How to solve Iift 2010 Question 20?

This question can be solved by applying the core concept explained in the solution above, followed by step-by-step calculation and elimination of options.

Which concept is tested in this question?

This question tests fundamental concepts commonly asked in IIFT exams.

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Iift 2010 Question 20

This is a detailed step-by-step solution for Iift 2010 Question 20. Understand the concept, common mistakes, and expert tips to solve similar questions in IIFT exam.

IIFT 2010

7! can be broken into prime factors.
7! = 1*2*3*4*5*6*7 = 2⁴ ∗ 3² ∗ 5 ∗ 7
Hence, the smallest perfect square which is divisible by 7! will 2⁴ ∗ 3² ∗ 5 ∗ 7 = 5040*5*7 = 176400

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