[ equations in integers ]

The number 1047, when divided by $A$, gives a remainder of 23, and when divided by $A+1$, it gives a remainder of 7. Find $A$.

#
Since 1047 gives a remainder of 23 when divided by $A$, then $1047-23=1024$ is divisible by $A$. Similarly, $1047-7=1040$ is divisible by $A+1$. Since $1024=2^{10}$, then $A=2^{n}$, where $n-$ is a natural number and $n \leq 10$. At the same time, $A>23$, so $n \geq$ 5. Among the numbers $2^{5}+1,2^{6}+1,2^{7}+1,2^{8}+1,2^{9}+1,2^{10}+1$, only $2^{6}+1=65$ is a divisor of the number 1040. Therefore, $A=2^{6}$.

## Answer

64.