How would I even do such a problem(without a calculator are anything like that)? I really don't have any idea on where to start for this problem. I know the answer is $1001$ from the  that I made to solve this problem. It might be worth noting that the integer pairs that produced equal remainders all produced remainders of $1$ except for $x=y=0$, which produced a remainder of $0$.
124071
Find the number of integer pairs 0 ≤ a, b ≤ 100 such that $a^{20}$ ≡ $b^{50} \pmod {101}$.  Here is the solution: "Since is prime, there exists a primitive root g in modulo 101. For some integers x and y, with 1 ≤ x, y ≤ 100, let a ≡ $g^x$ and b ≡ $g^y$. Hence, $g^{20x}$ ≡ $g^{50y}$ (mod 101). Therefore, $g^{20x−50y}$ ≡ 1 (mod 101), and 100|20x − 50y, or alternatively, 10|2x − 5y. Since 2|2x − 5y, we have 2|y. Also, 5|2x − 5y, so 5|x. Hence, the congruence holds for all x and y such that x is a multiple of 5 and y is a multiple of 2. There are 20 choices for x and 50 choices for y, so our answer is 20 · 50 = 1000."  In general, I'm very confused on the logical reasoning behind these steps. I'm a beginner in learning primitve roots and perhaps don't have enough background yet.  I'm confused why you can let a ≡ $g^x$ and b ≡ $g^y$.  Wouldn't that mean both $g^{x}$ and $g^{y}$ are both one Modulo 101? Finally, why does 100 have to divide 20x-50y. In fact, shouldn't it be 20x-50y dividing 100? I thought ord(a) of x always has to divide totient of x, and therefore 20x-50y|totient 101 which is 100.
70953
I believe that this may require the use of Fermat's Little Theorem. I rewrote it as $(a+1)^p - a^p \equiv 1 \pmod{p}$ because the right-hand side looks similar to Fermat's Little Theorem, but I was unable to figure out how I can get the left-hand side to become $a^{(p-1)}$.
1506
Which two numbers when added together yield $16$, and when multiplied together yield $55$.   I know the $x$ and $y$ are $5$ and $11$ but I wanted to see if I could algebraically solve it, and found I couldn't.   In $x+y=16$, I know $x=16/y$ but when I plug it back in I get something like $16/y + y = 16$, then I multiply the left side by $16$ to get $2y=256$ and then ultimately $y=128$. Am I doing something wrong?
119485
with -24 $\lt l \le$ 25.  Then I need to conclude that all final 2-digit numbers of the decimal expansion of squares are to be found among those of $0^2, 1^2, 2^2,...., 25^2$.  I'm thinking that I could maybe rewrite the division theorem proof for this but I don't know how/what to do with that. Any ideas?
115516
i have looked at/practiced several methods for solving ex: $7^{9999}$. i have looked at techniques using a)modulas/congruence b) binomial theorem c) totient/congruence d) cyclicity.   my actual desire would be a start to finish approach using totient/congruence. i have figured out how to work the individual steps but not how to combine them and in what order to be able to solve any "end digits" questions.
31861
Original problem  Show that there are exactly 16 pairs of integers $(x,y)$ such that $11x+8y+17=xy$.  My work  From case by case analysis I come to know that the equation will hold if and only if $x$ is odd and $y$ is even.  Also I found that $(8-x)|(11x+17)$ and $(11-y)|(8y+17)$  This is all what I have found.  Please see that my work are right or not.  This is a new kind of question which I have encountered so please help me is solving this problem.
110046
Even though this is a 3rd-grade math problem, people found it extremely hard. Any people have a solution, or algorithm is welcome. I'll try make a program base on the algorithm and see if it's correct. And, people are welcome to provide the best solution (least loop, recursion)  Enough said, here's the original: Fill in the blank the number from 1-9 to complete the equation. (probably distinct numbers each blank)    PS: Help me with tags. I'm not sure which tags should I use.  EDIT 1: After a few shortened, here's what I've got:  a + d - f + 12*e + 13*b/c + g*h/j = 87.   So, as you see, the priority does matter.  EDIT 2: Maybe my description is not clear enough, but I think it require to use every number   Like this: an array from [1:9], each fill in the blank remove one component. As Martigan point out, it's like: fill in the blanks using once and only once each number from 1 to 9. Simple math here: There are 9*8*7*6*5*4*3*2 = 362880 possibilitities. (You may correct this simple math if I'm wrong).  And b mod c should = 0 (as only integers appear.)  EDIT 3: I just asked my mom (who's an elementary teacher). She's confirmed that 3rd grade has studied about mathematic priority, so sorry Hagen von Eitzen. Your answer is half right, she'd give you an 7/10.  PS2: People, please share your wisdom and provide the math algorithm.  Sorry for my bad English.
115222
For example $151x - 294\equiv 44\pmod 7 $. How would I go about solving that?  The answer says to simplify it into the $ax\equiv b\pmod c $ form first, but I have no clue on how to get rid of the $294$... help?
90615
I have this problem: How many integers between $1$ and $10^4$ contain exactly one $8$ and one $9$?, but i dont know if $18, 19, 28, 29...$ serve, else it could be $8$ y $9 = 2$ numbers, help please.
59378
The following question was asked on a high school test, where the students were given a few minutes per question, at most:     Given that,   $$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$   and,   $$Q(x)=x^4+x^3+x^2+x+1$$   what is the remainder of $P(x)$ divided by $Q(x)$?     The given answer was:     Let $Q(x)=0$. Multiplying both sides by $x-1$:   $$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$   Substituting $x^5=1$ in $P(x)$ gives $x^4+x^3+x^2+x+1$. Thus,   $$P(x)\equiv\mathbf0\pmod{Q(x)}$$     Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?
114348
Problem Statement:- The number of zeroes at the end of the integer   $$100!-101!+\ldots-109!+110!$$   I am having a bit of a trouble in thinking how do I proceed. A little push in the right direction would be appreciated.   And if you are posting a full solution do use the spoiler tag, as sometimes I cant stop myself from seeing the whole solution and lose the chance of thinking through it by myself with a push in the right direction from you guys.  Also, I dont know whether I am using the right tag feel free to correct it if its wrong.
104049
Prove that there does not exist an integer solution for the diophantine equation $x^5 - y^2 = 4$.   It's obvious that $x$ and $y$ are of the same parity. We can also claim that if $x$ is odd, then it is $1 \pmod 4$. Also, if $x$ and $y$ are even, then $y \equiv 2 \pmod 4\text{ since } x^5 \text{ is a multiple of } 32$ and if $y$ were a multiple of $4$ then it would be at a distance of at least $16$ from $x^5$ or be equal to it. These are my observations.  How should I proceed with the proof? Please give hints.
68787
I actually do not have the basic idea on how to approach these type of questions....so please tell me a generalized method about all this too.  It came in RMO, and the question is:  What is the remainder when $2^{1990}$ is divided by $1990$ ?
54212
Find $a,b \in\mathbb{N}$ such that:      Remainder of $a$ divided by $b$ equals $5$       $\mathrm{lcm}(a,b)=12\gcd(a,b)$   I don't know how or where to start. The most similar problem I found was  Still I can't figoure out how to solve my problem, any help will be appreciated, thanks.
31562
I have a math problem, and I have no idea how to solve it. I first saw it over on the Code Golf StackExchange, . The poster there hinted at knowing that there is a proof, but no proof was provided.   We start with a pair of integers $a$ and $b$. We double one and add one to the other. We have the power to decide which to double and which to increment. We repeat this "doubling/+1" process, until the two integers are equal.  For example, starting with $(2, 5)$ we can double the 5 and increment the 2 to give $(3, 10)$. Then, we can double the three and increment the 10 for $(6, 11)$. We can double the six and increment the 11 for $(12, 12)$ - and now we have made the numbers equal.   Given any pair of integers, it it always possible to make them equal using these steps?  Partial Proof  All pairs of negative numbers will terminate. This relies on the fact that doubling 0 is 0. If we start with a pair of negative numbers, we can repeatedly add one to one of the numbers until it hits 0. While the other number will have been doubled several times, we can repeatedly decrement it until it too, hits 0.   $(-6, -3) \to (-12, -2) \to (-24, -1) \to (-48, 0) \to (-47, 0) \to (-46, 0) \to \cdots \to (0, 0)$
32207
The only solution that I have been able to come up with is $d=1$, but because $d&gt;2$, I can't seem to figure out how to find a solution.
26742
It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be?