# Quant Drill: Prime Your Mind For Factor Problems:

It’s July 20, which can be written as 7/20. But if you’re hoping to score 720 or better on the GMAT, you may want to write that as 7/(2 x 2 x 5). Why?

One of your greatest assets on the GMAT quantitative section should be your ability to break down numbers into their prime factors – to be able to analyze the constituent parts of a composite number. The GMAT loves to test your ability to do this through LCM/GCF problems, to reward you for this skill with division/multiplication problems that factor elegantly for those who see that potential, and to feature complex-looking word/exponent/algebra problems that cleverly assess prime factors in a convoluted way. Consider the example:

x is the product of each integer from 1 to 50, inclusive and , where k is an integer . What is the greatest value of k for which y is a factor of x?

(A) 0

(B) 5

(C) 6

(D) 10

(E) 12

Before we even attack this problem, let’s discuss a drill to help you master the art of prime factorization and maximize your recognition of prime factors – and we’ll then show you how that drill will greatly improve your ability to attack problems like the above.

**Prime Counting Drill:**

Count from 1 to 50, hitting each integer in that range, but only allow yourself to use prime numbers. For example, the first 10 numbers are not:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

But instead

1, 2, 3, (2^{^2}), 5, (2 x 3), 7, (2^3), (3^2), (2 x 5)

By doing this drill, you’ll force yourself to think in terms of prime factors and learn to quickly deconstruct a number into its prime, essential components. This ability – and this recognition – is hugely helpful on problems like the above. There, we’re asked to find out how large y can be and still be a factor of x. And since y is a placeholder value for , ultimately we’re being asked how many times 50! can be evenly divided by 100.

When approaching complex factor questions, prime factors are the name of the game. You should recognize that 100 is composed of the prime factors 2 x 2 x 5 x 5, so each time we can pair two 5s with two 2s we can divide by 100. And having performed the drill above you should see how we’ll obtain those 2s and 5s:

1, 2, 3, **(2 ^{^2)}**, 5,

**(2 x 3)**, 7, (

**2**, )

^{^3)}**3**,

^{^2)}**(2 x 5)**

As you’ll see here, there are 2s aplenty – within the first 10 terms of 50! we have eight factors of 2 and only two factors of 5. So 5s will be our limiter, and we’ll look for 5s the same way that we’ve been thinking about numbers throughout this post. We’ll break out any multiple of 5 into its prime factors so that we can pull out all the 5s to maximize the value of .

**5**

10 = 2 x **5**

15 = 3 x **5**

20 = 2 x 2 x **5**

25 =** 5 **x **5**

30 = 2 x 3 x **5**

35 = 7 x **5**

40 = 2 x 2 x 2 x **5**

45 = 3 x 3 x **5**

50 = 2 x **5** x **5**

There are 12 factors of 5 contained within 50!, and since we need 2 for each factor of 100 (and, as we’ve said, there are plenty of 2s to go around), there are 6 sets of 2 x 2 x 5 x 5, and the correct answer here is 6. And you can see, by training yourself to think in terms of prime factors, you have the full skill set available to you – plus the presence of mind to go with it – to attack a difficult question like this.

Recognizing and processing prime factors is a critical skill for high-level success on the GMAT, so you should train yourself to think through this lens. You may recall when you learned to type – for many of us, our initial typing class and its corresponding drills meant that nearly every time you heard a word, you’d mentally flip through a keyboard thinking of how to type it. For prime factorization you should think the same way; when you wake up on Sunday, July 24, you should almost immediately think: 7/(2 x 2 x 2 x 3). Sure, it’s annoying, but like it happened with typing this too will pass and the productive mindset will last. Hopefully it leads to something greater than 2 x 2 x 5 x 5 x 7…

## 12 comments

Srinivas Doijode on July 20th, 2011 at 3:34 am

x is the product of each integer from 1 to 50, inclusive and , where k is an integer . What is the greatest value of k for which y is a factor of x?

(A) 0

(B) 5

(C) 6

(D) 10

(E) 12

X is the product of all intergers from 1 to 50 inclusive.

so X = 50!

y=100^k = 10^(k+2)

Number of zero's in X is quotients of each of (50/5)+(50/25) = 10+2 = 12

hence X is of the form ABCD....X 10^12

hence k+2 = 12

k = 10

Option D.

What am I missing?

nadib002 on July 20th, 2011 at 9:07 am

@Srinivas

10^k = 10^2k, not 10^(k+2)

2k = 12; k = 6

nadib002 on July 20th, 2011 at 9:07 am

sorry, typo

100^k = 10^2k

bharat on July 20th, 2011 at 4:39 am

thank you!!...realy nice explanation.

bharat on July 20th, 2011 at 4:39 am

thank you.....realy nice explanation!!

Mike Rai on July 20th, 2011 at 3:57 pm

Srinivas,

Can you explain how you got this?

Number of zero's in X is quotients of each of (50/5)+(50/25) = 10+2 = 12

lak on July 24th, 2011 at 7:18 pm

hello ppl,

Even am unable to understand the foll:

Number of zero's in X is quotients of each of (50/5)+(50/25) = 10+2 = 12

Also ,the ques asks for the greatest value of k for which y is a factor of X,how does the value 6 prove it??

TX on July 25th, 2011 at 6:15 am

What did the author mean by "there are 6 sets of 2 x 2 x 5 x 5?"

vipin on July 25th, 2011 at 9:14 am

This is very straight question!

Question:

x is the product of each integer from 1 to 50, inclusive

and ,y=100^k where k is an integer .

What is the greatest value of k for which y is a factor of x?

Answer:

x is product of each integer from 1 to 50 i.e. x=50!

and y=100^k.

For "y" to be factor of "x" : x=zy

where z should be an integer.

So, x=z . 100^k

=> 50! = z . 100^k

=> z = 50!/(100^k)

Now "z" will be an integer if 100^k divides 50! without leaving any remainder.

So we have to find no. of zeroes i.e. 10s or (5x2)s in 50!.

Now obviously there will be more 2s than 5s in 50! so we will check only for 5s that will give us the no. of (5x2)s pairs which will give us no. of zeroes in 50!.

No. of 5s in 50! = (50/5)+(10/2) = 12

(to find no. of 5s in 50! : First divide 50 by 5 i.e. 10.

Then divide 10 (from above division) by 5 i.e. 2

So 10+2 = 12)

So back to our equation above : z = 50!/100^k

Since there are 12 zeroes or 10s in 50! so there will be six (6) 100s.

So k=6 for "z" to be an integer.

So Option C is the answer.

Hope this is explains............

vipin on July 25th, 2011 at 9:20 am

Sorry the calculation for 5s in 50! will be :

No. of 5s in 50! = (50/5)+(10/5) = 12

TX on July 29th, 2011 at 8:00 am

How did you come up with this idea? is there a formula for this?

Puneet Goyal on August 10th, 2011 at 11:25 pm

The formula is as follows:

keep dividing the number by 5, 5^2, 5^3...5^n as far as you get integer values.

In this case, we divide by 5 & 25. Dividing by 5^3=125 does not give us integer value (50/125 is non-integer) so we stop at dividing by 5 & 25.