IOQM 2023 Apple Mock Two

30 June 20233 minutesviews

This article contains spoilers for the mock test.

I recommend attempting the problems before proceeding to read the article. You can download the problems and the answer key using the buttons below.

This is the second IOQM Mock test from MOMC. The problems are taken from PuMaC 2019 and CHMMC 2013. All the credit and authorship of the problems belongs to their respective sources.

This time, there are no bonus problems since I only used two sources. I would like to discuss a beautiful algebra and a number theory problem from the mock. So if you are planning to attempt the mock, beware of the spoilers ahead.

Problem 24: (PUMaC 2019 Algebra A P3)\textbf{(PUMaC 2019 Algebra A P3)} Let QQ be a quadratic polynomial. If the sum of the roots of Q100(x)Q^{100}(x) (where Qi(x)Q^i(x) is defined by Q1(x)=Q(x),Qi(x)=Q(Qi1(x))Q^1(x)=Q(x), Q^i(x)=Q\left(Q^{i-1}(x)\right) for integers i2)\left.i \geq 2\right) is 88 and the sum of the roots of QQ is SS, find log2(S)\left|\log _2(S)\right|.

Solution: Such a strange problem. We are just given the sum of roots of QQ composed 100100 times to itself, and are asked to find the sum of roots of QQ. Denote the sum of roots of Qi(x)Q^i(x) by SiS_i. Let us try to come up with a relation between consecutive terms of the sequence SiS_i. Let Q(x)=a(xb)(xc)Q(x) = a(x-b)(x-c). Therefore:

Qi+1(x)=a(Qi(x)b)(Qi(x)c)Q^{i+1}(x) = a(Q^i(x)-b)(Q^i(x)-c)

Now Qi+1(x)Q^{i+1}(x) vanishes when Qi(x)=bQ^i(x) = b or cc. Thus Si+1S_{i+1} is the sum of all solutions to Qi(x)=b,cQ^i(x) = b,c. Now the key observation here is that, by Vieta, the sum of roots of a polynomial PP depends only on the coefficient of xd,xd1x^d, x^{d-1} where dd is the degree of PP. Clearly degree of Qi(x)Q^i(x) is atleast 22. Thus a change of b-b or c-c in the constant term doesn't make a difference in the sum of roots. Thus the sum of roots of Qi(x)b=0Q^i(x)-b = 0 or Qi(x)c=0Q^{i}(x)-c=0 is Si+Si=2SiS_i+S_i = 2S_i. Thus we have the simple recurrence Si+1=2SiS_{i+1} = 2S_{i}. Now it is given that S100=8S_{100} = 8. Thus S1S_1 is 8299=296\frac{8}{2^{99}} = 2^{-96}. Hence the answer is 9696.

Problem 25: (CHMMC 2013 Teams Round P10)\textbf{(CHMMC 2013 Teams Round P10)} Compute the lowest positive integer kk such that none of the numbers in the sequence {1,1+\{1,1+ k,1+k+k2,1+k+k2+k3,}\left.k, 1+k+k^2, 1+k+k^2+k^3, \cdots\right\} are prime.

Solution: Our main aim is to find a small positive integer kk satisfying the problem. This would be, by wishful thinking, the lowest such positive integer kk. Now let us try to factorize the terms in the sequence to show that they are composite. Let Tn=1+k++knT_n = 1+ k + \cdots + k^n. Firstly its easy to see that if nn is odd, then 1+k1+k divides TnT_n. Thus if n>1n>1 is odd then TnT_n is composite. Note that we need to separately ensure that 1+k1+k is composite.

Now for the case when nn is even, recall the identity: 1+k2+k4=(1+k+k2)(1k+k2)1 + k^2 + k^4 = (1+k+k^2)(1-k+k^2). Note that this can be extended as follows:

1+k2+k4++k4m=(1+k+k2++k2m)(1k+k2+k2m)1 + k^2 + k^4 + \cdots + k^{4m} = (1+k+k^2 + \cdots + k^{2m})(1-k + k^2 - \cdots + k^{2m})

The issue here is that we miss the odd exponent terms. But if kk itself is a perfect square then we can apply the above identity to show that TnT_n is composite for all even nn.

Suppose kk is a perfect square. Then TnT_n is composite for all n>1n>1. For n=1n=1 we require k+1k+1 to be composite. Thus 1,41,4 don't work and 99 is the smallest perfect square that works. Now its just a matter of manual computations to check that no k<9k<9 works.

Also check out P27 and P28, as they are quite nice as well.