Wednesday, October 28, 2009

Winner for Contest Giveaway - Pearl River Bangle

Updated 28 October 2009

Dear Bead Pot fans,

Our contest closed at midnight last night and the time has come for us to reveal the correct number of pearls that completes this bangle:

And the answer is:

View Pearl River (silver) here

We had more than 30 entries and NOBODY - yes, NOBODY - got the answer right!

So last night, we had a very very tough decision to make and finally, we have decided that the winner shall be based on the CLOSEST number of pearls given. We had 7 entries with 60 pearls and 5 entries with 66 pearls.

It was decided that 66 pearls will not be taken into consideration as that's way past the correct answer. Hence, the Pearl River Bangle (silver) will be given to the very FIRST email entry with the answer, 60 pearls.

CONGRATULATIONS to Karen Wong who sent in her answer on 21st October @ 8.15am. Karen, if you're reading this, I shall be in contact with you via email shortly. :)

Thank you all, for your participation and watch this space for more giveaways!!

See all previous giveaways


Dear Bead Pot fans,

October 09: This month's giveaway is somewhat different from the rest of the months. The Bead Pot is collaborating with Detailed Wears, one of our favourite online reviewers, in conjunction with Detailed Wears' 1st Year Anniversary.

We are making this a fun contest. So celebrate with us by participating in this contest!

Starting 20th October 2009 @ 10.00pm until 27th October 2009

Very simple!
1. Simply guess the correct number of pearls that complete this bangle!
2. Send your answer to
3. Optional: Feel free to leave me a comment so I know you have participated :)

1. Only 1 entry per participant
2. Only one winner will be picked
3. Winner will be announced and notified via email

So hurry and send in an answer today! This contest is only open for the next 7 days!

There will be giveaways from time to time. Watch this space!
See all giveaways

Thank you for all your support and wait for the good news!! *muaks*

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