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  1. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 Does anyone have all the papers for this year's comp please?
  2. M

    hi there, I just noticed your message now (1 year late!)... just goes to show how unfamiliar I...

    hi there, I just noticed your message now (1 year late!)... just goes to show how unfamiliar I am with this forum :p no, I'm a long way past year 12, I majored in maths at uni but like to keep in touch with the AMC since I use the Qs for maths tutoring
  3. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 Unfortunately that doesn't follow - take e.g. y = 2, z = 2 and x = -1
  4. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 Has anyone done Q29,30 on the senior paper by the way? I get 176 and 870 for these two Qs respectively...
  5. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 Yes indeed, I think I was reading the wrong half of your factorisation and 72 and copied it without thinking, I meant to write the (2+1)-1 part and leave out the (3+1) coming from 2^3 Sometimes my brain goes to sleep recently... must be because the end of...
  6. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 oops, I missed 15 = 1+2+3+4+5 Thanks for the very informative link... most importantly it defines: "the politeness of x equals the number of odd divisors of x that are greater than one" (emphasis added) So your answer to the original problem is right for...
  7. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 I agree almost verbatim with what you did for Q23 :-) For Q21, I started by writing x^5 = (x^2)^2 * x = (x+3)^2 * x = ..., maybe this gets to the answer a bit quicker??
  8. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 I think you've got the right answer for the wrong reason here... I agree that there are 7 ways of writing 2013 as the sum of at least 2 consecutive positive integers, but there certainly aren't 11 ways of writing 72 as such a sum. Also, your formula...
  9. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 I've just hit on the short and neat way of doing this Q :-) Whenever an odd number of goals have been scored, the teams must be one goal apart. If team X leads team Y in such a situation, the next 2 goals can be scored in 3 ways: XY, YX, YY The...
  10. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 972 is correct, and a nice way to picture things is with a trellis showing possible scores and the numbers of ways of reaching them. Apologies for the crappy image, but since you already have the working I think you can get the picture - the numbers are...
  11. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 Looks good. Here's another variation: Since the line divides the shaded area in half (this is only true because it also divides the whole rectangle in half, by the way), it must past through the midpoint of the line connecting the circles' centres...
  12. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 A slight miscalculation I think... make that 161 for Q27 :-)
  13. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 I agree with both Q28: 495 and Q29: 506. For Q28, a sum of 9 consecutive positive integers is always a multiple of 9 (it's 9 times the middle number). Similarly a sum of 11 consecutive positive integers is always a multiple of 11. For even numbers (when...
  14. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 For 26, 27 and 30 I get 512, 162 and 765 respectively
  15. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 On the junior paper, the easiest way to see that 18 is D is to observe that triangle DNC is equilateral, TM is parallel to DC, and AM is parallel to DN
  16. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 None of the questions are ever intended to be solved by trial and error, though a bit of experimenting with numbers can often set you on the right track :-) For the 37 & 18 question, note first that 37 x 3 = 111 which seems to be heading vaguely in the...
  17. M

    Australian Maths Competition

    Re: Australian Maths Competition 2013 Five 1s are enough: 1101101 = 29759 x 37 + 18
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