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Updated on: 16 Dec 2016, 06:00
Originally posted by EgmatQuantExpert on 16 Dec 2016, 05:33.
Last edited by Bunuel on 16 Dec 2016, 06:00, edited 2 times in total.
Last edited by Bunuel on 16 Dec 2016, 06:00, edited 2 times in total.
Renamed the topic.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Take a stab at the following GMAT like question on Number Properties. The official solution will be posted soon.
What is the greatest value of \(x\) such that \(15^x\) completely divides 300! ?
What is the greatest value of \(x\) such that \(15^x\) completely divides 300! ?
- A. 20
B. 54
C. 74
D. 148
E. 222
To practise ten 700+ Level Number Properties Questions attempt the The E-GMAT Number Properties Knockout
.Click here.
Updated on: 25 Dec 2016, 07:10
Originally posted by EgmatQuantExpert on 16 Dec 2016, 05:39.
Last edited by EgmatQuantExpert on 25 Dec 2016, 07:10, edited 2 times in total.
Last edited by EgmatQuantExpert on 25 Dec 2016, 07:10, edited 2 times in total.
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Alright so let's look at the official solution of the above question. Let's use a simple 3-step approach to solve questions similar to the above question.
Let's apply the above steps to solve this question and then we will look at a couple of questions where we can apply this learning.
We need to find the highest power of 15, so let's begin by doing
Step-1: Prime factorization of 15.
\(15 = 3*5 = 3^1*5^1\)
Step-2: Find the highest power of each of the prime factors in the factorial
Since 15 has two prime factors 3 and 5, we need to find the highest power of 3 and 5 separately in 300!. Now how do we do so? There are 2 ways based on the same principle.
Note: The question can be solved by finding the highest power of 5 alone. However, we don't advocate using this shortcut until you're 100% confident about these questions. The reason is there are certain complicated factors that you need to keep in mind to use this shortcut without making errors and the advantage gained is not big enough to take that risk. Hence, let's find the highest powers of both the prime factors 3 and 5.
Method 1
Keep dividing the factorial successively by 3 and keep adding the quotient till you don't have anything left to divide. Remember, successive division means dividing the quotient obtained at each step by the same divisor by which we start the division.
So, let's do it quickly.
\(\frac {300}{3} = 100\)
\(\frac {100}{3} = 33\)
\(\frac {33}{3} = 11\)
\(\frac {11}{3} = 3\)
\(\frac {3}{3} = 1\)
\(\frac {1}{3} = 0\)
As there is nothing left to divide, let's add the quotients to find the highest power of 3 in \(300!\)
Sum of quotients \(= 100+33+11+3+1+0 = 148\)
Similarly, let's find the highest power of 5 in 300! by using the same method.
\(\frac {300}{5} = 60\)
\(\frac {60}{5} = 12\)
\(\frac {12}{5} = 2\)
\(\frac {2}{5} = 0\)
As there is nothing left to divide, let's add the quotients to find the highest power of 5 in \(300!\)
Sum of quotients \(= 60+12+2+0 = 74\)
Method 2
Divide 300 by consecutive powers of 3, till you get 0 as a quotient and add all quotients. This method is based on the same principle as Method 1.
So, we have highest power of 3 in 300! = \(\frac {300}{3} + \frac {300}{3^2} +\frac {300}{3^3}+\frac {300}{3^4}+\frac {300}{3^5} = 100+33+11+3+1 = 148\)
So, we can conclude that the highest power of 3 in \(300!\) is \(3^{148}\)
Following a similar process to find the highest power of 5 in 300!, we get that the highest power is \(5^{74}\)
Step-3: Calculate how many such numbers can be created using the highest power of each of its prime factors
Let's try to figure how many 15's we can create using \(3^{148}\) and \(5^{74}\).
Since, \(15 = 3^1*5^1\), we can write \(300!\) as \(300!=3^{148}*5^{74}*k = (3*5)^{74} * 3^{74}*k =15^{74}*3^{74}*k\), where k is a positive integer
Or, in simple terms the highest power of 15 in 300 is \(15^{74}\).
Hence, answer is choice C.
Click here.
Regards,
Piyush
e-GMAT
- Step 1. Prime factorize the number whose highest power is to be found.
Step 2. Find the highest power of each of the prime factors in the factorial
Step 3. Calculate how many such numbers (whose highest power is to be found) can be created using the highest power of each of its prime factors.
Let's apply the above steps to solve this question and then we will look at a couple of questions where we can apply this learning.
We need to find the highest power of 15, so let's begin by doing
Step-1: Prime factorization of 15.
\(15 = 3*5 = 3^1*5^1\)
Step-2: Find the highest power of each of the prime factors in the factorial
Since 15 has two prime factors 3 and 5, we need to find the highest power of 3 and 5 separately in 300!. Now how do we do so? There are 2 ways based on the same principle.
Note: The question can be solved by finding the highest power of 5 alone. However, we don't advocate using this shortcut until you're 100% confident about these questions. The reason is there are certain complicated factors that you need to keep in mind to use this shortcut without making errors and the advantage gained is not big enough to take that risk. Hence, let's find the highest powers of both the prime factors 3 and 5.
Method 1
Keep dividing the factorial successively by 3 and keep adding the quotient till you don't have anything left to divide. Remember, successive division means dividing the quotient obtained at each step by the same divisor by which we start the division.
So, let's do it quickly.
\(\frac {300}{3} = 100\)
\(\frac {100}{3} = 33\)
\(\frac {33}{3} = 11\)
\(\frac {11}{3} = 3\)
\(\frac {3}{3} = 1\)
\(\frac {1}{3} = 0\)
As there is nothing left to divide, let's add the quotients to find the highest power of 3 in \(300!\)
Sum of quotients \(= 100+33+11+3+1+0 = 148\)
Similarly, let's find the highest power of 5 in 300! by using the same method.
\(\frac {300}{5} = 60\)
\(\frac {60}{5} = 12\)
\(\frac {12}{5} = 2\)
\(\frac {2}{5} = 0\)
As there is nothing left to divide, let's add the quotients to find the highest power of 5 in \(300!\)
Sum of quotients \(= 60+12+2+0 = 74\)
Method 2
Divide 300 by consecutive powers of 3, till you get 0 as a quotient and add all quotients. This method is based on the same principle as Method 1.
So, we have highest power of 3 in 300! = \(\frac {300}{3} + \frac {300}{3^2} +\frac {300}{3^3}+\frac {300}{3^4}+\frac {300}{3^5} = 100+33+11+3+1 = 148\)
So, we can conclude that the highest power of 3 in \(300!\) is \(3^{148}\)
Following a similar process to find the highest power of 5 in 300!, we get that the highest power is \(5^{74}\)
Step-3: Calculate how many such numbers can be created using the highest power of each of its prime factors
Let's try to figure how many 15's we can create using \(3^{148}\) and \(5^{74}\).
Since, \(15 = 3^1*5^1\), we can write \(300!\) as \(300!=3^{148}*5^{74}*k = (3*5)^{74} * 3^{74}*k =15^{74}*3^{74}*k\), where k is a positive integer
Or, in simple terms the highest power of 15 in 300 is \(15^{74}\).
Hence, answer is choice C.
To practise ten 700+ Level Number Properties Questions attempt the The E-GMAT Number Properties Knockout
Click here.
Regards,
Piyush
e-GMAT
General Discussion
16 Dec 2016, 05:38
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In order to give you a head-start let's look at the process of solving this question. After going through the process, apply the same to solve the question. We will be posting the official solution soon explaining how to use this process and we will add more practice questions. All the best!!!
In any question where we need to find the highest power of a number which can divide a factorial (product of first n natural numbers) or in other words, is a factor of the given factorial, all we need to do is
Try to apply this process on the above question to get the correct answer. Official solution will be posted soon along with a detailed solution.
Regards,
Piyush
In any question where we need to find the highest power of a number which can divide a factorial (product of first n natural numbers) or in other words, is a factor of the given factorial, all we need to do is
- 1. Prime factorize the number whose highest power is to be found.
2. Find the highest power of each of the prime factors in the factorial
3. Calculate how many such numbers (whose highest power is to be found) can be created using the highest power of each of its prime factors.
Try to apply this process on the above question to get the correct answer. Official solution will be posted soon along with a detailed solution.
Regards,
Piyush
