:\boldsymbol{{u}}\:+\boldsymbol{{a}}\:=\mathrm{944}\: \\ $$

Commented by York12 last updated on 14/Jul/23

and I hope if you can add some codes for text colour

$${and}\:{I}\:{hope}\:{if}\:{you}\:{can}\:{add}\:{some}\:{codes}\:{for} \\ $$$${text}\:{colour} \\ $$

Commented by Tinku Tara last updated on 14/Jul/23

for the equation with a^n latex is correct tap on 3 dots → get latex paste latex in mathjax to validate

$$\mathrm{for}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{with}\:{a}^{{n}} \:\mathrm{latex}\:\mathrm{is}\:\mathrm{correct} \\ $$$$\mathrm{tap}\:\mathrm{on}\:\mathrm{3}\:\mathrm{dots}\:\rightarrow\:{get}\:{latex} \\ $$$${paste}\:{latex}\:{in}\:{mathjax}\:{to}\:{validate} \\ $$

Commented by York12 last updated on 14/Jul/23

I was using overleaf okay sir I will try

$${I}\:{was}\:{using}\:{overleaf} \\ $$$${okay}\:{sir}\:{I}\:{will}\:{try} \\ $$

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

$$\mathrm{text}\:\mathrm{color}\:\mathrm{is}\:\mathrm{additional}\:\mathrm{latex}\:\mathrm{module}\:\mathrm{not} \\ $$$$\mathrm{supported}\:\mathrm{in}\:\mathrm{many}\:\mathrm{places}\:\mathrm{so}\:\mathrm{colors} \\ $$$$\mathrm{are}\:\mathrm{not}\:\mathrm{coded} \\ $$

Commented by York12 last updated on 14/Jul/23

okay sir thanks

$${okay}\:{sir}\:{thanks} \\ $$

Commented by York12 last updated on 18/Aug/23

sir cannot The generated Latex code be shorter

$${sir}\:{cannot}\:{The}\:{generated}\:{Latex}\:{code}\:{be}\:{shorter} \\ $$

Commented by York12 last updated on 18/Aug/23

canot you make it gendrate latex using asmath package

$${canot}\:{you}\:\:{make}\:{it}\:{gendrate}\: \\ $$$${latex}\:{using}\:{asmath}\:{package} \\ $$

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

$$\mathrm{The}\:\mathrm{generate}\:\mathrm{could}\:\mathrm{probably}\:\mathrm{be}\:\mathrm{shorter} \\ $$$$\mathrm{but}\:\mathrm{you}\:\mathrm{really}\:\mathrm{need}\:\mathrm{to}\:\mathrm{make}\:\mathrm{minor}\:\mathrm{modification} \\ $$$$\mathrm{only} \\ $$$$\mathrm{Microsoft}\:\mathrm{Equation}\:\mathrm{Editor}\:\mathrm{and}\:\mathrm{MathJax} \\ $$$$\mathrm{are}\:\mathrm{able}\:\mathrm{to}\:\mathrm{show}\:\mathrm{this}\:\mathrm{code}\:\mathrm{without} \\ $$$$\mathrm{modification} \\ $$

Commented by York12 last updated on 18/Aug/23

thanks

$${thanks} \\ $$

Commented by Tinku Tara last updated on 18/Aug/23

what is the error that you get in asmath?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{error}\:\mathrm{that}\:\mathrm{you}\:\mathrm{get}\:\mathrm{in} \\ $$$$\mathrm{asmath}? \\ $$

Commented by York12 last updated on 18/Aug/23

Ovearleaf was giving alerts for errors but it is okay now

$${Ovearleaf}\:{was}\:{giving}\:{alerts}\:{for}\:{errors} \\ $$$${but}\:{it}\:{is}\:{okay}\:{now} \\ $$

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