A four - digit natural number which is divisible by $7$ is given. The number obtained by writing the digits in reverse order is also divisible by $7$. Furthermore, both the numbers leave the same remainder when divided by $37$. Find the 4-digit number.
Answer:- $\boxed{7000,7007,7777}$
Let the number be $\overline {abcd}= 1000a+100b+10c+d \equiv 6a+2b+3c+d \equiv \overline {dcba}= 1000d+100c+10b+a \equiv 6d+2c+3b+a \equiv 0 \mod 7 \implies b+c \equiv 0 \mod 7 $ ( by subtracting) and adding we get $ 6(a-d)+(b-c)\equiv 0 \mod 7.$
Again we have $\overline {abcd}= 1000a+100b+10c+d \equiv a+26b+10 c+d \equiv \overline {dcba}= 1000d+100c+10b+a \equiv d+26c+10b+a \equiv x \mod 37 \implies 26(b-c)+10(c-b)= 16(b-c) \equiv 0 \mod 37 \implies b-c \equiv 0 \mod 37. $
Since $b,c$ are digits and $b-c \equiv 0 \mod 37 \implies b=c.$ But $b+c\equiv 0 \mod 7 \implies b,c$ are $7$ or $0.$
Since we have $ 6(a-d)+(b-c)\equiv 0 \mod 7 \implies a,d \in {7,0}.$ Note $a$ can't be $0$ because of the 4 digit number stuff.
So we get the following number $7000,7007,7777.$