If $r$ is the remainder when each of the numbers $1059,~1417$, and $2312$ is divided by $d$, where $d$ is an integer greater than $1$, then $d-r$ equals 
$\textbf{(A) }1\qquad \textbf{(B) }15\qquad \textbf{(C) }179\qquad \textbf{(D) }d-15\qquad  \textbf{(E) }d-1$
We are given these congruences:

$1059 \equiv r \pmod{d} \qquad$ (i)
$1417 \equiv r \pmod{d} \qquad$ (ii)
$2312 \equiv r \pmod{d} \qquad$ (iii)

Let's make a new congruence by subtracting (i) from (ii), which results in \[358 \equiv 0 \pmod{d}.\]
Subtract (ii) from (iii) to get \[895 \equiv 0 \pmod{d}.\]

Now we know that $358$ and $895$ are both multiples of $d$. Their prime factorizations are $358=2 \cdot 179$ and $895=5 \cdot 179$, so their common factor is $179$, which means $d=179$.

Plug $d=179$ back into any of the original congruences to get $r=164$. Then, $d-r=179-164= \boxed{\textbf{(B) }15}$. ~[jiang147369](https://artofproblemsolving.com/wiki/index.php/User:Jiang147369)