For how many positive values of $n$ are both $\frac n3$ and $3n$ four-digit integers?
Any help is greatly appreciated. I think the smallest n value is 3000 and the largest n value is 3333. Does this make sense?
[Math] For how many positive values of $n$ are both $\frac n3$ and $3n$ four-digit integers
elementary-number-theory
Related Solutions
It is easier to count the total directly: you have five possibilities for the first digit (any digit except $0$), and six each for the remaining three digits, so the total number is $5\times 6^3$. This agrees with your count ($6^4-6^3 = 6^3(6-1) = 6^3\times 5$), but I think it's easier to just count them directly.
You can use the same method for counting the possibilities of the first two digits: five possibilities for the first digit, six for the second, for a total of $5\times 6$; this is the same as your count, $6^2-6^1 = 6(6-1)$.
The rest is almost correct; again, you can be a bit briefer with the count of the last two digits: the first digit can be any of the six possibilities. If the first digit is 0, 2, or 4, then the second digit must be 0 or 4 to get a multiple of $4$; so for each of the three possibilities you have two $2$-digit combinations, giving $3\times 2=6$ possibilities. If the first digit is $1$, $3$, or $5$, then the second digit must be $2$, so you have only three more. The total here is the sum of the two, so we have $6+3=9$ possibilities.
What you forgot is that $00$ also works.
Since the choices for the first two digits are independent of the choices for the last two, so you multiply the totals. The total will then be $$5\times 6 \times 9 = 270.$$
The fact that the total number (1080) is larger than the count of those that are divisible by 4 should not be a surprise: there are lots of numbers among the 1080 4-digit numbers in which every digit is one of 0, 1, 2, 3, 4, and 5 that are not divisible by $4$. So I'm not sure what your last line is meant to represent.
First, note that $43120 = 2^\color{red}{4}\cdot 5^\color{blue}{1}\cdot 7^\color{green}{2}\cdot 11^\color{purple}{1}$.
Part A
The number of positive integer divisors is the product of one plus each exponent in the prime factorization. That is, $$d(43120) = \color{red}{(4+1)}\color{blue}{(1+1)}\color{green}{(2+1)}\color{purple}{(1+1)} = (5)(2)(3)(2) = 60$$
Part B
The number of positive integers coprime to $43120$ can be found using Euler's Totient function, $\phi(n)$. Since $\phi$ is multiplicative for coprime integers:
$$\phi(43120) = \phi(2^4)\phi(5)\phi(7^2)\phi(11)$$
Also, $\phi(p^n) = p^{n-1}(p-1)$ for prime integers $p$. Then:
$$\begin{align}\phi(43120) &= \left[2^3(2-1)\right]\left[5-1\right]\left[7^1(7-1)\right]\left[11-1\right]\\ &= 8\cdot4\cdot 42\cdot 10\\ &= 13440 \end{align}$$
Part C We want all exponents of $a^2$ to be multiples of three, and we want the smallest such exponents. $$\left(2^\color{red}{4}\cdot 5^\color{blue}{1}\cdot 7^\color{green}{2}\cdot 11^\color{purple}{1}\right)^2 = 2^\color{red}{8}\cdot 5^\color{blue}{2}\cdot 7^\color{green}{4}\cdot 11^\color{purple}{2}$$ Well, it is easy to see that we need to add $1$ to the first exponent, $1$ to the second, $2$ to the third and $1$ to the fourth. Thus, our integer $m$ is: $$m= 2^\color{red}{1}\cdot 5^\color{blue}{1}\cdot 7^\color{green}{2}\cdot 11^\color{purple}{1}$$
Part D
This is a similar exponent-related trick as in part c if I'm thinking it through correctly. I'll leave this one as an exercise to the reader. :)
Best Answer
Your answer makes sense.
Minimum 4 digit number is $1000$
Maximum 4 digit number is $9999$
$$max = 3n = 9999$$
$$n_{max}=3333$$
$$min=\frac{n}{3}=1000$$
$$n_{min}=3000$$
Keep in mind that n must be divisible by 3. So, the answer would be: $$\frac{3333-3000}{3}+1=112$$