please check:
The digit 3 is written at the right of a certain two-digit number thus forming a three-digit number. The new number is 372 more than the original two-digit number. What was the original two-digit number? my answer is 94?
Let x be the original two-digit number. Then, the new number is:
$\displaystyle 10x + 3$. Now, we know that $\displaystyle 10x + 3 = x + 372$
Solving this gives us:
$\displaystyle 9x = 369 \Rightarrow x = 41$
How did you arrive at 94? You should have easily noted that if you modify the number as stated you get 943, which is obviously not 372 more than 94.
Let the original number be 10x +y with $\displaystyle x\in\{1, 2, ..., 9\}$ and $\displaystyle y\in\{0, 1, 2, ..., 9\}$
Then the 3-digit-number is 100x+10y+3.
According to the wording of the question you get:
$\displaystyle 100x+10y+3-372=10x+y$
$\displaystyle 10x+y=41$
The only possible value for x = 4 and y = 1. Thus the original number is 41.