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STEP 2005 Solutions Thread

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Reply 60
Avatar for davros
davros
16
Original post by rath90
Hi there how you work out the first few terms of the expansion (1+(3/x))^-1 ?


Standard binomial expansion :smile:
Original post by Glutamic Acid
II/4:

1st part



2nd part



3rd part




For part III does it matter which values you choose?
Original post by Principia
For part III does it matter which values you choose?


In order to use the previous result you must choose values so that p+q+s=13,p+q+r=21,p+r+u=87 ,p+r+v-187 and p=7
Original post by brianeverit
In order to use the previous result you must choose values so that p+q+s=13,p+q+r=21,p+r+u=87 ,p+r+v-187 and p=7


I did that, but can it be any values that meet those criteria? i.e. there are then an infinite amount of solutions.
Original post by Principia
I did that, but can it be any values that meet those criteria? i.e. there are then an infinite amount of solutions.


Yes, the solution is not unique but you must check that your values satisfy all the criteria.
Reply 65
Avatar for tccheng
tccheng
Original post by SimonM
STEP III, Question 4

Spoiler



Very nice approach by Simon. In (ii), probably need to exclude period =1, i.e. a=b and k=2 (c=1).
Original post by SimonM
STEP II, Question 1

Spoiler



when comparing coefficients did you put this
Let P=x4+px3+qx2+rx+sP = x^4+px^3+qx^2+rx+s

P′=4x3+3px2+2qx+rP' = 4x^3+3px^2+2qx+r

into


Therefore P′(x)−2xP(x)=kx(x2−a2)(x2−b2)P'(x) - 2xP(x) = kx(x^2-a^2)(x^2-b^2)
An almost complete solution to STEP III Q 12:
Reply 68
Avatar for bobin
bobin
4
Original post by nuodai
Ah what the hell,
STEP I 2005 Question 1

Part (i)



Part (ii)




One of my students just came up with what I think is a really great way of solving part (b), which requires no splitting into cases, and in my opinion is more elegant than the board's published answer (and the answer already suggested on this thread):

Solution as follows:

The maximum sum is 45, by using 99999 (uniquely). To make 39, we must subtract a total of six from the digits of 99999, so use the symbol x to represent subtracting one from a digit, and / to represent moving on to the next digit in the number [e.g. x/x/x/x/xx would mean subtract one from the 1st, 2nd, 3rd, and 4th digits, and subtract two from the 5th digit; //xxx/x/xx would mean subtract nothing from the 1st and 2nd digits, then three from the 3rd digit, one from the 4th digit, and two from the 5th digit]. But now the problem just becomes how many ways are there of arranging 6 "x" symbols and 4 "/" symbols, i.e.10C4 = 210.
Reply 69
Avatar for duuuuu
duuuuu
Original post by sonofdot
STEP II 2005 Question 8

First Part


Sketch One


Sketch Two




if we consider the sketch part for x<0, Is it symmetric of x>0?

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