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ROME OR OSLO? ANDREW, JULIE, TO VISIT.
ROMEO ROSLO AND REW JULIET OVISIT
> ROMEO AND JULIET

WARM TOO FEW, THE SUBWORLD SNEEZES
WAR MTO OF EW THE SUB WORLDS NEEZES
> WAR OF THE WORLDS

AN I MALE? MYSELF, ARM, AND I?
ANIMAL EMYSEL FARM ANDI
> ANIMAL FARM

GO NEXT, GO! WITHOUT OTHER I DWINDLE. GO!
GONE XTGO WITH OUTO THE RID WIND LEGO
> GONE WITH THE WIND

WATER. SHIP, BOAT. STORM! DOWN SINK!
WATERSHIP BOATSTORM DOWN SINK
> WATERSHIP DOWN

BRAVERY IN A NEW AGE, WORLD'S HOPE.
BRAVE RYINA NEW AGE WORLD SHOPE
> BRAVE NEW WORLD

Puzzle 5

Using the BrainTracker grid below, how many words can you find? Each word must contain the central T and no letter can be used twice, however, the letters do not have to be connected. Proper nouns are not allowed, however, plurals are. Can you find the nine letter word?

Hint: Try breaking each cog size into prime factors.

Answer: 260 revolutions.

There are a number of ways of thinking about the solution, and we find this one the quickest way to find the answer.

The total number of teeth moved by Cog 1 will be wholly divisible by each cog in turn, therefore:

Revolutions x Cog 1 ÷ Cog2 is an integer
Revolutions x Cog 1 ÷ Cog3 is an integer
Revolutions x Cog 1 ÷ Cog4 is an integer

So we are after the first number of revolutions x 81 that is an integer after division by 52, 36 and 20.

Thus:

81 81 81
-- and -- and -- all need to be integers (and not fractions).
52 36 20

An easy way to do this would be to multiply by 52 x 36 x 20 = 37,440 revolutions, which would be a correct answer, but not necessarily the smallest answer.

A better way is to break each cog down into its prime factors, where Cog 1 has the largest number of teeth:

Cog 1 - 81 = 3 x 3 x 3 x 3

Cog 2 - 52 = 2 x 2 x 13
Cog 3 - 36 = 2 x 2 x 3 x 3
Cog 4 - 20 = 2 x 2 x 5

3 x 3 x 3 x 3 and 3 x 3 x 3 x 3 and 3 x 3 x 3 x 3 and these need to be integers
------------- ------------- ------------
2 x 2 x 13 2 x 2 x 3 x 3 2 x 2 x 5

Simplifying any fraction that can be simplified gives:

3 x 3 x 3 x 3 and 3 x 3 and 3 x 3 x 3 x 3 and these need to be integers
------------- ----- -------------
2 x 2 x 13 2 x 2 2 x 2 x 5

Multiplying throughout by 2 x 2 gives:

3 x 3 x 3 x 3 and 3 x 3 and 3 x 3 x 3 x 3 and these need to be integers
------------- -------------
13 5

Multiplying throughout 13 gives:

3 x 3 x 3 x 3 and 3 x 3 x 13 and 3 x 3 x 3 x 3 x 13 and these need to be integers
------------------
5

Multiplying throughout 5 gives:

3 x 3 x 3 x 3 x 5 and 3 x 3 x 13 x 5 and 3 x 3 x 3 x 3 x 13

They are all now integers, and we have therefore multiplied by 2 x 2 x 13 x 5.

2 x 2 x 13 x 5 = 260 revolutions. As required.

The easy way from above of 37,440 is exactly 144 times this answer.

Puzzle 8

A. The year 2013 uses four different digits. When did this last happen?

B. The digits in 2013 form a sequence (0, 1, 2, 3). When did this last happen (not necessarily the same digits)?

C. The digits in 2013 form a sequence (0, 1, 2, 3). When will this next happen (not necessarily the same digits)?

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