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Yeah I would've never got this in the exam...

Okay so (n + k)|n^2 n^2 = (n + k)M for some M integer. Take note: n^2 = (n - k)(n + k) + k^2 (n + k)M = (n - k)(n + k) + k^2 (n + k)(M + k - n) = k^2 Therefore (n + k)|k^2 If k <= sqrt(n) that means n + k > k^2 and thus a contradiction (Take note k, n are positive integers) Therefore k >...

Answers on moodle Edit: omg i know who you are

Why divide by 2! ? Theres no more repeats. EDIT: lel the word rat repeats...

Just to compare answers I got : \frac{8!}{3!2!2!} - \frac{6!}{2!} +4! .

Its funny because I only picked that up when writing my though pattern lel... Need help for the exam

What I did was : Words which contain map - words which containt map and rat (Also take note that rat can appear twice) + missing

The way I did it was that the worst case scenario is when we pick {26,27,...,50} since none are multiples of the other and by the pigeon hole principle we have to pick another number which is a multiple of one of the other numbers already picked.

I love how tense these type of threads are ahahah

I was more trying to say this implies is that the same thing?

If x is and element of A then x is also an element of C (a is a subset of C). I think instead of writing "or" I should've wrote "and" as it is an intersection. So the last step is not required and the proof pretty much is your proof. (Just got confused with and/or). Edit: If it was union...

That's much more straightfoward. But I still think my way is perfectly valid can't see anything wrong with it :) Thanks

Because we have: B^c \subseteq C $ and $ B \subseteq C $ If either of these weren't true then we would already have $ A \subseteq C.

B \cup B^c \subseteq C U \subseteq C \therefore C = U

Can someone check my logic? $ Let A, B, and C be sets. Prove that if $ A - B \subseteq C $ and $ A \cap B \subseteq C $ then $ A \subseteq C. $ $ Let $ x \in A - B \\ \Leftrightarrow \ x \ \in \ A \cap B^c \therefore $ Either $ x \in A $ \Leftrightarrow x \in C \therefore A \subseteq C...

Determine the number of cycles of length 4 that Q_n contains.

Paradoxica is more salty than the Pacific Ocean...

Definition of not cumulative : \forall \ m \ \geq 2 \ \exists \ n \ > m \ $\noindent $S_{n} \not\subset S_2 \cup S_3 \cup \cdots \cup S_m$$ Prove that the sequence S_2,S_3... where S_n = {multiples of n} is not a cumulative function. I let n = m + 1 is this right ?

ind the probability that a hand of 13 cards dealt from a standard pack will have exactly one card in at least one of the four suits. What is this even asking?

What is the negation of : S_n \subset S_2 \cup S_3 ... \cup S_m