Australian Maths Competition (3 Viewers)

sharin_3

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The amount of zeros a number ends in is the amount of 10s that are in the prime factorization. Now 10 = 2 x 5. And so when we prime factorize 2008! we need to count the amount of 2s and the amount of 5s.

Now there is clearly going to be more 2s than 5s, so to determine how many factors of 10 there are in the prime factorization of 2008! we must only find how many 5s are in the prime factorization.

Now clearly finding the prime factorization of 2008! would be a torturous task, so we will "look" at each term of 2008! ie 1, 2, 3, 4, 5, 6, . . . ,2007, 2008.

So how many multiples 5s are there between 1 and 2008, well





As m is an integer then there are 401.

Now some terms in the prime factorization will have 25 as a factor, and hence that terms will have 2 factors of 5. So we must count how many terms are multiples of 25. Now we already have counted 1 of those factors of 5 (out of those numbers that are multiples of 5^2), and so when we determine how many factors are multiples of 5^2, they will add 1 extra multiple of 5 into our count.





As n is an integer then there are 80.

But some of the factors of 2008! will be multiples of 5^3 = 125, so when we count these by the same logic as before they will contribute an extra 1 multiple of 5 into our count, so





As p is an integer then there are 16.

But some of the factors of 2008! will be multiples of 5^4 = 625, and so these will add an extra 1 multiple of 5 into our count. So,





As j is an integer there are 3 of these.

As 5^5 is greater than 2008, then there are no multiples of 5^5 up to 2008.



So in total the amount of factors of 5 in 2008! are



Therefore there are 500 factors of 10 in 2008! and hence 2008! ends in 500 zeros.




Good luck everyone for the competition today =)
THANKS A LOT! :guitar:
 

Kajixtatsu

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Re: How did people go in the maths comp today?

What'd you get for question 14?
That was unusual... because they had to be integers you were obviously including the squares... 0, 1, 4, 9, 16... 49 and four quadrants but that gives 32 , I don't know if I doubled up =| but I went with B) 30
 

gurmies

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Re: Australian Mathematics Competition

I got 48

Keep in mind a pair like 3, root[41] satisfied it
 
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khorne

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Re: Australian Mathematics Competition

I don't think this counts for a valid solution...But for the ascending question, I just divided the 4 digit number given by 6...and it worked...6 marks in 1 minute.

Personally, I don't care about the mark...I only take the test to steal the question booklet afterward =]
 
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gurmies

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Re: Australian Mathematics Competition

i don't think this counts for a valid solution...but for the ascending question, i just divided the 4 digit number given by 6...and it worked...6 marks in 1 minute.
lol
 

maths94

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Re: Australian Mathematics Competition

hey i did the intermidate paper ...wat did every1 get for the rabbit one? also wat did they get for q30 the train one...also q29 the pattern wat was there answer for that..
 

Official

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Re: Australian Mathematics Competition

hey i did the intermidate paper ...wat did every1 get for the rabbit one? also wat did they get for q30 the train one...also q29 the pattern wat was there answer for that..
Kurt already answered the rabbit one.
 

kurt.physics

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Re: Australian Mathematics Competition

hey i did the intermidate paper ...wat did every1 get for the rabbit one? also wat did they get for q30 the train one...also q29 the pattern wat was there answer for that..
The pattern was 6, 6 + 9, 6 + 9 + 12, 6 + 9 + 12 + 15, . . .

So we have to find the 11th term, which is the sum of the first 11 terms of the arithmetic sequence with a = 6, d = 3.







 

kurt.physics

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Re: Australian Mathematics Competition

I don't think this counts for a valid solution...But for the ascending question, I just divided the 4 digit number given by 6...and it worked...6 marks in 1 minute.

Personally, I don't care about the mark...I only take the test to steal the question booklet afterward =]
How did you think you went? =)
 

Aquawhite

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Re: Australian Mathematics Competition

I don't think this counts for a valid solution...But for the ascending question, I just divided the 4 digit number given by 6...and it worked...6 marks in 1 minute.

Personally, I don't care about the mark...I only take the test to steal the question booklet afterward =]
This!
 

occer

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Re: 2009 AMT Mathematics Competition

I pretty much gave up from 16 onwards... didnt even try the last 5 questions. I know, really bad.
Join the club.

Hopefully credit.
 

Winn

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Re: Australian Mathematics Competition

I don't think this counts for a valid solution...But for the ascending question, I just divided the 4 digit number given by 6...and it worked...6 marks in 1 minute.
OMGGG why didnt i think of that????!!!!
i reckon the maths comp committee purposely did that just to tease us.:(
 

Official

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Re: Australian Mathematics Competition

OMGGG why didnt i think of that????!!!!
i reckon the maths comp committee purposely did that just to tease us.:(
Heh, yeh I bet you they did lol. Darn if I found out about that then it would have saved me the 5minutes it took me to answer the question :mad:
 

Winn

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Re: Australian Mathematics Competition

Heh, yeh I bet you they did lol. Darn if I found out about that then it would have saved me the 5minutes it took me to answer the question :mad:
how do you actually do that question?
= =MAN i feel inferior to you guys lol.:hammer:
 

lpodnano

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Re: Australian Mathematics Competition

I only got one of the ten mark questions and then our teacher said we have 5 min left.
 

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