#Mark 3 Putnam questions

23 messages · Page 1 of 1 (latest)

inland rampartBOT
hearty hawk
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for question A1, why are you assuming that A,B,C have to be consecutive?

shell portal
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i used consecutive values as a minimal test to show the bound is tight and all the multiples of 9 can occur

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choosing a,b, c to be consecutive makes the difference between them as small as possible, which minimises (a-b)^2 + (a-c)^2 + (b-c)^2. that gives the smallest non-zero values of the expression

hearty hawk
shell portal
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yeah true

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it doesn't say that. if a= b =c, then using the factorisation the sum of squares is 0, so the whole expression is 0. after that, to find the smallest non-zero values, you minimise the sum of squares subject to it not being zero

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i checked the mark scheme, and my final proof was accurate, however these problems are worth 10 marks. i'm not sure how much my proof will score, given that it's quite brief

hearty hawk
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your conclusion seems off to me anyway - A=2, B=1, C=1 then A^3+B^3+C^3-3ABC = 4

shell portal
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hm

hearty hawk
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I don't know where you are getting the conclusion that 3 divides E from

shell portal
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i supposed 3 divides E, then used the expression 1/2(a+b+c)...
if 3 divides E, it'll have to divide through the 2 factors (a+b+c)(a^2+b^2 +c^2 -ab...)

hearty hawk
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right, so what if 3 doesn't divide E then

shell portal
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how many marks would you suggest i get for my proof

hearty hawk
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it's good that you said it out loud

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i have not marked for the putnam before

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i do not know how they grade i cannot answer that question

shell portal
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they grade for a concise, fully-accurate and well-proven solution, i think my working out was quite brief so it wouldn't hit the max marks

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however, it's a correct proof according to the mark scheme

shell portal
hearty hawk
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no, you will have to ask around

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but you will need to present your proofs more clearly regardless