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Uni Admission Maths test help?

q8.PNGq3.PNGq2.PNGHi, I've been trying to prepare for uni maths admission tests but half the content are topics that I haven't even covered or heard about in my A level syllabus. As a result I really need some urgent help, please post your method too, thank you.
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For 8. Consider n+n+1+n+2+n+3=4n+6n+n+1+n+2+n+3=4n+6. This is not a strictly divisible by 66 in this form ∀n∈Z\forall n \in \mathbb{Z}. However let us assume n=3kn=3k, for k∈Zk \in \mathbb{Z}, then our expression is 4(3k)+6=12k+6=6(2k+1)4(3k)+6=12k+6=6(2k+1) which is divisible by 6. Hope this helps. I am not willing to provide solutions to all the problems but if you need any hints message me.

I assume you are familiar with the notation used. If not just say.
(edited 7 years ago)
@Jas1947 A hint for question 3. What does squaring do and what can result from it. i.e. (−4)2=42(-4)^{2}=4^{2}
Reply 3
Avatar for Minnie16
Minnie16
OP
8
Original post by Cryptokyo
For 8. Consider n+n+1+n+2+n+3=4n+6n+n+1+n+2+n+3=4n+6. This is not a strictly divisible by 66 in this form ∀n∈Z\forall n \in \mathbb{Z}. However let us assume n=3kn=3k, for k∈Zk \in \mathbb{Z}, then our expression is 4(3k)+6=12k+6=6(2k+1)4(3k)+6=12k+6=6(2k+1) which is divisible by 6. Hope this helps. I am not willing to provide solutions to all the problems but if you need any hints message me.

I assume you are familiar with the notation used. If not just say.


I got to the part where 4n+6 but couldn't get any further, I understand that now. Thank you :smile:
Original post by Jas1947
I got to the part where 4n+6 but couldn't get any further, I understand that now. Thank you :smile:


For 5. Take the log of both sides i.e.log⁡3(25)≈log⁡3(33)\log_{3}(2^{5})\approx\log_{3}(3^{3})
Reply 5
Avatar for Minnie16
Minnie16
OP
8
Original post by Cryptokyo
@Jas1947 A hint for question 3. What does squaring do and what can result from it. i.e. (−4)2=42(-4)^{2}=4^{2}


Oh I see, x cannot equal -4 because you cannot square root a negative
Original post by Cryptokyo
For 8. Consider n+n+1+n+2+n+3=4n+6n+n+1+n+2+n+3=4n+6. This is not a strictly divisible by 66 in this form ∀n∈Z\forall n \in \mathbb{Z}. However let us assume n=3kn=3k, for k∈Zk \in \mathbb{Z}, then our expression is 4(3k)+6=12k+6=6(2k+1)4(3k)+6=12k+6=6(2k+1) which is divisible by 6. Hope this helps. I am not willing to provide solutions to all the problems but if you need any hints message me.

I assume you are familiar with the notation used. If not just say.

What is the symbol next to the n called?
Original post by Jas1947
Oh I see, x cannot equal -4 because you cannot square root a negative


You are half correct. You identified the incorrect solution but for the wrong reason. The reason is the square root function is only the positive answer as a function can only have a one-to-one or many-to-one mapping. i.e. y=∣x∣y=|\sqrt{x}|
Original post by Dynamic_Vicz
What is the symbol next to the n called?


I am not sure of its name but it means "for all". In LaTeX you would write \forall.

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