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

Correct Answer: 4195

4-Digit Number Problem: Maximizing the Number 

Let's denote the digits of the 4-digit number as \(a, b, c, d\) for the thousands, hundreds, tens, and units places respectively. We are given the following conditions:

• The sum of the digits at the thousands, hundreds, and tens places is 14: \[ a + b + c = 14. \]
• The sum of the digits at the hundreds, tens, and units places is 15: \[ b + c + d = 15. \]
• The tens place digit is 4 more than the units place digit: \[ c = d + 4. \]

We need to find the highest possible 4-digit number satisfying these conditions.

From the relation \(c = d + 4\), we can express \(c\) in terms of \(d\): \[ c = d + 4. \]

Now, let's substitute \(c\) into the other equations:

Equation 1: \(a + b + c = 14\)

Substituting \(c = d + 4\), we get: \[ a + b + (d + 4) = 14 \quad \Rightarrow \quad a + b + d = 10. \]

Equation 2: \(b + c + d = 15\)

Substituting \(c = d + 4\), we get: \[ b + (d + 4) + d = 15 \quad \Rightarrow \quad b + 2d = 11. \]

Our goal is to find the largest values for \(a, b, c, d\) while satisfying these equations.

Step 1: Solving for \(b\)

From the equation \(b + 2d = 11\), we solve for \(b\): \[ b = 11 - 2d. \]

Since \(b\) must be a digit (i.e., between 0 and 9), we have the condition: \[ 11 - 2d \geq 0 \quad \Rightarrow \quad d \leq 5. \]

Step 2: Solving for \(a\)

Now substitute \(b = 11 - 2d\) into the equation \(a + b + d = 10\): \[ a + (11 - 2d) + d = 10 \quad \Rightarrow \quad a = d - 1. \]

Since \(a\) must be a digit (i.e., between 1 and 9) and the leading digit, we also require: \[ d - 1 \geq 1 \quad \Rightarrow \quad d \geq 2. \]

Step 3: Combining Constraints

From the previous conditions, we now have: \[ 2 \leq d \leq 5. \]

To maximize the 4-digit number, we choose the highest possible value for \(d\), which is 5.

Step 4: Substituting \(d = 5\)

If \(d = 5\), then: \[ a = 5 - 1 = 4, \quad b = 11 - 2 \times 5 = 1, \quad c = 5 + 4 = 9. \]

Therefore, the number is 4195.

Step 5: Verification

• The sum \(a + b + c = 4 + 1 + 9 = 14\) (satisfied).
• The sum \(b + c + d = 1 + 9 + 5 = 15\) (satisfied).
• The relation \(c = d + 4 = 9 = 5 + 4\) (satisfied).

Conclusion

Thus, the highest possible 4-digit number is 4195, which satisfies all the given conditions.