Help with past trial question. (1 Viewer)

z3bra · May 24, 2017

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.

Sent from my iPhone using Tapatalk

BenHowe · May 24, 2017

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

$(i)\(b-1)^{2n}=\binom{2n}{0}b^{0}(-1)^{2n}+\binom{2n}{1}b^{1}(-1)^{2n-1}+...\\+\binom{2n}{2n-1}b^{2n-1}(-1)^{1}+\binom{2n}{2n}b^{2n}(-1)^{0}\vspace{0.5cm}\\ (ii)\ \text{This is equal to}\1-2nb+\binom{2n}{2}b^{2}-...-2nb^{2n-1}+b^{2n}\\ \text{Take the 1 and the -2nb over to the other side and replace}\ (b-1)\ \text{with}\ (a) \\a^{2n}+2nb-1=\binom{2n}{2}b^{2}-...-2nb^{2n-1}+b^{2n}\\ \text{You can now take a factor of b squared out of the RHS giving}\\a^{2n}+2nb-1=b^{2}(\binom{2n}{2}-...-2nb^{2n-3}+b^{2n-2})\\ \therefore \frac{a^{2n}+2nb-1}{b^{2}}=\binom{2n}{2}-...-2nb^{2n-3}+b^{2n-2}\\ b^{2}\ \text{divides}\ a^{2n}+2nb-1,\ \text{thus it is exactly divisible by}\ b^{2}\vspace{0.5cm}\\ (iii)\ \text{This is where I had trouble as it clear from inspection you want to set b to be a multiple of 3, so that b squared will be divisible by 9}\\ \text{So let}\ b=3, a=2\ \rightarrow 2^{2n}+6n=2^{159}+2\\ \text{I got n to be 79.5 which contradicts the initial condition where}\ n\in\mathbb{Z^{+}}$

Yeah so I would appreciate some feedback but I hope this helps

z3bra · May 24, 2017

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

$(i)\(b-1)^{2n}=\binom{2n}{0}b^{0}(-1)^{2n}+\binom{2n}{1}b^{1}(-1)^{2n-1}+...\\+\binom{2n}{2n-1}b^{2n-1}(-1)^{1}+\binom{2n}{2n}b^{2n}(-1)^{0}\vspace{0.5cm}\\ (ii)\ \text{This is equal to}\1-2nb+\binom{2n}{2}b^{2}-...-2nb^{2n-1}+b^{2n}\\ \text{Tale the 1 and the -2nb over to the other side and replace}\ (b-1)\ \text{with}\ (a) \\a^{2n}+2nb-1=\binom{2n}{2}b^{2}-...-2nb^{2n-1}+b^{2n}\\ \text{You can now take a factor of b squared out of the RHS giving}\\a^{2n}+2nb-1=b^{2}(\binom{2n}{2}-...-2nb^{2n-3}+b^{2n-2})\\ \therefore \frac{a^{2n}+2nb-1}{b^{2}}=\binom{2n}{2}-...-2nb^{2n-3}+b^{2n-2}\\ b^{2}\ \text{divides}\ a^{2n}+2nb-1,\ \text{thus it is exactly divisible by}\ b^{2}\vspace{0.5cm}\\ (iii)\ \text{This is where I had trouble as it clear from inspection you want to set b to be a multiple of 3, so that b squared will be divisible by 9}\\ \text{So let}\ b=3, a=2\ \rightarrow 2^{2n}+6n=2^{159}+2\\ \text{I got n to be 79.5 which contradicts the initial condition where}\ n\in\mathbb{Z^{+}}$

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.

Sent from my iPhone using Tapatalk

seanieg89 · May 24, 2017

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 · May 24, 2017

seanieg89 said:
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.

Sent from my iPhone using Tapatalk

BenHowe · May 25, 2017

seanieg89 said:
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

$\2^{2n}+6n-1=9r\ \text{such that}\ r\in\mathbb{R}$ ?

seanieg89 · May 25, 2017

z3bra said:
Can u please explain what u did ? I'm confused.

Sent from my iPhone using Tapatalk

-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 · May 25, 2017

BenHowe said:
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

$\2^{2n}+6n-1=9r\ \text{such that}\ r\in\mathbb{R}$ ?

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

BenHowe · May 25, 2017

seanieg89 said:
Z not R but yes. a|b means that there exists an integer c with ac=b, i.e. a divides b.

Ahhh ok cool. Thanks heaps.

Help with past trial question. (1 Viewer)

z3bra

New Member

BenHowe

Active Member

z3bra

New Member

seanieg89

Well-Known Member

z3bra

New Member

BenHowe

Active Member

seanieg89

Well-Known Member

seanieg89

Well-Known Member

BenHowe

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)