## FANDOM

41,554 Pages

30 December 2014
Christmas cracker 31 December 2008 688,800,000 2,147,483,644 Unchanged
Blue partyhat 340,100,000 2,147,483,647
Green partyhat 114,800,000 1,452,340,733
Purple partyhat 83,100,000 1,233,390,908
Red partyhat 129,400,000 1,738,594,833
White partyhat 183,300,000 2,147,483,644
Yellow partyhat 96,300,000 1,314,553,072
Pumpkin 5,300,000 149,804,919
Easter egg 4,300,000 49,183,614
Santa hat 14,800,000 115,823,577
Disk of returning 4,700,000 200,971,362
Half full wine jug 31,100,000 348,603,157
Fish mask 29 September 2012 4,555,462 601,926
Christmas tree hat 20 January 2013 2,295,576 12,651,188
Crown of Seasons 23 July 2013 8,307,542 4,044,842
Black Santa hat 20 January 2013 222,967,707 340,890,058
Cloak of Seasons 19 June 2014 7,614,236 27,631,528
Off-hand rubber turkey 3,590,829
Christmas scythe 72,191,580
Holly wreath 372,492,991

Calculations

From the old divisor obtained from the templates:

${div}_{\text{old}} = 15.3476$

We need to calculate a new divisor:

${div}_{\text{new}} = {div}_{\text{old}} \times \frac{\sum \left ( \frac{p}{q} \right )_{\text{new}}}{\sum \left ( \frac{p}{q} \right )_{\text{old}}}$

To calculate the new divisor, we need to find:

\begin{align} \sum \left ( \frac{p}{q} \right )_{\text{old}} & = \text{sum of ratios prior to change} \\ & = \frac{2,147,483,644}{688,800,000} + \frac{2,147,483,647}{340,100,000} + \frac{1,452,340,733}{114,800,000} + \dots + \frac{340,890,058}{222,967,707} + \frac{27,631,528}{7,614,236} \\ & = 211.91329563 \text{ (up to 8 d.p.)} \end{align}

And also:

\begin{align} \sum \left ( \frac{p}{q} \right )_{\text{new}} & = \text{sum of ratios prior to change} - \text{sum of removed ratios} + \text{sum of added ratios} \\ & = \sum \left ( \frac{p}{q} \right )_{\text{old}} - \text{sum of removed ratios} + \text{number of added items} \\ & = 211.91329563 - 0 + 4 \\ & = 215.91329563 \text{ (up to 8 d.p.)} \end{align}

Thus, the new divisor is:

\begin{align} {div}_{\text{new}} & = {div}_{\text{old}} \times \frac{\sum \left ( \frac{p}{q} \right )_{\text{new}}}{\sum \left ( \frac{p}{q} \right )_{\text{old}}} \\ & = 15.3476 \times \frac{215.91329563}{211.91329563} \\ & = 15.6373 \text{ (4 d.p.)} \end{align}