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HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)
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simpleetal
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Re: HSC 2015 4U Marathon - Advanced Level
the problem is trivial for primes, so i'll just consider the composite numbers
Case 1: n = a*b, where a>b>1. Then, since a(b-1)-1>0, then ab-1>a>b>1. Thus, b and a divides (ab-1)!.
Case 2: n= a*a, where a>1. Then, since the smallest integer multiple of a which is greater than a is 2a, we need a^2-1 > 2a for a^2 to divide (a^2-1)!. Thus to find out which ones don't divide we consider where a^2-1<2a, and by solving the quadratic and taking the positive component we get a<1+root 2, and since a>1, the only integer value lying between this is a=2, meaning that n=4 does not divide 3!.
Hence, the set is n= all primes & n=4.
will have a look at the combs q later this week when i'm not so busy
Sorry for the lack of latex, on the train home right now and latexing on phone is annoying.!)
the problem is trivial for primes, so i'll just consider the composite numbers
Case 1: n = a*b, where a>b>1. Then, since a(b-1)-1>0, then ab-1>a>b>1. Thus, b and a divides (ab-1)!.
Case 2: n= a*a, where a>1. Then, since the smallest integer multiple of a which is greater than a is 2a, we need a^2-1 > 2a for a^2 to divide (a^2-1)!. Thus to find out which ones don't divide we consider where a^2-1<2a, and by solving the quadratic and taking the positive component we get a<1+root 2, and since a>1, the only integer value lying between this is a=2, meaning that n=4 does not divide 3!.
Hence, the set is n= all primes & n=4.
will have a look at the combs q later this week when i'm not so busy
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Re: HSC 2015 4U Marathon - Advanced Level
Has anyone solved this yet?Or is it the wording of the question that is throwing people off? Here is the exact wording of the question.
Re: HSC 2015 4U Marathon - Advanced Level
lol apparently that perms and combs q is an imo question according to integrand......doubt it was posted as a legit question. maybe glitter can do it though
glittergal96
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Re: HSC 2015 4U Marathon - Advanced Level
Nope, its a semi-recent IMO Q3, which probably makes it v. hard indeed, and it's also in my weakest area of combinatorics. Will try it later if I am bored and have free time, but I'm not confident I would have much success.
Re: HSC 2015 4U Marathon - Advanced Level
Are you from Ruse? Did u ever go to tutoring in your high school years?
glittergal96
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Re: HSC 2015 4U Marathon - Advanced Level
Replace each of the nine integers with their value mod 5.
So the possible values are 0, 1, 2, 3, 4
Case I: All nine integers have the same value
Take any five of the integers - we must get a sum which is a multiple of 5
Case II: The nine integers have two different values
By the pigeonhole principle, one of the values must occur at least 5 times.
Take those five numbers.
Case III: The nine integers have three different values
################
Case IV: The nine integers have four different values
################
Case V: All 5 values are used at least once.
Take one of each value to get a multiple of 5
I'll leave it for someone else to contemplate cases 3 & 4
(I'm not saying this is the way to attack it - they are simply my thoughts so far)
A start on this question:
Replace each of the nine integers with their value mod 5.
So the possible values are 0, 1, 2, 3, 4
Case I: All nine integers have the same value
Take any five of the integers - we must get a sum which is a multiple of 5
Case II: The nine integers have two different values
By the pigeonhole principle, one of the values must occur at least 5 times.
Take those five numbers.
Case III: The nine integers have three different values
################
Case IV: The nine integers have four different values
################
Case V: All 5 values are used at least once.
Take one of each value to get a multiple of 5
I'll leave it for someone else to contemplate cases 3 & 4
(I'm not saying this is the way to attack it - they are simply my thoughts so far)
simpleetal
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Re: HSC 2015 4U Marathon - Advanced Level
Since no one is posting their soln, due to everyone being so busy, I'll post mine. there could be more elegant methods, but this was what i came up with when i did the question a couple of weeks back (didn't look for an elegant solution after solvign it)
When I say “class”, i mean numbers with the same remainder upon division by 5.
Case 1: There is at least 5 numbers in the same class
Then, by choosing 5 of this class we have a sum divisible by 5
Case 2: There are 2 classes present
This means that there are at least 5 numbers of the same class, which is case 1
Case 3: There are 3 different classes present
Either the set is eliminated by case 1, or there exist two subsets of at least three numbers with the same residue mod 5. Subtract the same integer from each element of the given 9 integer set, until there are 3 numbers with 0 mod5 (this would eventually occur since we had 2 subsets of elements with identical residues mod 5, and each subset consists of at least 3 elements). If, in this new set there exists two numbers whose residues mod 5 only differ in sign, (for instance 1 and -1) then we are done, by combining 3 numbers of 0 mod 5, and the two numbers whose remainders cancel each other out. If there are no such two numbers, then aside from zero, the only possible sets to consider at those consisting of the elements {0,1,-2}, {0,2,-1}, {0,2,1} and {0,-2,-1} mod5.
A set consisting of {0,1,-2} is congruent to {-1,0,2} (since -3 congruent to 2 (mod5))and {2,0,1}congruent to {0,-2,-1} in the mod 5 system so we only have to consider the first of each.
{0,1,-2}: we know that there are three zeroes in this set and at least three of either 1 or -2. If we have three of -2s, then we are done. Suppose that we have either one or two numbers of remainder with -2 (or 3) when div by 5. Then, this implies that there are at least two numbers with residue 1 mod 5, and so we can construct the set {0,1,-2,1,0}.
{2,0,1}: We know that there are three zeroes in this set also, and at least three of either 1 or 2. If there are more than 3 numbers with remainder 2 when div by 5, we can construct the set {0,2,1,2,0}. If there is less than 3 numbers iwth remainder 2 when div by 5, there are at least 3 numbers with remainder 1 when div by 5. So then we can construct the set {0,1,2,1,1).
NOTE: Subtracting an integer from each element of the set and constructing a subset divisible proves that such a subset exists in the original set, i.e. if in the modified set we were able to construct a subset (a,b,c,d,e) whose sum is idivisble by 5, then in the original set we can construct (a+n,b+n,c+n,d+n,e+n) whose sum is still 0 mod 5.
Case 4: There are 4 types of numbers present
Then, there must be at least three numbers of the same residue mod 5.. Whatever set of numbers you have, subtract an integer from each element until there are 3 or more numbers with residue 0 mod 5. In addition, there must be two other numbers whose remainders upon div by 5 have same magnitude but different directions in mod 5 system. Hence, by combining those 3 zeroes and the two numbers which cancel each other out, we have a sum of zero mod 5. Obviously, this means that you can construct a set of 5 with zero mod 5 from the original set before the subtraction occurred, since n and n+5m have the same residue mod 5. (I could too brief in my explanation here, if so then plz tell me)
Case 5: There are 5 different numbers in the set.
then, by choosing those five numbers, we have 1+2+3+4+5 (mod 5) = 0 (mod 5) and hence the sum is divisible by 5
Since no one is posting their soln, due to everyone being so busy, I'll post mine. there could be more elegant methods, but this was what i came up with when i did the question a couple of weeks back (didn't look for an elegant solution after solvign it)
When I say “class”, i mean numbers with the same remainder upon division by 5.
Case 1: There is at least 5 numbers in the same class
Then, by choosing 5 of this class we have a sum divisible by 5
Case 2: There are 2 classes present
This means that there are at least 5 numbers of the same class, which is case 1
Case 3: There are 3 different classes present
Either the set is eliminated by case 1, or there exist two subsets of at least three numbers with the same residue mod 5. Subtract the same integer from each element of the given 9 integer set, until there are 3 numbers with 0 mod5 (this would eventually occur since we had 2 subsets of elements with identical residues mod 5, and each subset consists of at least 3 elements). If, in this new set there exists two numbers whose residues mod 5 only differ in sign, (for instance 1 and -1) then we are done, by combining 3 numbers of 0 mod 5, and the two numbers whose remainders cancel each other out. If there are no such two numbers, then aside from zero, the only possible sets to consider at those consisting of the elements {0,1,-2}, {0,2,-1}, {0,2,1} and {0,-2,-1} mod5.
A set consisting of {0,1,-2} is congruent to {-1,0,2} (since -3 congruent to 2 (mod5))and {2,0,1}congruent to {0,-2,-1} in the mod 5 system so we only have to consider the first of each.
{0,1,-2}: we know that there are three zeroes in this set and at least three of either 1 or -2. If we have three of -2s, then we are done. Suppose that we have either one or two numbers of remainder with -2 (or 3) when div by 5. Then, this implies that there are at least two numbers with residue 1 mod 5, and so we can construct the set {0,1,-2,1,0}.
{2,0,1}: We know that there are three zeroes in this set also, and at least three of either 1 or 2. If there are more than 3 numbers with remainder 2 when div by 5, we can construct the set {0,2,1,2,0}. If there is less than 3 numbers iwth remainder 2 when div by 5, there are at least 3 numbers with remainder 1 when div by 5. So then we can construct the set {0,1,2,1,1).
NOTE: Subtracting an integer from each element of the set and constructing a subset divisible proves that such a subset exists in the original set, i.e. if in the modified set we were able to construct a subset (a,b,c,d,e) whose sum is idivisble by 5, then in the original set we can construct (a+n,b+n,c+n,d+n,e+n) whose sum is still 0 mod 5.
Case 4: There are 4 types of numbers present
Then, there must be at least three numbers of the same residue mod 5.. Whatever set of numbers you have, subtract an integer from each element until there are 3 or more numbers with residue 0 mod 5. In addition, there must be two other numbers whose remainders upon div by 5 have same magnitude but different directions in mod 5 system. Hence, by combining those 3 zeroes and the two numbers which cancel each other out, we have a sum of zero mod 5. Obviously, this means that you can construct a set of 5 with zero mod 5 from the original set before the subtraction occurred, since n and n+5m have the same residue mod 5. (I could too brief in my explanation here, if so then plz tell me)
Case 5: There are 5 different numbers in the set.
then, by choosing those five numbers, we have 1+2+3+4+5 (mod 5) = 0 (mod 5) and hence the sum is divisible by 5
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simpleetal
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Re: HSC 2015 4U Marathon - Advanced Level
If there are any things which i haven't quite made clear pls let me know, then I'll explain when i next log on
If there are any things which i haven't quite made clear pls let me know, then I'll explain when i next log on
Re: HSC 2015 4U Marathon - Advanced Level
I was hoping that the partial solution which I posted only an hour earlier would have prompted someone to join in before your final solution was posted.
simpleetal
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Re: HSC 2015 4U Marathon - Advanced Level
Just came back from a 3 day camp and saw this, so I may have made some silly mistakes. NOTE: the LAtex takes a bit of time to load, so please be patient
of this rectangular slice is given by similar triangles:$ \frac{a}{x}=\frac{k-h-j}{k-h}. \ a= \frac{x(k-h-j)}{k-h} \ $where x$= \frac{s(H-h)}{H}. \\ $So a$ =\frac{S(H-h)(k-h-j)}{H(k-h)}. )
Just came back from a 3 day camp and saw this, so I may have made some silly mistakes. NOTE: the LAtex takes a bit of time to load, so please be patient
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simpleetal
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Re: HSC 2015 4U Marathon - Advanced Level
}{H}. $ Summing the thin slices from j= k-h to 0 gives us:$ V_{khP}= \lim_{\delta j\rightarrow 0}\sum_{0}^{k-h} ab\delta j = \int_{0}^{k-h} \frac{s^2(H-h)(H-j-h)(k-h-j)}{H^2(k-h)}. \ $which equals to$ \frac{(k+2h-3H)(s^2))(Hh-kH-h^2+hk)}{6H^2}. \ $Then, add frustrum volume to get the volume of lower piece as $ \ \ V_{lower piece}=\frac{(k+2h-3H)(s^2))(Hh-kH-h^2+hk)}{6H^2} +\frac{Hs^2}{3}-\frac{(H-h)s^2(H-h)^2}{3H^2} )
Please notify me if there are silly algebra mistakes. Also, you have to wait for a bit for the latex to load (happened to me)
Just came back from a 3 day camp and saw this, so I made have made some silly mistakes.
of this rectangular slice is given by:$ \frac{a}{x}=\frac{k-h-j}{k-h}. \ a= \frac{x}{k-h} \ $where x$= \frac{s(H-h)}{H}. \\ $So a$ =\frac{S(H-h)(k-h-j)}{H(k-h)}. )
Please notify me if there are silly algebra mistakes. Also, you have to wait for a bit for the latex to load (happened to me)
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Re: HSC 2015 4U Marathon - Advanced Level
)$'$)
part does not make sense dimensionally, but you got the right integral and working out before that, so well done.
I think maybe there's a typo/error in the evaluation of the integral, since the
Please notify me if there are silly algebra mistakes. Also, you have to wait for a bit for the latex to load (happened to me)
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simpleetal
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Re: HSC 2015 4U Marathon - Advanced Level
ah yep that was a typo, now fixed. (forgot a bracket)I think maybe there's a typo/error in the evaluation of the integral, since the
Re: HSC 2015 4U Marathon - Advanced Level
Yep thought it was something like this (I didn't fully expand everything out when I did the Q).
simpleetal
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glittergal96
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Re: HSC 2015 4U Marathon - Advanced Level
The polynomial simplifies to n^3-18n^2-54=0.
So any solution must satisfy n^2(n-18)=54.
But the only squares that divide 54 are 1,9. So the only possible solutions are +-1,+-3. But at any of these values it is easy to see the second LHS factor now does not divide the RHS. Contradiction.
So no solutions exist.
The polynomial simplifies to n^3-18n^2-54=0.
So any solution must satisfy n^2(n-18)=54.
But the only squares that divide 54 are 1,9. So the only possible solutions are +-1,+-3. But at any of these values it is easy to see the second LHS factor now does not divide the RHS. Contradiction.
So no solutions exist.
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