We see that:
$$1000_3=1\cdot3^3\leq28 < 2\cdot3^3=2000_3 $$

Therefore $28$ is a $4$-digit number starting with a $1$ in base $3$. Only thing left is $28-1\cdot3^3=1$ which is $1$ in all number bases. Putting everything together: $$28=1\cdot3^3+1\cdot3^0=1000_3+1_3=1001_3$$ Answer: $1001_3 .$

We get:
$$7777_9=$$
$$=7_9\cdot1111_9=$$
$$=7\cdot(9^3+9^2+9^1+9^0)=$$
$$=7\cdot(10\cdot9^2+10)=$$
$$=70\cdot82=$$
$$=5740$$
Answer: $5740.$

We see that:
$$ 8\cdot9^2=648 < 667 < 9^3 $$
Therefore $667$ is a $3$-digit number starting with an $8$ in base $9$. The rest is done on the fly: $$667-8\cdot9^2=19=2\cdot9+1=21_9$$
Putting everything together: $667=800_9+21_9=821_9.$
Answer: $821_9.$

We convert to base $7$ on the fly:
$$43_8=4\cdot8+3=35=5\cdot7=50_7$$
Answer: $50_7.$

We want to use the fact that $9998$ almost equals $10^4$. So start by writing $10^4$ in base 5:
$$10^4=16\cdot5^4=(3\cdot5+1)\cdot5^4=310000_5$$
Use this to write $9998$ in base 5:
$$9998=10^4-2=310000_5-2_5=304443_5$$
Answer: $304443_5.$

Write $221_3$ in base $7$:
$$221_3=2\cdot9+2\cdot3+1=25=3\cdot7+4=34_7$$
Now do the addition:
$$54_7+34_7=$$
$$=(50_7+30_7)+(4_7+4_7)=$$
$$=8\cdot10_7+8\cdot1_7=$$
$$=9\cdot10_7+1_7=$$
$$=12_7\cdot10_7+1_7=$$
$$=121_7$$
Answer: $121_7.$

We do this on the fly:
$$135_6+44_8=(36+18+5)+(32+4)=95=19\cdot5=(3\cdot5+4)\cdot5=340_5$$
Answer: $340_5.$

Of course the player who starts wins. The starting player removes all $3$ coins from the single pile and wins the game.

Let the players be named A and B, and assume A starts. There're equally many coins in the two piles. Every move that player A makes in one of the pile, player B can copy by picking equally many coins in the other pile. When player A empties one pile, player B empties the other pile and wins the game. So the player who starts loses.

Let the players be named A and B, and assume A starts. The game starts with piles with $1,2$ and $3$ coins respectively, which we denote $(1,2,3)$. Player A has $1+2+3=6$ possible opening moves. Three of the opening moves is to empty one of the piles, leaving the game with two piles with unequal number of coins:
$$ (1,2,3)\overset{\mathrm{A}}{\rightarrow} (0,2,3), (1,0,3), \text{or } (1,2,0) $$
B can then make these two piles have an equal number of coins. The three cases are:
$$ (0,2,3)\overset{\mathrm{B}}{\rightarrow} (0,2,2) $$ $$ (1,0,3)\overset{\mathrm{B}}{\rightarrow} (1,0,1) $$ $$ (1,2,0)\overset{\mathrm{B}}{\rightarrow} (1,1,0) $$
The other three opening moves for A is not emptying a pile, leaving three piles, two of which have an equal number of coins:
$$ (1,2,3)\overset{\mathrm{A}}{\rightarrow} (1,1,3), (1,2,2), \text{or } (1,2,1) $$
B can then empty the pile which has different number of coins then the other two, leaving two piles with an equal number of coins. The three cases are: 
$$ (1,1,3)\overset{\mathrm{B}}{\rightarrow} (1,1,0) $$ $$ (1,2,2)\overset{\mathrm{B}}{\rightarrow} (0,2,2) $$ $$ (1,2,1)\overset{\mathrm{B}}{\rightarrow} (1,0,1) $$
So regardless of which of the six possible opening moves A chooses, B can then always make a move such that there're only two piles left with equally many coins. According to the argument in the solution of Exercise 2, B will win. So the player who starts loses.

Let the players be named A and B, and assume A starts. In this setup we can use what we've learned from Exercise 2 and 3. We can divide the piles into two sets. One set has two piles with $7$ coins in each, we call this set $G$. The other set has three piles with $1,2$ and $3$ coins respectively, we call this set $H$. The idea is that player B will consider this game of Nim as two separate games. Every time player A removes coins from set $G$, then player B should play in the next move as if only the set $G$ exists. From Exercise 2 we then know that player B will remove the last coin in set $G$. And similarly, every time player A removes coins from set $H$, then player B should play in the next move as if only set $H$ exists. From Exercise 3 we then know that player B will remove the last coin in set $H$. Player B will remove the last coin from both sets, and thereby remove the last coin from the entire game and win the game. So the player who starts loses.

A rule that tells us when someone will lose, must have certain features. In a game with a single pile, the starting player will always win. And from Exercise 4 we realise that adding two piles with equally many coins doesn't change the game. To figure out a reasonable guess for the rule we use the tip and write the number of coins in each pile in base 2:
$$ \text{Exercise 1 (winning):} \qquad 11_2 $$
$$ \text{Exercise 2 (losing):} \qquad \qquad 111_2, \qquad 111_2$$
$$ \text{Exercise 3 (losing):} \;\qquad 1_2,\quad,10_2,\quad 11_2,\quad $$
$$ \text{Exercise 4 (losing):} \quad 1_2,\,10_2,\; 11_2, \; 111_2, \; 111_2 $$
After some trial and error we might test summing the $1_2$-digits from each pile. 
$$ \text{Exercise 1 (winning):} 1 $$
$$ \text{Exercise 2 (losing):} 1+1=2$$
$$ \text{Exercise 3 (losing):} 1+0+1=2$$
$$ \text{Exercise 4 (losing):} 1 + 0 + 1 + 1 + 1 =4$$
We see that the Exercises, where someone is losing, all have an even sum. The same happens to be true also when we sum the $10_2$-digits and sum the $100_2$-digits from each pile. So a reasonable guess for the rule is that someone is losing when the sum of the $1_2$-digits, $10_2$-digits, etc, is even. Notice that adding two identical piles doesn't change if the sums are even or odd, this is something that we required. We also required that the rule worked for a single pile. Which it does because when a positive number is written in base $2$, at least one digit is odd.

Background
This problem has too many coins, to test all possible moves. This makes the problem difficult. When you're faced with a difficult problem, it's often a good idea to look at simpler version of that problem, for instance Exercise 1,2,3,4. Then in Exercise 5 we guessed a rule for when someone will lose the game. Now we test if that guess is correct.

Preparation
In this game of Nim, we've got three piles with $7$, $9$, and $14$ coins respectively. We write the number of coins in each pile in base $2$:
$$ \;\;7=0111_2 $$
$$ \;\;9=1001_2 $$
$$ 14=1110_2 $$ 
We notice that when the game starts the sum of the $1_2$-digits, $10_2$-digits, $100_2$-digits, and $1000_2$-digits, are even. During the game the number of coins will change, so we introduce variables $a,b,c$ for the number of coins in the three piles. Similarly as above, we write $a,b,c$ in base $2$, where the digits are also variables:
$$a=(a_3a_2a_1a_0)_2$$
$$b=(b_3b_2b_1b_0)_2$$
$$c=(c_3c_2c_1c_0)_2$$
Remember that the digits, for example $a_0$, is either $0$ or $1$. Let the players be named A and B, and assume A starts. To test our guess from Exercise 5 we will look at two cases. The first case is when it's players A:s turn and the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, are even. The second case is when it's players B:s turn and at least one the sum of the $1_2$-digits,$\dots$ ,$1000_2$-digits, is odd. 

First Case: Player A:s turn
Suppose, at some point in the game, it's player A:s turn and the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, are even. So the sums:
$$ a_0+b_0+c_0,\quad a_1+b_1+c_1,\quad a_2+b_2+c_2,\quad a_3+b_3+c_3$$
are even. Either all piles are empty and player A has lost. Or the piles are not empty and A must choose a pile. Say player A chooses the pile with $a$ coins and leaves behind $a'=(a_3'a_2'a_1'a_0')_2$ coins in the pile. The numbers $a$ and $a'$ are different and must have at least one digit different. This means at least one of the sums:
$$ a_0'+b_0+c_0,\quad a_1'+b_1+c_1,\quad a_2'+b_2+c_2,\quad a_3'+b_3+c_3$$
is odd. So at least one the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, is odd after A:s move.

Second Case: Player B:s turn
Suppose, at some point in the game, it's player B:s turn and at least one the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, is odd. So at least one of the sums:
$$ a_0+b_0+c_0,\quad a_1+b_1+c_1,\quad a_2+b_2+c_2,\quad a_3+b_3+c_3$$
is odd. We want to show that player B can make these sums even. This is done by first finding the largest integer $n$ such that $ (a_n+b_n+c_n) $ is odd. At least one of the digits $a_n, b_n, c_n$ is $1$. Say for instance that $a_n=1$. Player B then chooses the pile with $a$ coins and and leaves behind some appropriate number of coins $a'=(a_3'a_2'a_1'a_0')_2$ in the pile. Consider an arbitrary digit $a_x'$, for some $x\in\{0,1,2,3\}$. For $x > n$ we know $ (a_x+b_x+c_x) $ is even, so let $a_x'=a_x$. Let $a_n'=0 < a_n$. These choices makes $(a_n'+b_n+c_n) $ even and ensures that $a' < a$. The rest of the digits are chosen such that the sums:
$$ a_0'+b_0+c_0,\quad a_1'+b_1+c_1,\quad a_2'+b_2+c_2,\quad a_3'+b_3+c_3$$
are even. So the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, are even after B:s move.

Conclusion
The game starts with sum of the $1_2$-digits, $\dots$, $1000_2$-digits, being even. So if we look at the First Case, we see that player A is forced to make at least one the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, odd. Now looking at the Second Case, we that that player B can make the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, even again. This puts player A in the original position but with fewer coins in the game. Player B can continue with this strategy until all coins are gone, and B wins (all piles are $0$ means even sum of $1_2$-digits, $\dots$, $1000_2$-digits). So the player who starts loses.

Background
We start by remembering something about the old game. In the old game someone was losing when $2$ divided the sum of the $1_2$-digits, $\dots$, $1000_2$-digits. This is consistent with the property that we learned in Exercise 4: Adding $2$ extra identical piles don't change the outcome of the old game. 
There's a similar property in this new game were your allowed to choose from one or two piles. Adding $3$ extra identical piles don't change the outcome (do you see why?). And $3$ extra piles won't change if $3$ divides the sum of the $1_2$-digits, $\dots$, $1000_2$-digits. It's therefore a reasonable guess that someone is losing when $3$ divides the sum of the $1_2$-digits, $\dots$, $1000_2$-digits. 

Preparation
In this game of Nim, we've got four piles with $8$, $10$, $11$, and $15$ coins respectively. We write the number of coins in each pile in base $2$:
$$ \;\;8=1000_2 $$
$$ 10=1010_2 $$
$$ 11=1011_2 $$
$$ 15=1111_2 $$
We notice that when the game starts the sum of the $1_2$-digits is $2$, which is not divisible by $3$. During the game the number of coins will change, so we introduce variables $a,b,c,d$ for the number of coins in the four piles. Similarly as above, we write $a,b,c,d$ in base $2$, where the digits are also variables:
$$a=(a_3a_2a_1a_0)_2$$
$$b=(b_3b_2b_1b_0)_2$$
$$c=(c_3c_2c_1c_0)_2$$
$$d=(d_3d_2d_1d_0)_2$$
The sum of the $1_2$-digits, $\dots$, $1000_2$-digits are:
$$ S_0=a_0+b_0+c_0+d_0,\quad S_1=a_1+b_1+c_1+d_1,\quad S_2=a_2+b_2+c_2+d_2,\quad S_3=a_3+b_3+c_3+d_3$$
Let the players be named A and B, and assume B starts. To test our guess of when someone is losing, we will look at two cases.

First Case: Player A:s turn
Suppose, at some point in the game, it's player A:s turn and $3$ divides $S_0, \dots, S_3$. Either all piles are empty and player A has lost. Or the piles are not empty and player A must choose one or two piles. Say player A chooses the pile with $a$ coins and possibly from the pile with $b$ coins. After player A:s move there remains $a'=(a_3'a_2'a_1'a_0')_2$ coins from the first pile and $b'=(b_3'b_2'b_1'b_0')_2$ coins in the second pile. Consider the largest integer $n$ such that: $ a_n' \neq a_n $ or $ b_n' \neq b_n $. We know that $a' < a$ and $b'\leq b$, which implies: 
$$0\leq a_n'+b_n' < a_n+b_n \leq 2 $$
We have assumed that $3$ divides $S_n=(a_n+b_n+c_n+d_n)$. The inequality above then implies that $3$ doesn't divide $(a_n'+b_n'+c_n+d_n)$. So after A:s move, at least one of the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, is not divisible by $3$.

Second Case: Player B:s turn
Suppose, at some point in the game, it's player B:s turn and at least one of $S_0,\dots S_3$, is not divisible by $3$. We want to show that player B can make sure that $3$ divides the sum of the $1_2$-digits, $\dots$, $1000_2$-digits, after the move. Let $n$ be the largest integer such that $3$ doesn't divide $ S_n=(a_n+b_n+c_n+d_n) $. At least one of the digits $a_n, b_n, c_n,d_n$ is $1$. Say for instance that $a_n=1$. Pick the pile with $a$ coins and leave behind some appropriate number of coins $a'=(a_3'a_2'a_1'a_0')_2$ in the pile. Let $a_n'=0$ and for every $x>n$ let $a_x'=a_x$. This ensures that $a'< a$. Assume there exists a largest integer $m\leq n$ such that $(b_m+c_m+d_m)=1$. Either $b_m, c_m, d_m$ is $1$. Say for instance that $b_m=1$. Pick the pile with $b$ coins as a second pile and leave behind some appropriate number of coins $b'=(b_3'b_2'b_1'b_0')_2$ in the pile. If $m$ doesn't exist then let $b'=b$, and the argument will work anyway. Let $b_m'=0$ and for every $x > m$ let $b_x'=b_x$. This ensures that $b'\leq b$. Let's pick the rest of the digits so $3$ divides the sum:
$$ S_x'=a_x'+b_x'+c_x+d_x $$for all $x\in\{0,1,2,3\}$. For $x > n\geq m$ we've already defined $3$ to divide $S_x'=S_x$. For $x=m\leq n$ let $a_m'=0$. This implies that $3$ divides $S_m'$ because: 
$$ S_m'=a_m'+b_m'+(c_m+d_m)=a_m'+b_m'+(1-b_m)=0+0+1-1=0 $$
For $m < x\leq n$, we know $b_x'=b_x$ and $(b_x+c_x+d_x)\neq 1$ from the definition of $m$. It's possible to pick $a_x'=0$ or $1$, so $3$ divides $S_x'$ because:
$$ S_x'-a_x'=b_x'+c_x+d_x=b_x+c_x+d_x=0,2,3 $$
For $x < m$ it's possible to pick $a_x'+b_x'=0,1,$ or $2$, so $3$ divides $S_x'$. Now we've shown that $3$ divides $S_x'$ for all $x$. I.e. the that $3$ divides the sum of the $1_2$-digits, $\dots$, $1000_2$-digits.

Conclusion
The game starts with $3$ not dividing the sum of the $1_2$-digits. So if we look at the Second Case, we see that player B can make $3$ divide the sum of the $1_2$-digits, $\dots$, $1000_2$-digits. Now looking at the First Case, Player A is forced to put player B in the original position but with fewer coins. Player B can continue with this strategy until all coins are gone, and B wins (all piles are $0$ means $3$ divides the sum of $1_2$-digits, $\dots$, $1000_2$-digits). So the player who starts wins.

Use the fact that $9^2=81$. We then get:

$$ 101_{81}=81^2+1=9^4+1=10001_9 < 10101_9 $$

Answer: $10101_{9}$ is the biggest.

Start by subtracting $201_x$ from both sides. We get: $$130_x=50_x+90_x$$ Divide both sides by $10_x$. We get $$13_x=5_x+9_x$$ This means that $x+3=5+9$. The solution is $x=11$. Notice that all the digits in the original equation are valid in base $11$, thus the solution $x=11$ is valid.

Answer $x=11$.

We first notice that $7=10_7$ is not a divisor because $500103_{7}$ doesn't end with a zero. We continue to investigate if $6$ is a divisor. We expand our number: $500103_{7}=5\cdot7^5+7^2+3$. Use the fact that $7=6+1$ repeatedly: $$7^5=6\cdot7^4+7^4=6\cdot7^4+6\cdot7^3+7^3=\dots=6k+1$$where $k$ is some integer. We use that result while we expand our number: $$500103_{7}=5\cdot7^5+7^2+3=5\cdot(6k+1)+(6\cdot8+1)+3=5+1+3+6\cdot(5k+8)=9+6\cdot(5k+8)$$ which is divisible by $3$ but not by $2$ or $6$.

Answer: $500103_{7}$ is divisible by $3$, but not by $2,6,7$.

Use the fact that $3$ almost divides $8$. More specifically $8+1=3\cdot3$. We expand our number with this in mind: $$1101100101_8=(8^9+8^8)+(8^6+8^5)+(8^2+1)=9\cdot8^8+9\cdot8^5+65$$Because $65$ is not divisible by $3$, neither is $1101100101_8.$

Answer: No, $1101100101_8$ is not divisible by $3$.

Notice that $11111_2$ is the biggest $5$-digit number. Thus $11111_2+1$ must be the smallest $6$-digit number, which is $100000_2$.
Answer: $100000_2$.