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Joined: Sat May 02, 2009 6:14 am
Posts: 44
Between them 3 children have won a total of 24 boxes of gold coins:
7 boxes each containing 16 gold coins
7 boxes each containing 14 gold coins
10 boxes each containing 9 gold coins.

Part 1:
They decide to share what they have won equally, each child receiving the same number of coins
How many coins should they each receive ?

This part is fine, answer 100

Part 2:

They find they can share the coins equally WITHOUT OPENING ANY OF THE BOXES
Work out how they do this.

1 child takes.....boxes of 16 coins and .... boxes of 14 coins and .... boxes of 9 coins.
Another child takes ....boxes of 16 coins and ....boxes of 14 coins and...boxes of 9 coins.
The third child takes ....boxes of 16 coins and ...boxes of 14 coins and ....boxes of 9 coins.

Please can someone help work out Part 2

Many thanks.

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Joined: Fri Nov 17, 2006 8:54 pm
Posts: 1776
Location: caversham
Weigh them? Not maths but lateral thinking?

Algebra, box + x coins, etc.

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Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3826
greenvallley wrote:
Between them 3 children have won a total of 24 boxes of gold coins:
7 boxes each containing 16 gold coins
7 boxes each containing 14 gold coins
10 boxes each containing 9 gold coins.

Part 1:
They decide to share what they have won equally, each child receiving the same number of coins
How many coins should they each receive ?

This part is fine, answer 100

Part 2:

They find they can share the coins equally WITHOUT OPENING ANY OF THE BOXES
Work out how they do this.

1 child takes.....boxes of 16 coins and .... boxes of 14 coins and .... boxes of 9 coins.
Another child takes ....boxes of 16 coins and ....boxes of 14 coins and...boxes of 9 coins.
The third child takes ....boxes of 16 coins and ...boxes of 14 coins and ....boxes of 9 coins.

Please can someone help work out Part 2

Many thanks.

1 child takes.1.boxes of 16 coins and .6... boxes of 14 coins and .0... boxes of 9 coins.
Another child takes 2....boxes of 16 coins and .1...boxes of 14 coins and.6..boxes of 9 coins.
The third child takes ..4..boxes of 16 coins and .0..boxes of 14 coins and ..4..boxes of 9 coins.

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Joined: Sat May 02, 2009 6:14 am
Posts: 44
Wow thanks moved!
Did you just keep trying various combinations and finally found the correct for all 3?(we tried this way and just couldnt get them all to work)
Or this there a method to working these sorts of questions out?

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Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3826
There is a method for a 10 yr old, it involves trying but using a logical approach.

Start with 100 and take away the 16 as it is the biggest number with the fewest multiples

100 - 16 = 84. 84 is a multiple of 14, happily.

This leaves us with only 1 box of 14.

100 - 16 - 14 = 70, not a multiple of 9

100- 16 - 16 = 68, 68 -14 = 54, happily a multiple of 9.

We are then left with 4 boxes of 16 = 64 and 4 boxes of 9 = 36.

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Joined: Sat May 02, 2009 6:14 am
Posts: 44
Thank you for explaining.That's a hard question!

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