Question:
The probability that the sum of two integers $m$ and $n$, where $m, n \in \{1, 2, \dots, 50\}$, chosen randomly and independently, being divisible by 3 is
The probability that the sum of two integers $m$ and $n$, where $m, n \in \{1, 2, \dots, 50\}$, chosen randomly and independently, being divisible by 3 is
Show Hint
If the set of numbers was a perfect multiple of 3 (e.g. 51 elements), the probability would be exactly $\frac{1}{3} \approx 0.3333$.
Since we have slightly more elements in $R_1$ and $R_2$ than in $R_0$, the probability is slightly higher than $\frac{1}{3}$. This immediately eliminates options B and C.
Since we have slightly more elements in $R_1$ and $R_2$ than in $R_0$, the probability is slightly higher than $\frac{1}{3}$. This immediately eliminates options B and C.
Updated On: Jun 16, 2026
- 0.3336
- 0.3332
- 0.3333
- 0.3338
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem asks for the probability that the sum of two integers chosen independently from the first 50 natural numbers is divisible by 3.
We will use modulo 3 arithmetic to classify the numbers and find the total number of favorable combinations.
Step 2: Key Formula or Approach:
For the sum $m + n$ to be divisible by 3, the remainders of $m$ and $n$ when divided by 3 must sum to a multiple of 3.
The possibilities for the remainders are:
- Both are $0 \pmod 3$
- One is $1 \pmod 3$ and the other is $2 \pmod 3$
Step 3: Detailed Explanation:
• First, let us partition the set $\{1, 2, \dots, 50\}$ into three subsets based on their remainder modulo 3:
1) $R_0 = \{3, 6, 9, \dots, 48\} \implies |R_0| = 16$
2) $R_1 = \{1, 4, 7, \dots, 49\} \implies |R_1| = 17$
3) $R_2 = \{2, 5, 8, \dots, 50\} \implies |R_2| = 17$
• Since $m$ and $n$ are chosen independently, the total number of possible pairs is:
\[ \text{Total Pairs} = 50 \times 50 = 2500 \]
• For $(m + n)$ to be divisible by 3, we have three cases:
Case 1: Both $m, n \in R_0$.
The number of such pairs is:
\[ 16 \times 16 = 256 \]
Case 2: $m \in R_1$ and $n \in R_2$.
The number of such pairs is:
\[ 17 \times 17 = 289 \]
Case 3: $m \in R_2$ and $n \in R_1$.
The number of such pairs is:
\[ 17 \times 17 = 289 \]
• Summing the favorable pairs:
\[ \text{Favorable Pairs} = 256 + 289 + 289 = 834 \]
• The probability is given by:
\[ P = \frac{834}{2500} = 0.3336 \]
Step 4: Final Answer:
The probability is 0.3336.
The problem asks for the probability that the sum of two integers chosen independently from the first 50 natural numbers is divisible by 3.
We will use modulo 3 arithmetic to classify the numbers and find the total number of favorable combinations.
Step 2: Key Formula or Approach:
For the sum $m + n$ to be divisible by 3, the remainders of $m$ and $n$ when divided by 3 must sum to a multiple of 3.
The possibilities for the remainders are:
- Both are $0 \pmod 3$
- One is $1 \pmod 3$ and the other is $2 \pmod 3$
Step 3: Detailed Explanation:
• First, let us partition the set $\{1, 2, \dots, 50\}$ into three subsets based on their remainder modulo 3:
1) $R_0 = \{3, 6, 9, \dots, 48\} \implies |R_0| = 16$
2) $R_1 = \{1, 4, 7, \dots, 49\} \implies |R_1| = 17$
3) $R_2 = \{2, 5, 8, \dots, 50\} \implies |R_2| = 17$
• Since $m$ and $n$ are chosen independently, the total number of possible pairs is:
\[ \text{Total Pairs} = 50 \times 50 = 2500 \]
• For $(m + n)$ to be divisible by 3, we have three cases:
Case 1: Both $m, n \in R_0$.
The number of such pairs is:
\[ 16 \times 16 = 256 \]
Case 2: $m \in R_1$ and $n \in R_2$.
The number of such pairs is:
\[ 17 \times 17 = 289 \]
Case 3: $m \in R_2$ and $n \in R_1$.
The number of such pairs is:
\[ 17 \times 17 = 289 \]
• Summing the favorable pairs:
\[ \text{Favorable Pairs} = 256 + 289 + 289 = 834 \]
• The probability is given by:
\[ P = \frac{834}{2500} = 0.3336 \]
Step 4: Final Answer:
The probability is 0.3336.
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