Finding the minimum and maximum values of $q+r$ when $p=3^q\cdot 2^r$ and $100<p<1000$

Question

Question:

Given positive integers $p$, $q$ and $r$ with $p=3^q\cdot2^r$ and $100<p<1000$. The difference between maximum and minimum values of $(q+r)$, is _______.

My Approach:

As we are given that $100<p<1000$, on taking logarithm to the base $10$, we get:

$$\log100<\log p<\log 1000$$

$$2<\log(3^q\cdot2^r)<3$$

$$2<q\log 3+r\log 2<3$$

$$2<q(0.4771)+r(0.3010)<3$$

The only way I could think of to obtain the maximum and minimum values of $q$ and $r$ is to substitute them with natural numbers and look when the condition is satisfied. Is there any formal approach (other than substituting $q$ and $r$ with natural numbers) using which we can find the minimum and maximum values of $(q+r)$? If yes, it would be helpful if you could explain it.

On substituting different values of $q$ and $r$ in the condition arrived, the minimum and maximum values of $(q+r)$ comes out to be $5$ and $9$ respectively. And the answer to the above question is thus $9-5=4$. This is also the correct answer with respect to the answer key provided in my book.

Answer 1

Given positive integers $p$, $q$ and $r$ with $p=3^q\cdot2^r$ and $100<p<1000$. The difference between maximum and minimum values of $(q+r)$, is _______.

The answer is quite straightforward if you draw the lines \begin{align*} \frac q{\frac 2{\log 3}}+\frac r{\frac 2{\log 2}}&=1\Rightarrow \frac q{4.\cdots}+\frac r{6.6\cdots}=1\tag{1}\\ \frac q{\frac 3{\log 3}}+\frac r{\frac 3{\log 2}}&=1\Rightarrow \frac q{6.\cdots}+\frac r{9.9\cdots}=1\tag{2}\\ q+r&=10\tag{3}\\ q+r&=4\tag{4}\\ \end{align*} on $q-r$ axes neatly and see that the points $(5,0)$ and $(0,9)$ can serve the best values $5$ and $9$ respectively, for the points lie strictly in the region between the lines $(1)$ and $(2)$ (let us call it region R). Lines $(3)$ and $(4)$ are drawn just to see that the best values can't be outside what we have obtained, that is $q+r$ can't be $4,10$.

EDIT:

But since $q,r\in \mathbb N$, we need to take positive integral points or points in which both $q,r\in\mathbb N$. I think the best approach would be to see that at least some segment of the lines $q+r=5$ and $q+r=9$ lies inside the region R, and then to make sure that those segments have positive integral points on them. For example, take the points $(4,1)$ and $(1,8)$. If they don't suffice(Well, they do here. I am just trying to provide an algorithmic approach to solve bigger such problems.), keep taking points from the segments of the lines $q+r=5$ and $q+r=9$. If they all don't suffice, move to lines that lie more inside the region R. For example, $q+r=4$ and $q+r=8$.

Note that the positive integral points nearest to the lines $(1)$ and $(2)$ will give the limits nearest to $100$ and $1000$. One can apply the "distance from a line" formula if the distances of two such points from the lines $(1)$ and $(2)$ are to be compared, and the inequality is not apparent by the neat plot.

Answer 2

This is easiest to just reason directly.

For any $M$ that $2^M < 3^12^{M-1} < 3^22^{M-2} < ..... < 3^{M-1}2 < 3^M$.

So if $q+r = m$ and $100 < 3^q 2^r < 1000$ then

1. $100 < 3^q2^r \le 3^{q+r}$ and
2. $2^{q+r} \le 3^q2^r < 1000$

And as $3^4 < 100 < 3^5$ then minimum value that $q+r$ can be, by 1) is $5$. And as $2^9< 1000< 2^{10}$ the maximum value that $q+r$ can be, by 2) is $9$.

But this assumes that allows for $r$ or $q$ to be zero and the question specifically says positive. So we didn't actually do it right. But we can modify really easily.

1. $100 < 3^q2^r \le 3^{q+r-1}\cdot 2 < 3^{q+r}$ and $3^4 < 100 < 81\cdot 2=3^4\cdot 2 < 3^5$.
2. $2^{q+r} < 2^{q+r-1}\cdot 3 \le 3^q2^r < 1000$ and $256\cdot 3=2^8\cdot 3 < 2^9 < 1000 < 2^{10}$.

And our conclusions hold.

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Note: Had the question been $80 < p < 1025$ we'd have had the same answers of a minimum of $4$ and a maximum of $9$ because, although $80< 3^4 < 2^{10} < 1025$ we are required that $r,q$ be non-zero and $3^3\cdot 2 < 80 < 2^10 < 3\cdot 2^9$.

Note 2: If we didn't have the requirement that $q,r$ but positive and allow them to be negative we would have no minimum or maximum.

For any negative $r$ then if $q= \lceil \log_3(101\cdot 2^{|r|})\rceil$ we will have $3^q2^r \ge 3^{\log_3(101\cdot 2^{|r|})}2^{r}=101\cdot 2^{r}2^{|r|}= 101$ and $3^q2^r < 3^{\log_3(101\cdot 2^{|r|})+1}2^{r}=3\cdot 101< 1000$.

And $\lim_{r\to -\infty} r+\lceil \log_3(101\cdot 2^{|r|})\rceil=$

$\lim_{r\to -\infty} r+ \log_3(101\cdot 2^{-r})=$

$\lim_{r\to -\infty} r-r\log_3 2 + \log_3(101)=$

$\lim_{r\to -\infty} r(1+\log_2 3) + \log_3(101)= -\infty$

so there is no minimum.

A similar argument for negative $q$ will show there is no maximum.

Answer 3

Observe that for the minimum value of $q+r$, we must have $2*3^{q}>100$ because $q,r>0$ and $2^r\leq3^q ,\forall r\leq q$ which means that multiplying by $3^q$ will get us to surpass $100$ the fastest, which requires the least amount of exponentiation. Solving for $q$ and taking the next integer that satisfies the inequality gives us $r+q=1+4=5$.

Then for the maximum, we need that $2^r*3 < 1000$ because multiplying by $2^r$ gets us to surpass $1000$ the slowest, which means the most amount of exponentiation. Solving for $r$ and taking the next integer that satisfies the inequality yields $r+q=8+1=9$.

Then the maximum minus the minimum of $q+r$ is $9-5=4$.

My argument is essentially based upon the fact that $2^x$ grows slower than $3^x$.