The-a-2024-2044-2032-2040-2036-

Facebook Tweet Pin

Question Number 208876 by Tawa11 last updated on 26/Jun/24

The 𝚌a𝚕𝚎𝚗𝚍𝚊𝚛 𝚘𝚏 𝚝𝚑𝚎 𝚢𝚎𝚊𝚛 2024 𝚒𝚜 𝚝𝚑𝚎 𝚜𝚊𝚖𝚎 𝚏𝚘𝚛
𝙰.2044
𝙱.2032
𝙲.2040
𝙳.2036

Commented by mr W last updated on 26/Jun/24

$${none}\:{of}\:{them}. \\ $$

Commented by Tawa11 last updated on 26/Jun/24

Thanks sir. Answer should be 2052 sir? by just adding 28 to 2024?

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{Answer}\:\mathrm{should}\:\mathrm{be}\:\:\mathrm{2052}\:\mathrm{sir}? \\ $$$$\mathrm{by}\:\mathrm{just}\:\mathrm{adding}\:\mathrm{28}\:\mathrm{to}\:\mathrm{2024}? \\ $$

Commented by A5T last updated on 26/Jun/24

$${The}\:+\mathrm{28}\:{rule}\:{works}\:{only}\:{for}\:{leap}\:{years}. \\ $$

Commented by mr W last updated on 26/Jun/24

2024 is a leap year. the next year with same calender must also be a leap year. say it is 2024+4n. we have n years with 366 days and 3n years with 365 days. such that that year starts with the same weekday as 2024, 3n×365+n×366=1461n must be divisible by 7. we get n_(min) =7. that means after 4×7=28 years, i.e. the year 2052 has the same calender as 2024.

$$\mathrm{2024}\:{is}\:{a}\:{leap}\:{year}.\:{the}\:{next}\:{year} \\ $$$${with}\:{same}\:{calender}\:{must}\:{also}\:{be}\:{a} \\ $$$${leap}\:{year}.\:{say}\:{it}\:{is}\:\mathrm{2024}+\mathrm{4}{n}. \\ $$$${we}\:{have}\:{n}\:{years}\:{with}\:\mathrm{366}\:{days}\:{and} \\ $$$$\mathrm{3}{n}\:{years}\:{with}\:\mathrm{365}\:{days}.\:{such}\:{that} \\ $$$${that}\:{year}\:{starts}\:{with}\:{the}\:{same} \\ $$$${weekday}\:{as}\:\mathrm{2024}, \\ $$$$\mathrm{3}{n}×\mathrm{365}+{n}×\mathrm{366}=\mathrm{1461}{n}\:{must}\:{be}\: \\ $$$${divisible}\:{by}\:\mathrm{7}.\:{we}\:{get}\:{n}_{{min}} =\mathrm{7}.\:{that}\: \\ $$$${means}\:{after}\:\mathrm{4}×\mathrm{7}=\mathrm{28}\:{years},\:{i}.{e}.\:{the}\: \\ $$$${year}\:\mathrm{2052}\:{has}\:{the}\:{same}\:{calender}\:{as}\: \\ $$$$\mathrm{2024}. \\ $$

Commented by Tawa11 last updated on 26/Jun/24

$$\mathrm{Why}\:\mathrm{3n}\:\mathrm{years}\:\mathrm{in}\:\mathrm{365}\:\mathrm{day}\:\mathrm{sir} \\ $$

Commented by Tawa11 last updated on 26/Jun/24

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$

Commented by mr W last updated on 26/Jun/24

a leap year has 366 days, a normal year has 365 days. among 4n years, there are n leap years, the other 3n years are normal years.

$${a}\:{leap}\:{year}\:{has}\:\mathrm{366}\:{days},\:{a}\:{normal} \\ $$$${year}\:{has}\:\mathrm{365}\:{days}. \\ $$$${among}\:\mathrm{4}{n}\:{years},\:\:{there}\:{are}\:{n}\:{leap}\: \\ $$$${years},\:{the}\:{other}\:\mathrm{3}{n}\:{years}\:{are}\:{normal} \\ $$$${years}. \\ $$

Answered by A5T last updated on 26/Jun/24

2024 January- x day of the week 2025 ∥ - x+2 day 2026 ∥ - x+3 2027 ∥ -x+4 2028 ∥ -x+5 2029 ∥ -x+7 ≡ x(mod 7) But 2029 is not a leap year, so you continue with this process until you get a leap year(One gets 2052). [Add +2 if the previous year was a leap year, +1 otherwise].

$$\mathrm{2024}\:{January}-\:{x}\:{day}\:{of}\:{the}\:{week} \\ $$$$\mathrm{2025}\:\:\:\:\:\:\:\:\shortparallel\:\:\:\:\:\:\:\:\:\:\:-\:\:{x}+\mathrm{2}\:{day} \\ $$$$\mathrm{2026}\:\:\:\:\:\:\:\:\shortparallel\:\:\:\:\:\:\:\:\:\:\:-\:{x}+\mathrm{3} \\ $$$$\mathrm{2027}\:\:\:\:\:\:\:\:\shortparallel\:\:\:\:\:\:\:\:\:\:\:-{x}+\mathrm{4} \\ $$$$\mathrm{2028}\:\:\:\:\:\:\:\:\shortparallel\:\:\:\:\:\:\:\:\:\:\:-{x}+\mathrm{5} \\ $$$$\mathrm{2029}\:\:\:\:\:\:\:\:\shortparallel\:\:\:\:\:\:\:\:\:\:\:-{x}+\mathrm{7}\:\equiv\:{x}\left({mod}\:\mathrm{7}\right) \\ $$$${But}\:\mathrm{2029}\:{is}\:{not}\:{a}\:{leap}\:{year},\:{so}\:{you}\:{continue}\: \\ $$$${with}\:{this}\:{process}\:{until}\:{you}\:{get}\:{a}\:{leap}\:{year}\left({One}\right. \\ $$$$\left.{gets}\:\mathrm{2052}\right). \\ $$$$\left[{Add}\:+\mathrm{2}\:{if}\:{the}\:{previous}\:{year}\:{was}\:{a}\:{leap}\:{year},\right. \\ $$$$\left.+\mathrm{1}\:{otherwise}\right]. \\ $$

Commented by Tawa11 last updated on 26/Jun/24

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by A5T last updated on 26/Jun/24

If y is a leap year, then the number of leap years from y to y+4q(excluding) is q x+q+4q≡x(mod 7) ⇒5q≡0(mod 7)⇒q≡0(mod 7)⇒4q≡0(mod 28) So, always add multiples of 28 to get a leap year starting from the same day.

$${If}\:{y}\:{is}\:{a}\:{leap}\:{year},\:{then}\:{the}\:{number}\:{of}\:{leap} \\ $$$${years}\:{from}\:{y}\:{to}\:{y}+\mathrm{4}{q}\left({excluding}\right)\:{is}\:{q} \\ $$$${x}+{q}+\mathrm{4}{q}\equiv{x}\left({mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow\mathrm{5}{q}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right)\Rightarrow{q}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{4}{q}\equiv\mathrm{0}\left({mod}\:\mathrm{28}\right) \\ $$$${So},\:{always}\:{add}\:{multiples}\:{of}\:\mathrm{28}\:{to}\:{get}\:{a} \\ $$$${leap}\:{year}\:{starting}\:{from}\:{the}\:{same}\:{day}. \\ $$

Commented by Tinku Tara last updated on 26/Jun/24

In leap years you need to also consider the fact year divible by 100 are not leap years however years divisble by 400 are leap years. 1900 not a leap year 2000 a leap year

$$\mathrm{In}\:\mathrm{leap}\:\mathrm{years}\:\mathrm{you}\:\mathrm{need}\:\mathrm{to}\:\mathrm{also} \\ $$$$\mathrm{consider}\:\mathrm{the}\:\mathrm{fact}\:\mathrm{year}\:\mathrm{divible}\:\mathrm{by} \\ $$$$\mathrm{100}\:\mathrm{are}\:\mathrm{not}\:\mathrm{leap}\:\mathrm{years}\:\mathrm{however} \\ $$$$\mathrm{years}\:\mathrm{divisble}\:\mathrm{by}\:\mathrm{400}\:\mathrm{are}\:\mathrm{leap}\:\mathrm{years}. \\ $$$$\mathrm{1900}\:\mathrm{not}\:\mathrm{a}\:\mathrm{leap}\:\mathrm{year} \\ $$$$\mathrm{2000}\:\mathrm{a}\:\mathrm{leap}\:\mathrm{year} \\ $$

Commented by A5T last updated on 26/Jun/24

Yes, if a year divisible by 100 but not by 400 is between y and y+4q, then the number of leap years from y to y+4q(excluding) is q−1 ⇒x+q−1+4q≡x(mod 7)⇒5q≡1(mod 7) ⇒q≡3(mod 7)⇒4q=4(7k+3)=28k+12 So,12 is the smallest we add in that case.

$${Yes},\:{if}\:{a}\:{year}\:{divisible}\:{by}\:\mathrm{100}\:{but}\:{not}\:{by}\:\mathrm{400} \\ $$$${is}\:\boldsymbol{{between}}\:{y}\:{and}\:{y}+\mathrm{4}{q},\:{then}\:{the}\:{number}\:{of} \\ $$$${leap}\:{years}\:{from}\:{y}\:{to}\:{y}+\mathrm{4}{q}\left({excluding}\right)\:{is}\:{q}−\mathrm{1} \\ $$$$\Rightarrow{x}+{q}−\mathrm{1}+\mathrm{4}{q}\equiv{x}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{5}{q}\equiv\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow{q}\equiv\mathrm{3}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{4}{q}=\mathrm{4}\left(\mathrm{7}{k}+\mathrm{3}\right)=\mathrm{28}{k}+\mathrm{12} \\ $$$${So},\mathrm{12}\:{is}\:{the}\:{smallest}\:{we}\:{add}\:{in}\:{that}\:{case}. \\ $$

Answered by BaliramKumar last updated on 27/Jun/24

determinant (((year),(Remainder),(Adding)),((2023),(((23)/4) =^R 3),(+11)),((2022),(((22)/4) =^R 2),(+11)),((2021),(((21)/4) =^R 1),(+6)),((1900),(((00)/4) =^R 0),(±6)),((2024),(Leap),(+28)),((2080),(Leap),(+40)),((2072),(Leap),(+40)),((1972),(Leap),(+28)))

$$ \\ $$$$\:\:\:\:\:\:\begin{array}{|c|c|c|c|c|c|c|c|c|}{\mathrm{year}}&\hline{\mathrm{Remainder}}&\hline{\mathrm{Adding}}\\{\mathrm{2023}}&\hline{\frac{\mathrm{23}}{\mathrm{4}}\:\overset{\mathrm{R}} {=}\:\mathrm{3}}&\hline{+\mathrm{11}}\\{\mathrm{2022}}&\hline{\frac{\mathrm{22}}{\mathrm{4}}\:\overset{\mathrm{R}} {=}\:\mathrm{2}}&\hline{+\mathrm{11}}\\{\mathrm{2021}}&\hline{\frac{\mathrm{21}}{\mathrm{4}}\:\overset{\mathrm{R}} {=}\:\mathrm{1}}&\hline{+\mathrm{6}}\\{\mathrm{1900}}&\hline{\frac{\mathrm{00}}{\mathrm{4}}\:\overset{\mathrm{R}} {=}\:\mathrm{0}}&\hline{\pm\mathrm{6}}\\{\mathrm{2024}}&\hline{\mathrm{Leap}}&\hline{+\mathrm{28}}\\{\mathrm{2080}}&\hline{\mathrm{Leap}}&\hline{+\mathrm{40}}\\{\mathrm{2072}}&\hline{\mathrm{Leap}}&\hline{+\mathrm{40}}\\{\mathrm{1972}}&\hline{\mathrm{Leap}}&\hline{+\mathrm{28}}\\\hline\end{array} \\ $$$$\:\:\:\:\:\: \\ $$

Facebook Tweet Pin

The-a-2024-2044-2032-2040-2036-

Leave a Reply Cancel reply