x is the product of each integer from 1 to 50, inclusive and y=100^k,

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?

(A) 0
(B) 5
(C) 6
(D) 10
(E) 12

Kudos for a correct  solution .

GMATinsight · Accepted Answer

x = 1*2*3*4*5*......*49*50 = 50!
y = 100^k

100 = 10^2 = 2^2 * 5^2

On prime factorization of x, power of 5 will always be greater than power of 2

CONCEPT: In every Factorial value (x!), On prime factorization, the power of Bigger Prime number > the power of Smaller Prime number

Also,  Power of any Prime Number in any factorial can be calculated by following understanding

Power of prime x in n! =   +   +   +   + ... and so on
Where,
  = Greatest Integer less than or equal to x and the explanation of terms is as follows
  = No. of Integers that are multiple of x from 1 to n
  = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step
  = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step
And so on.....

Where   is greatest Integer value of (n/x) less than or equal to (n/x)
i.e.   =   = 33
i.e.   =   = 11 etc.

Power of 5 in x =   +   +  ... = 10 + 2 + 0 = 12
Power of 2 in x =   +   +  ... = 25 + 12 + 6 + 3 + 1 = 47

i.e. x = 50! = 2^47 * 5^12 * ... = 10^12 * ... = 100^6 * ...

i.e. x is divisible by a 6th power of 100 at the most
i.e. y = 100^k = 100^6
i.e. k = 6

Answer: option C

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

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 100^k, 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 100^k.

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…

shailendra79s · Answer

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
n5+n52+n53+...+n5k, where k must be chosen such that 5k≤n
x = 1*2*3....*50 = 50!
No. of trailing zeros in 50! = 50/5 + 50/5^2 = 10+2 = 12
100^k = 10^2k → k = 12/2 = 6
Answer: C

RMD007 · Answer

The question is basically asking us to find how many 100's can be formed with product 1 to 50.

100 = 5^2*2^2

We have many 2^2 in the product of 1 to 50. So we have to find how many 5^2 we have in the product.
Total number of 5's we have are 10+2 =12. So 5^2 available are - 6.

Hope it helps..

Bunuel  , please correct my understanding if it is wrong.

Abhishek009 · Answer

x = 50! 
 y = 10^{2k}

Highest power of k will be -

10 = 2*5

So,  10^2k = 2^{2k}5^{2k}

Now, 50! will contain 50/5 = 10 , 10/5 = 2

So, k = 6 , As 10^12 = 10^{2k}, Answer must be (C)

bumpbot · Answer

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x is the product of each integer from 1 to 50, inclusive and y=100^k,

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