CAT Quantitative Aptitude Questions | CAT Number Systems - Factorial questions

The question is about basic factorial. We need to find out the possible values of 'n', when n! is a multiple of a^x but not b^x. Dealing with factorials of a number is a vital component in CAT Number Theory. Dealing with factorials of a number is a vital component in CAT Number Theory in the CAT Exam. A range of CAT questions can be asked based on this simple concept in the CAT exam. Factorials are also useful in Permutation Combination questions in CAT. Make use of 2IIMs Free CAT Questions, provided with detailed solutions and Video explanations to obtain a wonderful CAT score. If you would like to take these questions as a Quiz,head on here to take these questions in a test format, absolutely free.

Question 4: How many values can natural number n take, if n! is a multiple of 2²⁰ but not 3²⁰?

1. 11
2. 21
3. 16
4. 5

Explanatory Answer

Method of solving this CAT Question from Number Theory - Factorial: Finding the highest power of 3 that divides n! is easy. Find the smallest n such that n! is a multiple of 3^k might not be easy. Think about that.

The highest power of 2 that will divide n! = [\\frac{n}{2}\\)] + [\\frac{n}{4}\\)] + [\\frac{n}{8}\\)] + [\\frac{n}{16}\\)]........ and so on. So, let us try to find the smallest n such that n! is a multiple of 2²⁰.

If n = 10, the highest power of 2 that will divide n! = [\\frac{10}{2}\\)] + [\\frac{10}{4}\\)] + [\\frac{10}{8}\\)] = 5 + 2 + 1 = 8

If n = 20, the highest power of 2 that will divide n! = [\\frac{20}{2}\\)] + [\\frac{20}{4}\\)] + [\\frac{20}{8}\\)] + [\\frac{20}{16}\\)] = 10 + 5 + 2 + 1 = 18

If n = 24, the highest power of 2 that will divide n! = [\\frac{24}{2}\\)] + [\\frac{24}{4}\\)] + [\\frac{24}{8}\\)] + [\\frac{24}{16}\\)] = 12 + 6 + 3 + 1 = 22

{Here we can also see that each successive number is just the quotient of dividing the previous number by 2. As in, [\\frac{12}{2}\\)] = 6, [\\frac{6}{2}\\)] = 3, [\\frac{3}{2}\\)] = 1. This is a further short-cut one can use.}

So the lowest number of n such that n! is a multiple of 2²⁰ is 24.

Now, moving on to finding n! that is a multiple of 3. The highest power of 3 that will divide n! = [\\frac{n}{3}\\)] + [\\frac{n}{9}\\)] + [\\frac{n}{27}\\)] + [\\frac{n}{81}\\)] and so on.

When n = 20, the highest power of 3 that can divide 20! = [\\frac{20}{3}\\)] + [\\frac{6}{3}\\)] = 6 + 2 = 8
When n = 35, the highest power of 3 that can divide 35! = [\\frac{35}{3}\\)] + [\\frac{11}{3}\\)] + [\\frac{3}{3}\\)] = 11 + 3 + 1 = 15
When n = 45, the highest power of 3 that can divide 45! = [\\frac{45}{3}\\)] + [\\frac{15}{3}\\)] + [\\frac{5}{3}\\)] = 15 + 5 +1 = 21

The lowest number n such that n! is a multiple of 3²⁰ is 45.

When n takes values from 24 to 44, n! will be a multiple of 2²⁰ and not 3²⁰. n can take 21 values totally.

The question is "How many values can natural number n take, if n! is a multiple of 2²⁰ but not 3²⁰?"

Hence the answer is "21 values"

Choice B is the correct answer.