There are 1001 pens and 910 pencils to be distributed among students.\nEach student must receive the same number of pens and the same number of pencils.\nWhat is the maximum possible number of students?

Introduction:
This question is a direct application of the highest common factor (H.C.F.) concept in a distribution context. It tests whether you can use H.C.F. to maximise the number of groups (students) while ensuring equal distribution of items (pens and pencils) to each student.

Given Data / Assumptions:

• Total pens = 1001.
• Total pencils = 910.
• Each student must receive the same number of pens.
• Each student must also receive the same number of pencils.
• No pen or pencil should be left undistributed.

Concept / Approach:
If each student gets the same number of pens and pencils, then the number of students must divide both 1001 and 910 exactly. To maximise the number of students, we must find the greatest number that divides both totals, which is the H.C.F. of 1001 and 910.

Step-by-Step Solution:
Step 1: Compute HCF(1001, 910). Step 2: Factorise 1001 = 7 * 11 * 13. Step 3: Factorise 910 = 7 * 13 * 10 = 2 * 5 * 7 * 13. Step 4: Common prime factors are 7 and 13. Step 5: Therefore, H.C.F. = 7 * 13 = 91. Step 6: Hence, the maximum number of students possible is 91.

Verification / Alternative check:
Check divisibility: 1001 ÷ 91 = 11 pens per student. 910 ÷ 91 = 10 pencils per student. Both are integers, meaning all pens and pencils are distributed equally among 91 students with no remainder left.

Why Other Options Are Wrong:
910 or 1001: These would require each student to receive fewer than one item of the other type, which is impossible with whole items. 1911: This is greater than both 1001 and 910, so it cannot divide either total. 70: Although 70 divides 910, it does not divide 1001 exactly.

Common Pitfalls:
Some students mistakenly compute the least common multiple or simply take a random common divisor instead of the greatest. Others may not factorise properly and miss the correct H.C.F. Using prime factorisation or the Euclidean algorithm systematically avoids such errors.

Final Answer:
The maximum number of students among whom the pens and pencils can be distributed equally is 91.