Question

The number of integers lying between 1000 and 10000 such that the sum of all the digits in each of those numbers becomes 30 is

• A. 84
• B. 96
• C. 45
• D. 75

Hint:

When the required sum of digits is "high" (close to the maximum sum), transform the variables by letting \( y_i = (\text{max digit}) - d_i \). This flips the problem to a "low sum" problem, often removing the need to use the Principle of Inclusion-Exclusion for upper bounds.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

Integers between 1000 and 10000 are 4-digit numbers. Let the number be \( d_1 d_2 d_3 d_4 \). We need to find the number of solutions to \( d_1 + d_2 + d_3 + d_4 = 30 \) subject to the digit constraints: \( 1 \le d_1 \le 9 \) and \( 0 \le d_2, d_3, d_4 \le 9 \).
Step 2: Key Formula or Approach:

A substitution method using \( y_i = 9 - d_i \) is very efficient for digit sum problems where the sum is close to the maximum possible sum (which is 36 for four digits).
Step 3: Detailed Explanation:

Let \( y_i = 9 - d_i \) for \( i = 1, 2, 3, 4 \). Substituting \( d_i = 9 - y_i \) into the sum equation: \[ (9 - y_1) + (9 - y_2) + (9 - y_3) + (9 - y_4) = 30 \] \[ 36 - (y_1 + y_2 + y_3 + y_4) = 30 \] \[ y_1 + y_2 + y_3 + y_4 = 6 \] Now, let's determine the constraints on \( y_i \): 1. For \( d_1 \): \( 1 \le d_1 \le 9 \implies 1 \le 9 - y_1 \le 9 \implies 0 \le y_1 \le 8 \). 2. For \( d_2, d_3, d_4 \): \( 0 \le d_i \le 9 \implies 0 \le 9 - y_i \le 9 \implies 0 \le y_i \le 9 \). Since the sum of \( y_i \) is only 6, it is impossible for any \( y_i \) to exceed 6. Therefore, the upper bound constraints (\( y_1 \le 8 \) and \( y_i \le 9 \)) are automatically satisfied by any non-negative solution. The problem reduces to finding the number of non-negative integral solutions to: \[ y_1 + y_2 + y_3 + y_4 = 6 \] Using the stars and bars formula with \( n = 6 \) and \( r = 4 \): \[ \text{Number of solutions} = \binom{n+r-1}{r-1} = \binom{6+4-1}{4-1} \] \[ = \binom{9}{3} \] \[ = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \] \[ = 3 \times 4 \times 7 = 84 \]
Step 4: Final Answer:

There are 84 such integers.