Challenge Question of the Week
Welcome to Quantitative Challenge Question week-02, here’s the weekly challenge question of GMAT, GRE and SAT exams. You can see the answer explanation below the question.
Quantitative Challenge Question week-02
What is the maximum possible value of integer n, if 200! is divisible by 5n ?
Perhaps you know the term factorial in Math. If not, let me help. If P is a positive integer, P! is equals to the product of all positive integers from 1 to P inclusive. For instance, 4! = 1 × 2 × 3 × 4 = 24
Know you have refreshed this concept, let’s solve the question.
According to the given condition:
The process is simple, but conceptual.
If you divide 200 by 5, you’ll get multiples of 5 that are less than 200 inclusive. And that are 40 integers.
(For imagination: it will be 5, 10, 15, 25, …… and 200)
Notice that few integers in this multiples of 5 can be breakdown into two different 5s.
In order to find number of integers that can be broken down into two different 5s, you can simply divide 200 by 52, which is equals to 25.
After doing so, we’ll get 8 integers that would have at least two 5s.
(For imagination: it will be 25, 50, 75, …… and 200)
Notice that in first division of 200 by 5, we get 40 integers; this shows 40 integers have one 5 in its prime factorization. These integers have cancelled out one 5 by dividing from 5.
Rough Working for novice students:
Here's some rough working to clear your ambiguity in this: If, let's say, you divide 25 by 51, you'll get 5 integers; this shows that there are 5 integers less than 25 inclusive that have one 5 if broken down into prime factorization. These integers are: 5, 10, 15, 20 and 25 But notice that if we divide 25 by 52, you'll get 1 integer; this shows that there is 1 integer less than 25 inclusive that has two 5s, and that is 25. And if you divide 25 by 53, you wouldn't get integer equals to or greater than 1; this shows there's no integer exists less than 25 inclusive that has three 5s if broken down into prime factorization. (Concept of prime factorization is discussed in FREE Study plan for beginners).
Now if you divide 200 by 53, which is equals to 125, you’ll still get answer of 1 point something (more specifically 1.6). This means that there is 1 integer less than 200 inclusive that can be broken down into three 5s, and that is 125.
It is important to note that the integer 125 was existed in multiples of 51 = 5, in multiples of 52 = 25 as well as in multiples of 53 = 125.
But let’s take an inter 125. When it was divided by 5, it one of the three 5s was removed. After that when it was divided by 52, its second 5 also removed. But the last 5 was left over. So when we divided 200 by 53, the last 5 also cancelled out. Therefore,
Total number of 5s in 200! = 40 + 8 + 1 = 49 Answer
Therefore choice B is right answer.
→ In integers theory, hard difficulty level questions can simply be solved through prime factorization, which is the best ever short-cut to answer several questions.
→ If the result of positive integer a divided by positive integer b is non-integer greater than 1, this shows there are integral part of this non-integer result multiples of b exists less than a inclusive. For instance, the result of 20 divided by 7 is 2 point something (non-integer), this shows there are 2 multiples of 7 exists less than 20 inclusive. So ignore the fractional part and only take integer part.
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