<s>[INST] The digits of a 3-digit integer are reversed to form a new integer of greater value. Find the new integer if their product is 91,567. The digits of a 3-digit integer are reversed to form a new integer of greater value. The product of this new integer and the original integer is 91,567. What is the new integer?
 [/INST] Let the original $3$ digit integer be $I$. 
Let $I$ = $100a + 10b + c$, with $0 \leq a,b,c \leq 9$ and $c\geq a$
Then the new integer, $J = 100c + 10b + a$
We are given that $IJ = 91567 = (100a+10b+c)(100c+10b+a)$
So let's factorise $91567 = 7 \cdot 103 \cdot 127$
Then, since $I$ and $J$ are both integers, one of them is equal to $127$, and the other is equal to $7\cdot 103$. Since we are given that $J > I$, it follows that $J = 7 \cdot 103 = 721$ and $I = 127$
Now we must solve the following equations:
$721 = 100c + 10b + a$
$127 = 100a + 10b + c$
We can deduce that $a = 1$
Then we get $72 = 10c +b$
and $27 = 10b + c$
This implies:
$c = 27-10b, 72 = 270 -100b + b$
Then we get $b =2$ and $c = 7$
So the new integer is $721$
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