If I am given a number, for example 6, what's the mathematical algorithm for finding the smallest positive integer with 6 factors. It will be 12, but I need a formula which can give me the answers for higher number of factors such as 18.   I'm confused with this. A clear explanation would be appreciated.
123012
While practicing for a Mathematics competition in the future, I was going through past problems which were used. One of the problems in the category of Arithmetic/Number Theory went as follows:     Consider all positive integers having exactly 12 positive integral factors. Determine the sum of all of the three smallest of these integers.   The solution was $216$.  Initially, I brute-forced the problem, guessing numbers I thought would have 12 factors, and checking them. While this can yield results, it yields them far too slowly for a timed contest.  After some thought, I discovered a solution which worked for the numbers I tested, though it doesn't work for prime numbers and is very time consuming for large numbers. The solution goes as follows:  Assume you are given a number of factors $n$. To create a number with $n$ factors, we can factor $n$. If we use the example $12$ for $n$, we get $(3)\cdot(4)$ and $(2)\cdot(6)$. We then factor these factors, resulting in $(3)\cdot(2)\cdot(2)$ (We get this twice but we only need to use it once) Using these factors, we can create formulas to generate numbers with $12$ factors. To do this we use the formula  $$a^x b^y c^z$$ where x is one of the factors -1, y is another factor -1, and z is the third. (This also works with more than three factors, it just needs to be extended.) For 12, this yields us  $$a^1 b^1 c^2 \\ a^3 b^2 \\ a^5 b^1$$  Each of these formulas can generate a number with 12 factors if we input prime numbers into $a$, $b$, and $c$. In this case, we want to create the lowest numbers with 12 factors, so we can use the first prime numbers $2$, $3$, and $5$, and arrange them in such a way that we get the smallest value, ie. put the smallest numbers into the larger exponents. We then simply plug in the values and test. (The lowest possible combination produces $n$ instead of a number with $n$ factors  $$2^2 \cdot 3^1 \cdot 5^1 = 12 \\ 2^2 \cdot 3^1 \cdot 7^1 = 84\\ 2^3 \cdot 3^2 = 72 \\ 2^2 \cdot 3^1 \cdot 5^1 = 60$$ This means that $60$, $72$, and $84$ should be the smallest three numbers with $12$ factors. Let's check!  $60$ has the factors $(1,2,3,4,5,6,10,12,15,20,30,60)$, 12 of them!  $72$ has the factors $(1,2,3,4,6,8,9,12,18,24,36,72)$, again, 12 of them  $84$ has the factors $(1,2,3,4,6,7,12,14,21,28,42,84)$ again, 12!  What about the sum of these numbers, does it fit the solution?   $60+72+84 = 216$! Amazing!  While this solution works, it is rather inefficient and time consuming, and requires composite numbers. tl;dr: Is there a solution that exists such that it works on prime numbers or that requires less time? Will this method of calculating numbers by creating prime factors in this way always work? Why does this work?
89238
Given any $30$ positive integers we can find some integers so their sum is divided by $30$.  I don't have any idea where to start. First I would like just some hints, then If possible I would like do discuss different ways to solve the problem.
100491
Find the smallest integer $n$ such that $\frac{11}{15}&gt;\frac{m}{n}&gt;\frac{7}{10}$ and $m$ is an integer. There are multiple ways to solve this problem. We can get rid of the fractions by multiplying by $30n$ and do casework, we can convert to decimals and do casework, heck, we can graph and do casework. I know that this problem is really simple, but I am just looking for a way to solve this problem without casework.
5019
What is the best approach to calculating the largest prime factor of a number?  I'm thinking the most efficient would be the following:   Find lowest prime number that divides cleanly Check if result of division is prime If not, find next lowest Go to 2.   I'm basing this assumption on it being easier to calculate the small prime factors. Is this about right? What other approaches should I look into?  Edit: I've now realised that my approach is futile if there are more than 2 prime factors in play, since step 2 fails when the result is a product of two other primes, therefore a recursive algorithm is needed.  Edit again: And now I've realised that this does still work, because the last found prime number has to be the highest one, therefore any further testing of the non-prime result from step 2 would result in a smaller prime.
87746
Does applying the Collatz conjecture algorithm to negative integers always result in sequences ending with $-1$ and then repeating themselves an infinite number of times? It is so for the first several seed values.  After hours of thinking and research, I don't know how to answer the question.
92212
While playing around with programming, a problem suddenly came across my mind.      Given $n$ let's say $2^{64}$ (64-bit unsigned integer). Find a positive integer $x$, $1 \leq x \leq n$, such that $x$ has maximum divisors.   My attempt was: Let $n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}.$ The number of divisors is given by: $$(a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)$$  We need to find the maximum of these products. Since $(a_1 + 1)_{max} \implies {a_1}_{max}$, the product becomes: $a_1 \times a_2 \times .... \times a_k$ Let $P = a_1 \times a_2 \times .... \times a_k$. The maximum of $P$ means $a_i = a_j$ where $1 \leq i, j \leq k$.   So I guess my question is how can we find a number that is less than $n$ which has this property?     Thanks,
61987
Given: A circle with nine distinct positive integers. n is a fixed positive integer.  So these positive integers are arranged in the circle in such a way that the product of any selected two NON-ADJACENT numbers in the circle is a multiple of n. The product of two ADJACENT numbers in the circle is NOT a multiple of n. What is the smallest posible value for n.
81464
Say I have 12 and have to factorize it in 2 factors. Now possible factors are 1, 12 2, 6 3, 4   Now 3,4 is such that value of maximum factor i.e. 4 is minimized among all pairs.  How to find the solution in generic case where a number N is to be divided in K factors minimizing maximum value.
37253
Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$. I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1.
105135
Find the order of $8/9, 14/5,48/28$ in the additive group of $\mathbb{Q}/\mathbb{Z}$    JUNK  Guessing asking for $8/9 * k = 0 $ (being a whole integer) for smallest integer so for this one is 9 . another guess would be the smallest element of $Q/Z$ being something like $9/8$  $14/5$ would be 5  $48/28=12/7$ would be 7?
102092
If $a+ib$, $c+id$, $e+if$ are three complex numbers, than can we tell which one is greater or smaller between them? If yes, then how and if no then why not?   Can somebody give explanation on this.... I will be grateful to him.
54300
Yeah so I'm kind of stuck on this problem, and I have two questions.   Is there a way to define a number mathematically so that it cannot be a multiple of $7$? I know $7k+1,\ 7k+2,\ 7k+3,\ \cdots$, but that will take ages to prove for each case.  Is this a proof? $$(7k + a)^3 \equiv (0\cdot k + a)^3 \equiv a^3 \bmod 7$$    Thank you.
24187
Hello everybody! The problem you see above is a combinatorial problem that I could not solve. :(  Since we are to choose distinct numbers from $\{1,2,3 \ldots K\}$, the numbers must be in increasing order because they are all positive. I think that the difficulty of this problem is in the fact that the Special Sequence can be of any length. We can go about choosing say $n$ elements in $\binom{K}{n}$ ways but again how do we control that? Talking about a recursive solution, I think that is the way to go but I can't get how to do that (that is get a recursive formula for that).  Any help in solving this problem would be much appreciated. Thanks.
58309
I am asked to find all positive integers $n$ such that $\phi(n)=6$, and to prove that I have found all solutions.  The way I am tackling this is by constructing all combinations of prime powers such that when passed to Euler's function will yield $6$.  For example, I know that $\phi(7)=6$, $\phi(3^2)=6$, and $\phi(2)=1$. Therefore, the numbers should be $7$, $7\cdot2=14$, $3^2=9$, and $3^2\cdot2=18$.  I believe that there cannot possibly be any others because of the way the $\phi$ function is defined. What do you guys think?
68761
I found the following question and can't make much headway with it.  Show that there exist infinitely many positive integers $n$ such that $2^n \equiv n \mod p$ where $p$ is a odd prime...  I started by writing the number as $n=(p-1)k+c$ which implies $$2^c \equiv c-k \mod p$$ where I used Fermat's Little theorem. Now if $k$ is of the form $pm$ for some integer $m$ we get $n=(p-1)(p)m+c$   $$2^c \equiv c \mod p$$ and we have reduced $n$ to some smaller integer $c$ but I can't figure out how to find the smallest number $c$ which would then generate infinitely many $n$.  Any help in finding a possible approach to this would be highly appreciated.
72479
Find all triples of positive integers (a, b, c) such that  $$ \left( 1 + \frac{1}{a} \right)\left( 1 + \frac{1}{b} \right)\left( 1 + \frac{1}{c} \right) = 2 $$  I tried to expand, yielding $ a+b+c+ab+bc+ca+1 = abc $, but I couldn't go anywhere from here.   This problem clearly involves using the fact that $a, b, c$ are positive integers, since there are three variables and only one equation. It would be great if someone also addressed this condition and maybe tried to solve the problem for other requirements, like when $a, b, c$ are positive or negative integers.  Finally - can anyone post or link other such problems which involves a mix of algebra manipulation and number theory. I would really appreciate it.  (Here is the full paper: )
79818
This is probably pretty basic, but to save me an hour or so of grief can anyone tell me how you can work out the number of bits required to represent a given positive integer in Java?  e.g. I get a decimal 11, (1011). I need to get the answer, 4.  I figured if I could work out how to set all the bits other than the most significant bit to 0, and then >>> it, I'd get my answer. But... I can't.