What is the largest number, $x$, which when divided by $22$ has quotient equal to remainder? I could solve this problem using a brute force mentality but I am looking for an elegant logical solution. So far I have tried to narrow down the possibilities and have found that:
$x > 22$ as any number, $c < 22$ when divided by $22$ has a quotient $= 0$ and a remainder $\not=$ to the quotient.
$x \thinspace\%\thinspace 22 \not=0$ so $x$ must be odd for remainder $\not= 0$.
What is the largest number, $x$, which when divided by $22$ has a quotient equal to the remainder?
 A: Say we have a number $x$. Using the division algorithm, we can write $x=22q+r$, where $q$ is the quotient upon dividing $x$ by $22$, and $0\leq r<22$ is the remainder. Since we are aiming to have $q=r$, we want to find the largest $x$ such that $x=22r+r=23r$. The largest such $x$ is then $x=23\cdot 21=483$.
A: The set of numbers with quotient equal to remainder upon division by $22$ is 
$$\{0,\;22\cdot1+1,\; 22\cdot2+2,\;\ldots\;,\;22\cdot21+21\}.$$
Of these, the largest is clearly $22\cdot21+21=483.$
A: The others gave very good answers, I'd just like to point out a natural way of interpreting this question. If the question asked for the number 10 instead of 22, you could immediately notice that it's asking for the largest two-digit number with equal digits (that is 99), because the remainder after division by 10 is the last digit, the rest is the quotient. So in this case, it's just asking for the largest two-digit number in base 22, which is $([21][21])_{22}=21\times 22+21=483$.
A: You have that any number $23k$ satisfies  $$\frac{23k}{22}=k+\frac{k}{22}$$ Since $k$ must be less that $22$ you take $k=21$
A: The Largest Remainder Can Be Obtained Diving By 22 Is 21 So largest No. x is 
21*22+22 = 483. Quotient Can't Be Largest Than Reminder. Hence 483 Is Right Answer.
A: If $\,x\!\div\! d\,$ has $\rm\color{#c00}{q}$uotient $\color{#c00}{= \rm r}$emainder then $ x = \color{#c00}q d\! +\! r = \color{#c00}rd\! +\! r = r(d\!+\!1)\,$ for $\, 0\le \color{#0a0}{r\le d\!-\!1}$
Therefore $\,x\,$ is maximal $\iff \color{#0a0}{r=d\!-\!1}\iff x = \underbrace{(d\!-\!1)(d\!+\!1)}_{\large d^2 -1 }\! = (d\!-\!1)\, d + d\!-\!1$
