STEP 1 2015 Solutions Thread

Alright guys, Dom Staff has requested a separate thread for discussing STEP 1 solutions,so I thought I would jump the gun, hope it all went well for everyone and good luck for STEP II and III if you need it :wink:

Q1: Zacken
Q2: Zacken for part iii, solutions should be StrangeBanana
Q3: Krollo
Q4: Gawain for correct final part StrangeBanana
Q5: Zacken
Q6: mikelbird
Q7: Necrofantasia ranges for a and values of M(a) are StrangeBanana
Q8: StrangeBanana

Q9: Krollo
Q10:
Q11: DFranklin

Q12: qwepoizxcmnb
Q13: Necrofantasia

Solutions given for all questions by MathHysteria

(edited 8 years ago)

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Reply 1

Necrofantasia

Here are my rubbish solution outlines that are worse than the examiners reports since I don't remember the questions themselves
Q2)

Spoiler

Reply 2

Jai Sandhu

Original post by Necrofantasia

Added.

Reply 3

Necrofantasia

Q7)

Spoiler

Someone is going to come along and post a proper solution. These are just placeholders while we have nothing pretty much.

(edited 8 years ago)

Reply 4

Jai Sandhu

Original post by Necrofantasia

Q7)

Spoiler

Someone is going to come along and post a proper solution. These are just placeholders while we have nothing pretty much.

I agree, I am sure Shamika and DFranklin will put something up but yes for the mean time.

Reply 5

Gawain

Original post by Necrofantasia

A much easier way of doing it is to write out the sum of n^k forward and backwards (like the proof for the normal sum of n) beneath one another.

Now since x^k+y^k is divisible by x+k, by adding up the two sums, each term is divisible by n+1. (So 1^k + n^k and 2^k and (n-2)^k are both divisible by n+1).

Now do the same thing but index shift back one to start with 0^k.
Now the sum of each paired term divisible by n (0^k and n^k, 1^k and (n-1)^k )

Since by index shifting we haven't changed the actual sum of n^k we only have to divide by a factor of two to show that the sum of n^k is divisible by (n)(n+1)/2 (or the sum of n).

I'll write it up neatly when I get a hold of the paper but this is a pretty standard technique when considering sums (both the index shift and overlaying the series) although I haven't seen it come up for divisibility questions.

Induction might also be possible but I admit I can't see exactly how.