The maximum number of students among who 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and same number of pencils is

Options

A. 91
B. 910
C. 1001
D. 1911

Answer

To solve the problem of distributing 1001 pens and 910 pencils among students such that each student receives the same number of pens and the same number of pencils, we need to find the highest number of students that can receive an equal share of both items. This can be determined by finding the Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of the two numbers.

### Step-by-Step Solution:

**Step 1: Find the prime factorization of 1001.**
- We can divide 1001 by 7 (since 1001 is odd, we skip 2):
- \(1001 ÷ 7 = 143\)
- Now, we factor 143:
- \(143 ÷ 11 = 13\)
- Thus, the prime factorization of 1001 is:
\[ 1001 = 7 × 11 × 13 \]

**Step 2: Find the prime factorization of 910.**
- We can divide 910 by 10:
- \(910 ÷ 10 = 91\)
- Now, we factor 91:
- \(91 ÷ 7 = 13\)
- Thus, the prime factorization of 910 is:
\[ 910 = 10 × 91 = 2 × 5 × 7 × 13 \]

**Step 3: Identify the common factors.**
- From the prime factorizations:
- \(1001 = 7 × 11 × 13\)
- \(910 = 2 × 5 × 7 × 13\)
- The common prime factors are 7 and 13.

**Step 4: Calculate the HCF.**
- The HCF is the product of the common prime factors:
\[ HCF = 7 × 13 = 91 \]

**Step 5: Conclusion.**
- The maximum number of students among whom 1001 pens and 910 pencils can be distributed such that each student gets the same number of pens and the same number of pencils is:
\[ \text{HCF} = 91 \]

### Final Answer:
The maximum number of students is **91**.

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