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# Twisting and Turning

Ben from the UK, Arkadiusz from the Costello School in the UK and Ved from WBGS in the UK worked out how the rope ends up after the series of moves

Having explored the use of twisting and turning I used the example given to

help me solve the first problem. I found an answer of $-\frac8{21}$ having gone through the sequence of

$1,2,3,-\frac13,\frac23,1 \frac23,2 \frac23,-\frac38,\frac58,1 \frac58,2

\frac58$ to finally get to $-\frac8{21}$

Arkadiusz and Ved also worked out how to get back to $0$. This is Arkadiusz' work:

Mohit from Burnt Mill Academy, Harlow in the UK untangled ropes starting from a different state:

TANGLE :

$-\frac{11}{30}$

UNTANGLE:

$-\frac{11}{30}+1=\frac{19}{30}, \\

\frac{19}{30} \rightarrow -\frac1x=-\frac{30}{19},\\

-\frac{30}{19}+1=-\frac{11}{19},\\

-\frac{11}{19}+1=\frac8{19},\\

\frac8{19}\rightarrow -\frac1x=-\frac{19}8, \\

-\frac{19}8+1=-\frac{11}8, \\

-\frac{11}8+1=-\frac38, \\

-\frac38+1=\frac58,\\

\frac58\rightarrow-\frac1x= -\frac85,\\

-\frac85+1=-\frac35, \\

-\frac35+1=\frac25,

\frac25\rightarrow-\frac1x=-\frac52, \\

-\frac52+1=-\frac32,\\

-\frac32+1=-\frac12,\\

-\frac12+1=\frac12, \\

\frac12\rightarrow-\frac1x=-2, \\

-2+1=-1, \\

-1=1=0$

T,R,T,T,R,T,T,T,R,T,T,R,T,T,T,R (T,T)

Arkadiusz described a method to untangle ropes starting from any position:

We keep twisting until we get a positive value, then we turn and repeat over and over until we get zero.

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Ben from the UK, Arkadiusz from the Costello School in the UK and Ved from WBGS in the UK worked out how the rope ends up after the series of moves

Twist, twist, twist, turn, twist, twist, twist, turn, twist, twist, twist, turn

Ben wrote:Having explored the use of twisting and turning I used the example given to

help me solve the first problem. I found an answer of $-\frac8{21}$ having gone through the sequence of

$1,2,3,-\frac13,\frac23,1 \frac23,2 \frac23,-\frac38,\frac58,1 \frac58,2

\frac58$ to finally get to $-\frac8{21}$

Arkadiusz and Ved also worked out how to get back to $0$. This is Arkadiusz' work:

Mohit from Burnt Mill Academy, Harlow in the UK untangled ropes starting from a different state:

TANGLE :

$-\frac{11}{30}$

UNTANGLE:

$-\frac{11}{30}+1=\frac{19}{30}, \\

\frac{19}{30} \rightarrow -\frac1x=-\frac{30}{19},\\

-\frac{30}{19}+1=-\frac{11}{19},\\

-\frac{11}{19}+1=\frac8{19},\\

\frac8{19}\rightarrow -\frac1x=-\frac{19}8, \\

-\frac{19}8+1=-\frac{11}8, \\

-\frac{11}8+1=-\frac38, \\

-\frac38+1=\frac58,\\

\frac58\rightarrow-\frac1x= -\frac85,\\

-\frac85+1=-\frac35, \\

-\frac35+1=\frac25,

\frac25\rightarrow-\frac1x=-\frac52, \\

-\frac52+1=-\frac32,\\

-\frac32+1=-\frac12,\\

-\frac12+1=\frac12, \\

\frac12\rightarrow-\frac1x=-2, \\

-2+1=-1, \\

-1=1=0$

T,R,T,T,R,T,T,T,R,T,T,R,T,T,T,R (T,T)

Arkadiusz described a method to untangle ropes starting from any position:

We keep twisting until we get a positive value, then we turn and repeat over and over until we get zero.

Two brothers were left some money, amounting to an exact number of pounds, to divide between them. DEE undertook the division. "But your heap is larger than mine!" cried DUM...

The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 ï¿½ 1 [1/3]. What other numbers have the sum equal to the product and can this be so for any whole numbers?

Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.