Determine the largest integer $n$ which has at most three digits and equals the remainder when $n^2$ is divided by $1000$. I've determined that the last digit can only be 1, 5, or 6, but if I try to continue, I meet some problems with $n^2$.
124252
Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.  So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod 1000$. I  don't know how to proceed since 1000 seems to be a bit high.
35146
Can anyone tell me what is the maximum number of consecutive composite numbers possible? I mean can I get say 1000 consecutive natural numbers. Is there any general theorem that when I have a n-digit number there will always be p consecutive composite numbers?
23726
Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.  What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of . Can anyone help me?
30245
The title says it all. For what natural numbers $n$ does $2^n$ divide $3^n - 1$? By substituting values I can see that this happens for $n = 1, 2, 4$ but are there more? I am not able to prove/disprove.
55623
The question is to determine the number of positive integers up to $2000$ that are not divisible by $2,3$ or $5$ but are divisible by $7$. The answer is supposed to be $76$ but not sure how it was derived  I know that if the question was how many integers are not divisible by $2,3,5$ or $7$ then the answer would be $458$ and I know how to derive this.
113454
I have this really hard problem that I am working on and I just don't seem to get it. The question is: let $n$ be a positive integer; prove that there does not exist any integer $m$ such that $3n^2+3n+7=m^3$.  Please explain in detail otherwise I might not understand. I am also open for any methods that might help. Thank you. Please help.
109068
Find all positive integers $n$ such that $n$ is divisible by all the positive integers less than or equal to $\sqrt{n}$  My thought: If n is a positive integer, let d(n) denote the number of positive divisors of n and I will find n such that $d(n) \ge 2\sqrt{n} - 1$
39264
Can anyone help me find the first digit of $2015^{2015}$?  It is easy to find the last digit but I have no idea for the first digit.
4042
Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.  Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.  If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.  And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.
55599
Given a number $n$. I need to find the largest $q$ such that $q^2$ divides $n$. I need the fastest method to find $q$. $q$ can be any number prime or composite.  At present I am factorizing the number $n$ to find the highest number $q$. I need a better approach which does not involve to factorize the number $n$ completely or by some other approach which is less than $O(\sqrt n)$. Constraint: $n\leq 10^{18}$  some test cases:   if $n=180$ then $q=6$  if $n=17$ then $q=1$ if $n=10000$ then $q=100$
56985
What is the shortest way to compute the last 3 digits of $17^{256}$ ?  My solution: \begin{align} 17^{256} &amp;=289^{128} \\ &amp;=(290 - 1)^{128}\\ &amp;=\binom{128}{0}290^{128} -  ... +\binom{128}{126}290^2 - \binom{128}{127}290 + \binom{128}{128} \end{align}  Computed the last 3 terms whose last 3 digits gave the last 3 digits of $17^{256}$.  Is there any shorter method to do this(which requires much less computation) ?
79321
When we divide $n^5+5$ by $n+5$, we get a remainder $-620$, i.e., $n^5+5=K(n+5)-620$ , now how to proceed further?
11305
How many integers from 1-999 do not have any repeated digits?  The answer is explained in this , but why is the last set 9*9*8? Why not 9*9*9?
39946
Find all $n&gt;1$ such that $\dfrac{2^n+1}{n^2}$ is an integer.   I know that $n$ must be odd, then I don't know how to carry on. Please help. Thank you.
69792
I'm solving a problem which asks me to find how many 5-digit numbers with distinct digits are there which are also divisible by $3$. I am familiar that there are $9\cdot9\cdot8\cdot7\cdot6=27216$ five-digit numbers with distinct digits. I know that a natural number $n$ is divisible by $3$ iff the digits sum up to a number divisible by $3$. Now, brute-forcing by a computer algorithm, i found that the result is $20928$. However, I am interested, if there is a mathematical solution for this. I know how to find amount of numbers divisible by $4$ or $5$. We just fix some amount on the last two (or last one) digit. But criterion for divisibility by $3$ is very different from those and I have no idea how to start. Eventually, I would like to find a general formula for $n$-digit numbers.  Edit: Equivallently we want to find how many solution does the equation $a+b+c+d+e=3k$ for $a,b,c,d,e\in\{0,1,2,...,9\}$, $a\neq 0$ also $a,b,c,d,e$ are all distinct and $k\in\mathbb{Z}$ (the highest possible $k$ is in this case $k=11$, the smallest being $k=5$)  Edit (2): According to the ideas from the answer. Make $A=\{0,3,6,9\},B=\{1,4,7\}, C=\{2,5,8\}$. Let's go through the cases that in the number $n$ there are $0,1,2,3,4$ digits from $A$.  i) Suppose none of $A$ is in $n$. We can't form a number divisible by three. We have to take 5 numbers, that is $3$ from $A$ and $2$ from $B$ or $2$ from $A$ and $3$ from $B$. None of the expressions $3(3k+1)+2(3k+2)$ and $2(3k+1)+3(3k+2)$ are divisible by three.  ii) Let $1$ digit from $A$ be in $n$. The amount of ways to choose the digit is $5$. Now we are left to choose $4$ digits. The only possible way is to choose $2$ from $A$ and $2$ from $B$. The amount of ways to do this is ${3\choose 2}^2=9$ which together gives $5\times 9$ ways which can be permuted and alltogether $5\times 9\times 5!=5400$ possible ways.  iii) Let $2$ digits from $A$ be in $n$. The amount of ways to choose the digit is ${5\choose 2}=10$. We are left with $3$ digits. Either all of them are from $A$ or all of them are from $B$. That's $2$ possible ways and all together $20$ possible ways which we permute to obtain $20\times 5!=2400$ possible ways.  iv) Let $3$ digits from $A$ be in $n$. The amount of ways to choose the digits from $A$ is ${5\choose 3}=10$. We are to choose $2$ digits. We must take exactly one from $B$ and exactly one from $C$. The amount of ways to do this is ${3\choose 1}^2=9$. Alltogether $9\times 10$ possible ways which we permute to get $90\cdot5!=10800$ possible ways  v) In the last case where all four digits from $A$ are present, there is no way how to make the number divisible by $3$.  Summing this all up gives $5400+2400+10800=18600$ which is less than what my algorithm yielded. We also have to subtract these which begin with $0$. Can anyone spot the mistake? (If i did any?)
122227
Can anyone please help me?  1) Find the last digit of $7^{12345}$   2) Find the last 2 digits of $3^{3^{2014}}$.  Attempt: 1)  By just setting the powers of $7$ we have $7^1 = 7$, $7^2=49$, $7^3=343$, $7^4 = 2401$, $7^5 = 16807$, $7^6 = 117649$, $\dots$  After the power of $4$, the last digits will repeat. Then by noticing the pattern the digits will end in $7,9,3$ and $1$.  Then we can divide the exponent $(12345)$ by $4$ since this is the cycle that makes it repeat. Then $12345 : 4$ has remainder $1$. So $7^1 = 7$ is the unit digit to $7^{12345}$.   So the last digit is $7$.   I know how to do it like this, the problem does not state how to find the last digit, but I know it has something do do with Euler's theorem.  for part $2$) I don't know how to start. Can anyone please help me?   Thank you for the help.
108470
I tried remainder theorem but that does not simplify it.   I tried factorizing $1446$ as $2\cdot3\cdot241$ and got remainders when numerator is divided by $2,3$ and $241$ individually but then I did not know what to do next? Stuck!