Easy -2.0 This is purely computational arithmetic following a given algorithm with no problem-solving, proof, or conceptual understanding required. Students simply substitute values and perform basic integer division operations step-by-step. It's significantly easier than typical A-level questions which require mathematical reasoning.
3 The following algorithm (J. M. Oudin, 1940) claims to compute the date of Easter Sunday in the Gregorian calendar system.
The algorithm uses the year, y, to give the month, m, and day, d, of Easter Sunday.
All variables are integers and all remainders from division are dropped. For example, 7 divided by 3 is 2 remainder 1 . The remainder is dropped, giving the answer 2.
$$\begin{aligned}
& c = y / 100 \\
& n = y - 19 \times ( y / 19 ) \\
& k = ( c - 17 ) / 25 \\
& i = c - ( c / 4 ) - ( c - k ) / 3 + ( 19 \times n ) + 15 \\
& i = i - 30 \times ( i / 30 ) \\
& i = i - ( i / 28 ) \times ( 1 - ( i / 28 ) \times ( 29 / ( i + 1 ) ) \times ( ( 21 - n ) / 11 ) ) \\
& j = y + ( y / 4 ) + i + 2 - c + ( c / 4 ) \\
& j = j - 7 \times ( j / 7 ) \\
& p = i - j \\
& m = 3 + ( p + 40 ) / 44 \\
& d = p + 28 - 31 \times ( m / 4 )
\end{aligned}$$
For example, for 2008:
\(\mathrm { y } = 2008\)
\(\mathrm { c } = 2008 / 100 = 20\)
\(n = 2008 - 19 \times ( 2008 / 19 ) = 2008 - 19 \times ( 105 ) = 13\), etc.
Complete the calculation for 2008.
3 The following algorithm (J. M. Oudin, 1940) claims to compute the date of Easter Sunday in the Gregorian calendar system.\\
The algorithm uses the year, y, to give the month, m, and day, d, of Easter Sunday.\\
All variables are integers and all remainders from division are dropped. For example, 7 divided by 3 is 2 remainder 1 . The remainder is dropped, giving the answer 2.
$$\begin{aligned}
& c = y / 100 \\
& n = y - 19 \times ( y / 19 ) \\
& k = ( c - 17 ) / 25 \\
& i = c - ( c / 4 ) - ( c - k ) / 3 + ( 19 \times n ) + 15 \\
& i = i - 30 \times ( i / 30 ) \\
& i = i - ( i / 28 ) \times ( 1 - ( i / 28 ) \times ( 29 / ( i + 1 ) ) \times ( ( 21 - n ) / 11 ) ) \\
& j = y + ( y / 4 ) + i + 2 - c + ( c / 4 ) \\
& j = j - 7 \times ( j / 7 ) \\
& p = i - j \\
& m = 3 + ( p + 40 ) / 44 \\
& d = p + 28 - 31 \times ( m / 4 )
\end{aligned}$$
For example, for 2008:\\
$\mathrm { y } = 2008$\\
$\mathrm { c } = 2008 / 100 = 20$\\
$n = 2008 - 19 \times ( 2008 / 19 ) = 2008 - 19 \times ( 105 ) = 13$, etc.\\
Complete the calculation for 2008.
\hfill \mbox{\textit{OCR MEI D1 2008 Q3 [8]}}