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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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Sy123

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Re: HSC 2013 4U Marathon

A good Putnam problem that can be done with elementary knowledge



 
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Sy123

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Re: HSC 2013 4U Marathon

Is the answer 1-1/n!-n/(n+1)! ?
I got: 1 - 1/n! + n/(n+1)! (only after a manipulation of my answer, I did not get it directly).

What method did you use?
 

JJ345

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Re: HSC 2013 4U Marathon

Yes, sorry I made a mistake when typing it out it was meant to be 1 - 1/n! + n/(n+1)!
 

seanieg89

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Re: HSC 2013 4U Marathon

A good Putnam problem that can be done with elementary knowledge



No-one else seems to be answering it, so:



by telescoping the two products separately. As N tends to infinity, this expression tends to 2/3 (divide numerator and denominator by N^2 and all "non-leading" terms will die.)

So the infinite product converges to 2/3.
 

shongaponga

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Re: HSC 2013 4U Marathon

Slightly advanced for 3U students + 3U marathon forum seems to be dead so i'll just post it here:

 
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Sy123

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Re: HSC 2013 4U Marathon

Slightly advanced for 3U students + 3U marathon forum seems to be dead so i'll just post it here:





From inspection,




and substituting in appropriately.

Thus



=================



 
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Re: HSC 2013 4U Marathon

The angles of a triangle are such that the largest is one right angle in excess of the smallest. Given that the lengths of the sides of the triangle form an arithmetic sequence, find the ratios of the side lengths
 
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JJ345

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Re: HSC 2013 4U Marathon





From inspection,




and substituting in appropriately.

Thus



=================



If not using the calculator at all, the only logic I can think of is--> we know 2^10 (1024) is slightly greater than 10^3
So (2^10)^100 should be slightly greater than (10^3)^100=10^300--> So it should have slightly more than 300 digits, I would estimate 310(is that in give or take 50 range?)

I'm thinking of finding it through using logs--> But i would probably need a calculator at some stage for that.
 
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Sy123

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Re: HSC 2013 4U Marathon

If not using the calculator at all, the only logic I can think of is--> we know 2^10 (1024) is slightly greater than 10^3
So (2^10)^100 should be slightly greater than (10^3)^100=10^300--> So it should have slightly more than 300 digits, I would estimate 310(is that in give or take 50 range?)
Yep that is quite accurate of an estimate, the correct value is 302

Its better than my method though which was: (10^(log(2) * 1000) ), (using a calculator we get immediately that log(2) = 0.3016....) but my intention was for the answer to contain an attempt using integration to rationally approximate the value of log(2).
I was able to get like 340 using that lol

EDIT: It should be 302 since, 10^1 has 2 digits, 10^3 has 4 digits, thus 10^(301) should have 302 digits.
 
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Drongoski

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Re: HSC 2013 4U Marathon

log102 = 0.3010 ....

.: log10 21000 = 301.02999 ..

.: answer = 301 + 1

(301 is the whole number part before the mantissa of the log of the number and is 1 less than the number of digits of the number excluding the fractional part)
 
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seanieg89

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Re: HSC 2013 4U Marathon

In a class consisting of n mathematics students, every student belongs to exactly one friendship group and each friendship group has at least 2 students.

Let S(n) be the number of possible arrangements of friendship groups.

Find a recursion relation for S(n) and calculate S(n) for n=1 to 6.
 
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