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If x is the product of numbers 1 through 150....

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by theCodeToGMAT » Thu Sep 26, 2013 10:09 am
To find "y" we need power of 5 in 150!
so,
150/5 = 30
30/5 = 6
6/5 = 1
1/5 = 0

Answer = 30 + 6 + 1 = 37
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by [email protected] » Thu Sep 26, 2013 11:58 am
Hi fourteenstix,

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: D

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Rich
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by helen.xia@mbawatch » Thu Sep 26, 2013 12:02 pm
The product of 1 to 150 can be written as 150!. To find how many 5's can be divided into 150! I listed all the numbers in the 150! pattern that are divisible by 5:
5, 10, 15, 20.....150

going over those number the first time, there are 30 numbers.

Of those 30 numbers, 25, 50, 75, 100, and 150 are divisible by 5 twice. 125 is divisible by 5 three times.

Therefore: 30 + 5 + 2 = 37.

I guess I don't have the best approach to this problem..can you confirm the answer?
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by GMATGuruNY » Thu Sep 26, 2013 12:16 pm
x = 150!.
5^y = the number of 5's that can be divided into x.
Count how many times EACH POWER OF 5 can be divided into 150:

150/5¹ = 30.
The calculation above indicates that 150! includes 30 multiples of 5¹.
Each multiple of 5¹ supplies at least one 5.

150/5² = 150/25 = 6.
The calculation above indicates that 150! includes 6 multiples of 5², each of which supplies an additional 5.

150/5³ = 150/125 = 1.
The calculation above indicates that 150! includes one multiple of 5³, supplying one more 5.

Thus, the maximum number of 5's that can be divided into x = 30+6+1 = 37.

The correct answer is D.

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by rahul.sehgal@btgchampion » Sat Oct 05, 2013 10:08 am
theCodeToGMAT wrote:To find "y" we need power of 5 in 150!
so,
150/5 = 30
30/5 = 6
6/5 = 1
1/5 = 0

Answer = 30 + 6 + 1 = 37

Can you help me understand how you did this ? I mean, why you divide this way ? I might be missing a trick here.
Best Regards,
Rahul Sehgal
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by theCodeToGMAT » Sat Oct 05, 2013 10:33 am
rahulsehgal wrote:
Can you help me understand how you did this ? I mean, why you divide this way ? I might be missing a trick here.
Refer the Mitch's solution above your post.. the method is exactly the same..
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by gmattesttaker2 » Sun Jun 08, 2014 9:53 pm
[email protected] wrote:Hi fourteenstix,

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: D

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

Thanks for the explanation. Here:
5, 10, 15....these all have just one 5
we have 30 5's.

I have tried to understand this as follows:

For 25, 50, 75, 100, 125, 150 we have one more 5 in each. Hence, we have 6 5's here.

For 125, we have one more 5.

Hence, total = 30 + 6 + 1 = 37

Is this correct?

Thanks a lot for your help.

Best Regards,
Sri
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by [email protected] » Sun Jun 08, 2014 10:26 pm
Hi Sri,

Yes, your way of organizing this question is correct. One of the "design elements" of most GMAT questions is that they can be solved in a variety of ways (even the "math" can be done in more than one way). This design rewards flexible thinkers; it also means that you won't necessarily get "stuck" if you're not sure how to solve a particular question - there's more than 1 way to the solution, so try to do something.

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