10 Year Anniversary Thread - problems for you to solve

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DFranklin
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Today is Mr M's 10th Year Anniversary on TSR, and it was mine yesterday.

So I thought I'd make a little fun problem thread.

(1) 10! = 3,628,800, so 10! has 2 trailing zeroes. How many trailing zeroes are there in 2007!?

(2) If we write the natural numbers in order, then when we have write 20 digits, we have written 1,2,3,4,5,6,7,8,9,10,11,12,13,14 ,1. The sum of the digits of these numbers is 61. If instead we stop after writing 2008 digits, what is the sum of the digits?

(3) Without using a calculator, what's the value of \dfrac{1+2009^4+2010^4}{1+2009^2  +2010^2}?

(4) Show that there is no function f from Z+ = {0,1,2, ...} s.t. f(f(n)) = n + 2011 for all n.

(5) Exhibit a set S of 2012 distinct positive integers <= 99999 s.t. no 3 elements of S form an AP.

(6) I have a set of 2013 dominoes. Show that by placing dominoes
off center from the ones below, it is possible to build a tower 2013 dominoes high with the top domino 3 dominoes
out from the bottom one. Bonus: can you find the maximum distance to 2 dp?

(7) Suppose we have 2 balls of mass M, m, with M = 2014 m. Ball m lies between ball M and a wall. We now roll ball M towards the wall. Assuming all collisions are perfectly elastic, how many times do the balls touch each other before the larger ball changes direction?

(8) Show that the set of real numbers satisfying \sum_1^{30} \frac{k}{x-k} \ge 3/13 is a union of disjoint intervals of total length 2015

(9) A positive integer is 00-free if its 00 doesn't appear in its binary representation. So 13 = 1101_2 is 00-free but 9 = 1001_2 isn't. If we sort all the 00-free numbers in ascending order, how many binary digits has the 2016th 00-free number?

(10) How many ways are there of making up £20.17 from an infinite supply of pennies, 10 pence pieces, pound coins and £5 notes?

Some of these are harder than others, and I confess some have been cribbed from IMO/BMO questions of the past. I hope there are no mistakes.
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Mr M
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Only five minutes before I have to go to work so I'll pick an easy one.

Question 3

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1+a^2+(a+1)^2=2(a^2+a+1) and 1+a^4+(a+1)^4=2(a^2+a+1)^2 so the answer must be 2009^2+2009+1
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atsruser
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(Original post by DFranklin)
Today is Mr M's 10th Year Anniversary on TSR, and it was mine yesterday.

So I thought I'd make a little fun problem thread.

(1) 10! = 3,628,800, so 10! has 2 trailing zeroes. How many trailing zeroes are there in 2007!?
I know in principle how to do this so I'll leave the details to someone else but:

Hints:

Spoiler:
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1. 2x5=10
2. How often do factors of 2 and 5 appear in 2005! ?




(3) Without using a calculator, what's the value of \dfrac{1+2009^4+2010^4}{1+2009^2  +2010^2}?
Done this, somewhat longwindedly.

(6) I have a set of 2013 dominoes. Show that by placing dominoes
off center from the ones below, it is possible to build a tower 2013 dominoes high with the top domino 3 dominoes
out from the bottom one. Bonus: can you find the maximum distance to 2 dp?
Hmm. Not sure. Maybe:

Spoiler:
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Harmonic series?




(8) Show that the set of real numbers satisfying \sum_1^{30} \frac{k}{x-k} \ge 3/13 is a union of disjoint intervals of total length 2015
I get the general idea, but I'm having trouble seeing how to get the lengths of the intervals out nicely. More thought required.

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I'm guessing that they'll form a GP, but don't see how, if so.

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atsruser
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(Original post by Mr M)
Only five minutes before I have to go to work so I'll pick an easy one.

Question 3

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1+a^2+(a+1)^2=2(a^2+a+1) and 1+a^4+(a+1)^4=2(a^2+a+1)^2 so the answer must be 2009^2+2009+1

I've done this but by hacking out the details - are those identities obvious? Maybe I'm being stupid.
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Mr M
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(Original post by atsruser)
I've done this but by hacking out the details - are those identities obvious? Maybe I'm being stupid.
I expanded and then factorised the denominator and then knew that at least part of the denominator was likely to be a factor of the numerator so it came out quickly.
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Mr M
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I'll do Number 1 but I feel a bit of a fraud as I've seen similar questions before.

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We need to consider multiples of 5 as a multiple of 5 x an even number will provide a trailing zero and there is no shortage of even numbers.

The quotient of 2007/5 is 401 so there are 401 multiples of 5.

But there is more to do.

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25 = 5 x 5 and we have only accounted for one of these 5s so far.

The quotient of 2007/25 is 80.

And we also need to consider higher powers of 5.

The quotient of 2007/125 = 16.

The quotient of 2007/625 = 3.

Total number of trialling zeroes = 401 + 80 + 16 + 3 = 500 which is a pleasingly nice answer!



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DFranklin
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[quote=atsruser]Correct on q6, incorrect on q8. Q8 is adapted from an IMO question and tbh it's the one I'm most likely to have messed up).
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DFranklin
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(Original post by atsruser)
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Correct on q6, incorrect on q8. Q8 is adapted from an IMO question and tbh it's the one I'm most likely to have messed up.
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DFranklin
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Some hints:

Q5: Think in trinary. (Also Cantor Set may be helpful for those who've encountered it).
Q6: You can use an induction argument to show how far out N dominoes can extend. Then you just have a series to estimate/evaluate.
Q7: I've posted this before: You can form+solve a 2nd order recurrence relation, or there's a (hard to see) geometric argument.
Q8: Sketch the curve and you can see what the intervals have to be. The lengths of the intervals (with some adjustments) are the roots of a suitably chosen polynomial, and we know we can link the sum of the roots to a certain coefficient of the polynomial.
Q9: Form a recurrence relation to find a formula for the number of n (binary) digit 00-good numbers. The rest is mechanical.
Q10: Find a formula for the number of ways of making up N pence from 1 and 10 p coins. Use this to find a formula for the number of ways of making up N pence from 1p, 10p and £1 coins. And then once more...
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