How to solve Shonk Sequences? A Shonk sequence is a sequence of positive integers in which


*

*each term after the first is greater than the previous term, and

*the product of all the terms is a perfect square


For example: 2, 6, 27 is a Shonk sequence since 6>2 and27>6 and 2*6*27 = 324 or 18^2
a. If 12, x, 24 is a Shonk sequence, what is the value of x?
b. If 28, y, z, 65 is a Shonk sequence, what are the values of y and z?
c. Determine the length of the longest Shonk sequence, each of whose terms is an integer between 1 and 12, inclusive. Your solution should include an example of this longest length, as well as justification as to why no longer sequence is possible.
d. A sequence of four terms a,b,c,d is a super-duper-Shonkolistic sequence (SDSS) exactly when a,b,c,d and a,b,c and b,c,d are a Shonk sequence. Determine the number of pairs (m, n) such that m, 1176, n, 48400 is an SDSS.

This question was taken directly from the Grade 9 Fryer Contest for the University of Waterloo. Permission has been granted to discuss these questions online as the contests are officialy over.
I'd like to know the answers and the steps of the solution. Thanks.
 A: Although you don't show any work in your question text, I assume you have at least tried to answer them yourself. If you haven't, I suggest you do this first before looking at the solutions hidden below.
For question (a):

 For $12, x, 24$ to be a Shonk sequence, note that $12 \times 24 = 2^5 \times 3^2$. Thus, $x$ must be $2$ times a perfect square. The only such value between $12$ and $24$ is $18$, so $x = 18$.

For question (b):

 For $28, y, z, 65$ to be a Shonk sequence, $28 \times 65 = 2^2 \times 5 \times 7 \times 13$. Thus, $yz$ must be $5 \times 7 \times 13$ times a perfect square. $y$ and/or $z$ must contain a factor of $13$, with only $39 = 3 \times 13$ and $52 = 4 \times 13$ being between $28$ and $65$. As neither value has a factor of $5$ or $7$, this means the other unknown must be a multiple of $35$, with only $35$ being in the range. Therefore, $35$ is one value, and the other one must be $13$ times a perfect square, so it must be $52$, giving that $y = 35$ and $z = 52$.

For question (c):

 It's easiest to consider the longest possible Shonk sequence by examining the prime factors decomposition of $12!$ to determine the minimum factors which can't be included. For the power of $2$, it is $\lfloor \frac{12}{2} \rfloor + \lfloor \frac{12}{4} \rfloor + \lfloor \frac{12}{8} \rfloor = 6 + 3 + 1 = 10$. Using similar calculations for the other prime factors gives that $12! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11$. To get a perfect square, each power must be even, so at least one factor of $3$, plus the factors of $7$ and $11$, can't be included. As no single integer from $1$ to $12$ contains $2$ or more of these factors, at least $3$ numbers must be excluded. As excluding $3, 7, 11$ satisfies the conditions, this shows that the $9$ remaining values of $1, 2, 4, 5, 6, 8, 9, 10, 12$ forms the longest possible Shank sequence.

For question (d):

 For an SDSS, with $a,b,c,d$ and $a,b,c$ and $b,c,d$ being Shonk sequences means that $abcd$, $abc$ and $bcd$ are each perfect squares. As such, $\frac{abcd}{abc} = d$ and $\frac{abcd}{bcd} = a$ must also be perfect squares. In addition, then, $\frac{abc}{a} = bc$ must be a perfect square too. For $m, 1176, n, 48400$ to be a SDSS, $m$ must be a perfect square. Since $\sqrt{1176} = 34.2\ldots$, then $m$ is $i^2$ for $1 \le i \le 34$ resulting in $34$ such values. Since $1176 = 2^3 \times 3 \times 7^2$, for $1176 \times n$ to be a perfect square requires that $n$ be $6$ times a perfect square $i^2$ with $i \gt 2 \times 7 = 14$. As $\sqrt{\frac{48400}{6}} = 89.8\ldots$, then $n$ is $6 \times i^2$ for $15 \le i \le 89$ resulting in $75$ such values. In total, there are $34 \times 75 = 2550$ pairs of $(m,n)$ values which work.

