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Help with past trial question. (1 Viewer)

z3bra

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Can someone please show me working for all parts of this question. Especially stuck on part ii and iii. How do I use part I in part ii.


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BenHowe

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Hey I'll try to help but I haven't done this for a while.



Yeah so I would appreciate some feedback but I hope this helps :)
 
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z3bra

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Hey I'll try to help but I haven't done this for a while.



Yeah so I would appreciate some feedback but I hope this helps :)
Thanks so much. Same I can't seem to figure out part iii.


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seanieg89

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9|2^{2n}+6n-1

Set n=79, so

9|2^{158}+6*79-1

=> 9|2^{159}+6*158-2
=> 9|2^{159}+946
=> 9|2^{159}+1+9*105
=> 9|2^{159}+1.
 

z3bra

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9|2^{2n}+6n-1

Set n=79, so

9|2^{158}+6*79-1

=> 9|2^{159}+6*158-2
=> 9|2^{159}+946
=> 9|2^{159}+1+9*105
=> 9|2^{159}+1.
Can u please explain what u did ? I'm confused.


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BenHowe

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9|2^{2n}+6n-1

Set n=79, so

9|2^{158}+6*79-1

=> 9|2^{159}+6*158-2
=> 9|2^{159}+946
=> 9|2^{159}+1+9*105
=> 9|2^{159}+1.
Aaah ok. So my mistake was not looking at it for all multiples of 9 rather than trying to find a single value.

So you've just done

?
 

seanieg89

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Can u please explain what u did ? I'm confused.


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-First line is the result in the previous part, with a=2,b=3.
-Then I fix n=79 to get a large power of 2 in my expression.
-Then I multiply by 2 to get the correct power of 2 in my expression.
-Then I subtract 9*105 from my expression. This preserves divisibility by 9.
 

seanieg89

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Aaah ok. So my mistake was not looking at it for all multiples of 9 rather than trying to find a single value.

So you've just done

?
Z not R but yes. a|b means that there exists an integer c with ac=b, i.e. a divides b.
 

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