\(x=150!\). We need to find the power of 5 in prime factorization of 150!.

150/5 + 150/5^2 + 150/5^3 = 30 + 6 + 1 = 37 (check here: everything-about-factorials-on-the-gmat-85592.html).

Answer: D.
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It basically asks for the number of 5s in 150!

150/5 + 150/25 + 150/125 = 30 + 6 + 1. Hence 37 Option d)
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total number of 5 is 150/5=30
among 30 there are 25 50 75 100 125 150

contain 1,1,1,1 ,2 , 1 the number of 5 more

total 30+7

d
very hard

1) To paraphrase the question, we need to find all the prime factors 5 of the number 150!
2) 150/5=30; 150/25=6; 150/125=1. The total number of 5's is 30+6+1=37

Posted from my mobile device

This is same as finding trailing zero - right.

Yes. The number of trailing zeros is equal to the number of power of 5 in n!.
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5^3 < 150 < 5^4

Hence, the total number of 5 in 150!:

150/5^1 + 150/5^2 + 150/5^3 = 30 + 6 + 1 = 37

So, y = 37
Ans: D.
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It is asking the number of 5 when the multiplication is written in terms of prime factors

So, y = [150/5] + [150/25] + [150/125] = 30 + 6 + 1 = 37

Answer D
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Spelled out a little more:

1) \(x\) = product of integers from 1 to 150
\(x\) = 150 * 149* 148 . . .* 3 * 2 *1:
That is, \(x\) = 150!

2) \(5^{y}\) is a factor of 150! What is the greatest possible value of \(y\)?

Using \(\frac{150}{5^{y}}\), consider each power \(y\), of 5. Do not worry about remainders.

\(\frac{150}{5^1}\) = 30
(5 divides into 150 thirty times)

\(\frac{150}{5^2}\) = 6
(25 divides into 150 six times)

\(\frac{150}{5^3}\) = 1
(125 divides into 150 only once. Ignore the remainder.)

\(5^4 = 625\) -- too large to divide into 150 as a factor.

3) Add the results: 30 + 6 + 1 = 37

Answer D

Once you know the theory and method, questions such as this one are pretty straightforward. The stats here might indicate that the suggestion below is indispensable.

Bunuel wrote:
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Hi All,

Since we're multiplying a big string of numbers together, this question comes down to "prime factorization"....we need to "find" all of the 5s that exist in this string of numbers. As a hint, some numbers have MORE THAN one 5 in them.

To start, we know that there are 30 multiples of 5 in the string from 1 to 150, so that's 30 5s right there.

Now, we need to think about numbers that have more than one 5 in them....

5, 10, 15....these all have just one 5

25, 50, 75, 100, 150...these all have TWO 5s; we already counted one of the 5s in each, so we have to now add the other one to the total = +5 more

125....this has THREE 5s; we already counted one of the 5s, so we have to now add the other two to the total = +2 more

30 + 5 + 2 = 37 fives.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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x is the product of the integers from 1 to 150, inclusive means x = 150!

5^y is a factor of 150! means, \(\frac{150!}{5^y}\) \(= I\), where "I" is an integer

We need to calculate the no. of 5s in 150!

= \(\frac{150}{5} + \frac{150}{25} + \frac{150}{125}\)

= \(30 + 6 + 1\)

= \(37\)

(D)
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The product of the integers from 1 to 150, inclusive, is 150!. To determine the number of factors of 5 within 150!, we can use the following shortcut in which we divide 150 by 5, and then divide the quotient of 150/5 by 5 and continue this process until we can no longer get a nonzero integer as the quotient.

150/5 = 30

30/5 = 6

6/5 = 1 (we can ignore the remainder)

Since 1/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 150!.

Thus, there are 30 + 6 + 1 = 37 factors of 5 within 150!.

Answer: D
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The answer must be D, i.e. 5 will have a total power of 37 in 150!.
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Bunuel EMPOWERgmatRichC
generis

When is it possible to get extra factors of 5 (or other numbers)? I recall doing a similar problem where there is an exception to this rule of just dividing by increasing amounts. I can't find it, but I think there was some extra factors because it involved adding 2 numbers (probably factorials), which added up to have 2 additional factors of 5 or something.

Hi energetics,

In simple terms, you would look for squares, cubes, quads, etc. that divide evenly into the larger product.

For example, in this prompt:
25 = (25)(1) = 5^2
50 = (25)(2) = (5^2)(2)
75 = (25)(3) = (5^2)(3)
...
125 = (25)(5) = (5^2)(5) = 5^3
Etc.

IF a question asks you to combine two numbers (through addition or multiplication), then there will almost always be some additional factors to account for (otherwise, why would the question ask you to do the additional 'math work'?). Remember that NOTHING about a GMAT question is ever 'random' - each question was written by a human to 'test' you on specific concepts, so one of the most important questions you can ever ask yourself when you're working through a Quant or Verbal question is "why was I given this information (because I was given it for some specific reason)?"

GMAT assassins aren't born, they're made,
Rich
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