Highest possible 4-digit number

Let the 4-digit number be $abcd$.

• $a + b + c = 14$
• $b + c + d = 15$
• $c = d + 4$

Subtracting the first equation from the second: $(b + c + d) - (a + b + c) = 15 - 14 \implies d - a = 1 \implies d = a + 1$

Since $c = d + 4$, we have $c = a + 5$. To maximize the 4-digit number $abcd$, we want to maximize the thousands digit $a$.

Constraints on digits ($0 \le a, b, c, d \le 9$ and $a \ge 1$): $c = a + 5 \le 9 \implies a \le 4$.

Let's test $a = 4$:

• $d = a + 1 = 5$
• $c = a + 5 = 9$
• $b = 14 - a - c = 14 - 4 - 9 = 1$

Digits are $a = 4, b = 1, c = 9, d = 5$. Checking constraints:

• $a+b+c = 4+1+9 = 14$
• $b+c+d = 1+9+5 = 15$
• $c - d = 9 - 5 = 4$

All conditions satisfied. The highest number is 4195.