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

We are asked to find the largest 4-digit number such that:

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

Let’s 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:

We want to maximize the 4-digit number \( abcd \).
That means we should try to take the largest possible value for the leftmost digit \( a \)

Since \( d = a + 1 \), and digit values can only range from 0 to 9, the maximum valid value for \( a \) such that \( d \leq 9 \) is:

\[ a + 1 \leq 9 \Rightarrow a \leq 8 \] But we also want \( c \) (tens digit) to be as large as possible to maximize the number. 
Try \( a = 4 \), which is a good balance.

Try \( a = 4 \) ⇒ \( d = a + 1 = 5 \)

From equation (1):

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

From equation (2):

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

Now maximize the number

We want to pick the values of \( b \) and \( c \) such that \( b + c = 10 \), and the number \( abcd = 4bcd \) is largest.

That happens when \( c = 9 \) (tens digit max)
Then \( b = 1 \), since \( 1 + 9 = 10 \)

So the number is:

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

✅ Final Answer: 4195