Item
 Base date
 Base price
 Price on adjustment date
 Comments

Christmas cracker
 31 December 2008
 688,800,000
 2,147,483,557
 Unchanged

Blue partyhat
 340,100,000
 2,147,483,647

Green partyhat
 114,800,000
 1,845,813,270

Purple partyhat
 83,100,000
 1,514,338,96

Red partyhat
 129,400,000
 2,100,264,784

White partyhat
 183,300,000
 2,147,483,598

Yellow partyhat
 96,300,000
 1,607,195,89

Pumpkin
 5,300,000
 177,606,966

Easter egg
 4,300,000
 82,884,794

Blue h'ween mask
 12,800,000
 169,489,376

Green h'ween mask
 10,600,000
 129,808,159

Red h'ween mask
 17,300,000
 234,599,317

Santa hat
 14,800,000
 149,818,733

Disk of returning
 4,700,000
 226,927,013

Half full wine jug
 31,100,000
 320,604,356

Fish mask
 29 September 2012
 4,555,462
 1,446,274

Christmas tree hat
 20 January 2013
 2,295,576
 26,024,637

Crown of Seasons
 –
 –
 8,307,542
 Added item

Calculations
From the old divisor obtained from the templates:
 $ {div}_{\text{old}} = 15.1547 $
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{2,147,483,557}{688,800,000} + \frac{2,147,483,647}{340,100,000} + \frac{1,845,813,270}{114,800,000} + \dots + \frac{26,024,637}{2,295,576} \\ & = 260.57227963 \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} \\ & = 260.57227963  0 + 1 \\ & = 261.57227963 \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}}} \\ & = 15.1547 \times \frac{261.57227963}{260.57227963} \\ & = 15.2129 \text{ (4 d.p.)} \end{align} $
