Question

A natural number n lies between 100 and 400, and the sum of its digits is 10. The probability that n is divisible by 4, is ________

• A. \( \frac{7}{27} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{2}{9} \)
• D. \( \frac{1}{3} \)

Hint:

For problems involving properties of digits, it's often best to break down the problem into cases based on the first digit. This makes the counting systematic and less prone to errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a probability problem that combines number theory concepts. We need to find the total number of integers satisfying the given conditions (the sample space) and then find how many of those are divisible by 4 (the event space). The probability is the ratio of these two numbers.
Step 2: Key Formula or Approach:
1. Find all numbers n such that \( 100<n<400 \) and the sum of digits is 10. 2. Count the total number of such n. This is the size of the sample space, S. 3. Among these numbers, identify those divisible by 4. A number is divisible by 4 if the number formed by its last two digits is divisible by 4. 4. Count the number of such divisible numbers. This is the size of the event space, E. 5. Probability P = E/S.
Step 3: Detailed Explanation:
Let the number be n = htu (hundreds, tens, units digit). We are given \( 100<n<400 \), so \( h \in \{1, 2, 3\} \). The sum of digits is \( h+t+u = 10 \). Finding the Sample Space (S):

Case h=1: \( t+u = 9 \). The pairs (t,u) can be (0,9), (1,8), ..., (9,0). There are 10 such numbers.
Case h=2: \( t+u = 8 \). The pairs (t,u) can be (0,8), (1,7), ..., (8,0). There are 9 such numbers.
Case h=3: \( t+u = 7 \). The pairs (t,u) can be (0,7), (1,6), ..., (7,0). There are 8 such numbers.
Total number of possible values for n is \( S = 10 + 9 + 8 = 27 \). Finding the Event Space (E): We need to find which of these 27 numbers are divisible by 4. We check the number formed by the last two digits, 'tu'.

Case h=1 (t+u=9): Numbers are 109, 118, 127, 136, 145, 154, 163, 172, 181, 190. The last two digits 'tu' are 09, 18, 27, 36, 45, 54, 63, 72, 81, 90. Divisible by 4 are: 36 and 72. So, the numbers are 136 and 172. (2 numbers)
Case h=2 (t+u=8): Numbers are 208, 217, 226, 235, 244, 253, 262, 271, 280. The last two digits 'tu' are 08, 17, 26, 35, 44, 53, 62, 71, 80. Divisible by 4 are: 08, 44, 80. So, the numbers are 208, 244, 280. (3 numbers)
Case h=3 (t+u=7): Numbers are 307, 316, 325, 334, 343, 352, 361, 370. The last two digits 'tu' are 07, 16, 25, 34, 43, 52, 61, 70. Divisible by 4 are: 16, 52. So, the numbers are 316 and 352. (2 numbers)
Total number of numbers divisible by 4 is \( E = 2 + 3 + 2 = 7 \). Calculating the Probability: Probability \( P = \frac{\text{Event Space}}{\text{Sample Space}} = \frac{E}{S} = \frac{7}{27} \). Step 4: Final Answer:
The probability that n is divisible by 4 is \( \frac{7}{27} \).