Question

For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

Answer 1

Determine the maximum 4-digit integer satisfying the following criteria:

• The sum of the thousands, hundreds, and tens digits equals 14.
• The sum of the hundreds, tens, and units digits equals 15.

Represent the number as:

\[ \text{Digits: } a \quad b \quad c \quad d \] Where:
\( a = \) Thousands digit
\( b = \) Hundreds digit 
\( c = \) Tens digit 
\( d = \) Units digit

Given Conditions:

\[ a + b + c = 14 \quad \text{(1)} \] \[ b + c + d = 15 \quad \text{(2)} \]

Subtract Equation (1) from Equation (2):

\[ (b + c + d) - (a + b + c) = 15 - 14 \Rightarrow d - a = 1 \Rightarrow d = a + 1 \]

Objective:

To maximize the 4-digit number \( abcd \), prioritize the highest value for the leftmost digit \( a \).

Given \( d = a + 1 \) and that digits range from 0 to 9, the maximum permissible value for \( a \) such that \( d \leq 9 \) is:

\[ a + 1 \leq 9 \Rightarrow a \leq 8 \] To further maximize the number, the tens digit \( c \) should be as large as possible. A balanced choice for \( a \) is 4.

Set \( a = 4 \) ⇒ \( d = a + 1 = 5 \)

From equation (1):

\[ a + b + c = 14 \Rightarrow 4 + b + c = 14 \Rightarrow b + c = 10 \quad \text{(3)} \]

From equation (2):

\[ b + c + d = 15 \Rightarrow b + c + 5 = 15 \Rightarrow b + c = 10 \quad \text{(confirms equation 3)} \]

Maximize the number using \( b \) and \( c \)

We need to select \( b \) and \( c \) such that \( b + c = 10 \) and the number \( abcd = 4bcd \) is maximized.

This occurs when \( c = 9 \) (maximum tens digit).
Consequently, \( b = 1 \), as \( 1 + 9 = 10 \).

The resulting number is:

\[ a = 4,\quad b = 1,\quad c = 9,\quad d = 5 \Rightarrow \boxed{4195} \]

✅ Final Answer: 4195