There is a number, the second digit of which is smaller than its first digit by 4, and if the number was divided by the digit's sum, the remainder would be 7.
Actually I know the answer is 623 I found it by using computer program which checks the conditions for all numbers but I wanted to know if there is a mathematical way to slove this problem.
On
You have $6$ possibilities for the two first digits: $a_1a_2=40,51,62,73,84,95$ and you can verify that the number cannot have two digits. Then you try with $3$ digits keeping in mind that the number minus $7$ is a multiple of $a_1+a_2+a_3$ so you have $$\frac{a_1a_2a_3-7}{a_1+a_2+a_3}=entero$$ (1) $\dfrac{400+a_3-7}{4+a_3}=\dfrac{393+a_3}{4+a_3}$. You have to verify with the nine possible values of $a_3$; in other words you must to see if some of the nine following quotiens is an integer: $$\dfrac{393}{4},\dfrac{394}{5},\dfrac{395}{6},\dfrac{396}{7},\dfrac{397}{8},\cdots,\dfrac{402}{13}$$
(2)Proving now with $51a_3$ you find $\dfrac{510+a_3-7}{6+a_3}=\dfrac{503+a_3}{6+a_3}$ and for $a_3=1$ you find an apparent solution since $\dfrac{504}{7}=72$. However the number $511$ is exactement divisible by $5+1+1$ so it must be discarded.
(3) Proving with $62a_3$ by the same procedure you'll find your given solution $623$.
One-digit case is impossible, since $4\not \equiv7 \mod 4$
Two-digit case: write number as $10(a+4)+a$.
$$10(a+4)+a \equiv 7 \mod (2a+4)$$
$$11a\equiv -33 \mod 2a+4$$
$$a+3 \equiv 0 \mod 2a+4$$
$$2(a+3) \equiv 2 \not \equiv 0 \mod 2a+4$$
Therefore, two-digit is impossible.
Three-digit case: Write number as $100(a+4)+10a+b$.
$$100(a+4)+10a+b\equiv 7 \mod (2a+b+4)$$
$$110a+b+400 \equiv 7 \mod (2a+b+4)$$
$$108a+396\equiv7 \mod (2a+b+4)$$
$$108a+389\equiv 0\mod (2a+b+4)$$
When $a=1$,
$$497\equiv 0 \mod b+6$$
Since $497=7\times 71$, $b+6=7$. However, we "do not want that", since the modulo must be greater than 7. "Remainder = 7" $\implies$ "Modulo > 7"
When $a=2$,
$$605 \equiv 0 \mod b+8$$
$605 = 5 \times 11^2$. Therefore, we can take $b+8=11 \implies b=3$.
Therefore our final answer is $\fbox{623}$.