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# If x is the product of the integers from 1 to 150, inclusive

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Intern
Joined: 04 Jul 2013
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If x is the product of the integers from 1 to 150, inclusive [#permalink]

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20 Oct 2013, 07:13
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Difficulty:

35% (medium)

Question Stats:

65% (00:35) correct 35% (01:00) wrong based on 1002 sessions

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If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39
[Reveal] Spoiler: OA
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Joined: 02 Sep 2009
Posts: 44319
Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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20 Oct 2013, 07:17
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bgribble wrote:
If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

$$x=150!$$. 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).

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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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20 Oct 2013, 07:18
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bgribble wrote:
If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

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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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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20 Oct 2013, 07:18
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Bunuel wrote:
bgribble wrote:
If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

$$x=150!$$. 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).

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Hope it helps.
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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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30 Apr 2015, 00:28
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bgribble wrote:
If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

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
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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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25 Jan 2017, 09:09
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

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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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23 Oct 2017, 07:46
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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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23 Oct 2017, 07:49
divyakesharwani wrote:
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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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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23 Oct 2017, 08:11
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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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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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23 Oct 2017, 09:32
bgribble wrote:
If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

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

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If x is the product of the integers from 1 to 150, inclusive [#permalink]

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23 Oct 2017, 10:55
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bgribble wrote:
If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

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

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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Re: If x is the product of the integers from 1 to 150, inclusive [#permalink]

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14 Mar 2018, 19:11
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.

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Re: If x is the product of the integers from 1 to 150, inclusive   [#permalink] 14 Mar 2018, 19:11
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