Find value of $n$ with given conditions

Question

The 4-digit positive number $n$'s digit sum is $20$.

The sum of the first two digits is $11$, the sum of the first and the last digit as well.

The first digit is the last digit $+3$.

What is the number $n$?

Answer 1

HINT:

For $\displaystyle(abcd)_{10}, a+b+c+d=20, a+b=11\iff c+d=9$

$\displaystyle a+b=a+d\iff b=d$

$\displaystyle a=d+3\implies a=b+3, (b+3)+b=11$

Hope you can take it home from here

Answer 2

One solution is the number 7454.

The solution can potentially start 29, 38, 47, 56, 65, 74, 83, 92. Since the sum of the first and second digit is the same as the sum of the first and last digit, the possible solutions look like this: 29?9, 38?8, 47?7, 56?6, 65?5, 74?4, 83?3, 92?2. But the constraint that the first digit be 3 more than the last rules out all these possibilities, leaving only 74?4. Since 7 + 4 + 4 = 15, which is 5 short of 20, that means that 5 must go in the wildcard spot, giving us 7454.