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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes

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Math Expert V
Joined: 02 Sep 2009
Posts: 64073
n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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2
7 00:00

Difficulty:   65% (hard)

Question Stats: 15% (00:37) correct 85% (02:16) wrong based on 47 sessions

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Competition Mode Question

n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes at the end. If n is a positive integer and 1 ≤ n ≤ 1000. How many values of n are possible ?

A. 8
B. 7
C. 6
D. 4
E. 1

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Director  V
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Re: n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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9
1
(n+1) must be a multiple of $$125=5^3$$ and n<=1000, so that the stated conditions in the stimulus (or question) are satisfied. For example: $$n! =124!$$ and $$(n + 1)! =125! =125*n! = 5*5*5*n!$$ will convert three unused factors of 2 to form $$5^3*2^3=1000$$

So, (n + 1) has 8 potential values: 125, 250, 375, ....,1000. However, we need to eliminate (n + 1) = 625 since $$(n+1)! =625! =625*624! =5*5*5*5*n!$$ will convert four unused factors of 2 to form $$5^4*2^4=10,000$$, instead of three factors of 2.

No of possible values of n are 8 - 1 = 7

##### General Discussion
Manager  B
Joined: 22 Sep 2014
Posts: 157
Location: United States (CA)
n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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125= 5^3 has 3 0s

1000/125=8

Since 625=5^4 has 4 0s

8-1=7

Originally posted by newyork2012 on 27 Feb 2020, 00:19.
Last edited by newyork2012 on 28 Feb 2020, 00:04, edited 1 time in total.
CrackVerbal Quant Expert P
Joined: 12 Apr 2019
Posts: 587
Re: n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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2
The number of zeroes at the end of a number depends on the highest power of 10 in that number. The highest power of 10 is in turn dependent on the highest power of 5 in that number.
Now, both n! has all the same numbers as (n+1)! except for (n+1) itself.

As per the question, n! has x zeroes at the end while (n+1)! has (x+3) zeroes at the end. This means that the highest power of 5 in n! is $$5^x$$ and in (n+1)! is $$5^{x+3}$$. Therefore,
n! = $$5^x$$ * (other numbers which are part of n!)
(n+1)! = $$5^{x+3}$$ *(other number which are part of (n+1)!)

Now we know that (n+1)! = (n+1) * n!

When we substitute the values of n! and (n+1)!, we have,

$$5^{x+3}$$ *(other number which are part of (n+1)!) = (n+1) * $$5^x$$ * (other numbers which are part of n!)
Note that the other numbers which are part of (n+1)! and n! are same and hence can be cancelled out. $$5^x$$ can be cancelled out as well. We are left with (n+1) = $$5^3$$ which means that n = 124.
This is the only value possible for n. The correct answer option is E.

A quick way of checking your answer is by finding out the highest powers of 5 in 124! and in 125!. The highest power of 5 in 124! is 28 and the highest power of 5 in 125! is 31. Clearly, the two of these differ by 3 which is what the question says.

Hope that helps!
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Re: n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes at the end. If n is a positive integer and 1 ≤ n ≤ 1000. How many values of n are possible ?

A. 8
B. 7
C. 6
D. 4
E. 1

Number of zeroes are contributed by 2,5 pairs in the prime factorization. Since there will be less exponents of 5 we can concern ourselves only with exponents of 5.

n! have x zeroes and n+1! have (x+3) zeroes. These 3 extra zeroes in (n+1)! will be contributed by extra 5's. Whenever there would be 5^3 added 3 extra 5's will be added.

Therefore n=124; n+1=125 will be the first such pair. Then we can just add 125 and find other values until we get 625 in which case the number of extra 5's will increase by 4 everytime.

Therefore n=124, 249, 374, 499 => 4 values [Next one will be 624,625 but the difference would be x+4 zeroes]

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Re: n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes at the end. If n is a positive integer and 1 ≤ n ≤ 1000. How many values of n are possible ?

A. 8
B. 7
C. 6
D. 4
E. 1

$$(n + 1)!$$ factorial can add $$3$$ zeros if that number is a multiple of $$5^3$$ or cube of $$5$$ multiples ($$10^3$$ or $$15^3$$ so on ....)
--> Number of possible values of $$(n + 1)$$ = number of multiples of $$5^3$$ till $$1000$$ = {$$125, 250, 375, 500, 625, 750, 875, 1000$$}
--> Possible values of n = {$$124, 249, 374, 499, 624, 749, 874, 999$$} = $$8$$ values

Option A
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Re: n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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1
Quote:
n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes at the end. If n is a positive integer and 1 ≤ n ≤ 1000. How many values of n are possible ?

A. 8
B. 7
C. 6
D. 4
E. 1

Find trailing zeros: sum quotients of n!/5^(1,2,…n) in which 0≤n≤1000

n!=24, sum quotients: 24/5=4
n+1!=25, sum quotients: 25/5=4 + 25/25=1, total=5

n!=124, sum quotients: 124/5+124/25=24+4=28=x
n+1!=125, sum quotients: 125/5+125/25+125/125=25+5+1=31=x+3

n!=125, sum quotients=31
n+1!=126, sum quotients=31

n!=149, sum quotients=37-1-1=35
n+1!=150, sum quotients=37

n!=249, sum quotients=62-1-1-1=x
n+1!=250, sum quotients=62=x+3

n!=374, sum quotients=93-1-1-1=x
n+1!=375, sum quotients=93

n!=624, sum quotients=93-1-1-1-1
n+1!=625, sum quotients=156

find positive multiples of 125 ≤ 1000:
1000-125/125+1=8

find positive multiples of 625 ≤ 1000 = 1

find difference multiples of 125 and 625:
8-1=7

Ans (B)
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n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes at the end. If n is a positive integer and 1 ≤ n ≤ 1000. How many values of n are possible ?

A. 8
B. 7
C. 6
D. 4
E. 1

As 1 ≤ n ≤ 1000, number of 0's in n! are decided by $$5^x$$ where 1≤x≤4(since 5^5 = 3125 would always result in 0 number of 0's on dividing 1000 by it). However, 5^4 have to be checked since it has an additional factor which results in 4 extra 0's wherein we are asked about only 3.
So, total number of 0's can be found as follows"
$$\frac{n!}{5^1}$$
$$\frac{n!}{5^2}$$
$$\frac{n!}{5^3}$$
$$\frac{n!}{5^4}$$
where each of the additional power of 5 would contribute an additional '0' in case of $$(n+1)!$$

It has to be noted that a difference of 3 0's would only be possible if $$divisor ≥ 5^3$$ is counted i.e. n has to be of 3 digits so $$n+1$$ precisely should be a multiple of $$5^3$$ or $$5^4$$.

Hence multiples of $$5^3$$ are 125, 250, 375 ..... 1000, a total of 8($$\frac{1000}{125}$$; including a common multiple of $$5^4$$ i.e. 625).
For example -
n + 1 = 125 so n = 124
Number of 0's in 124! = $$\frac{124}{5^1} + \frac{124}{5^2} + \frac{124}{5^3} + \frac{124}{5^4}$$
= 24 + 4 + 0 + 0 = 28
&
Number of 0's in 125! = $$\frac{125}{5^1} + \frac{125}{5^2} + \frac{125}{5^3} + \frac{125}{5^4}$$
= 25 + 5 + 1 + 0 = 31

Similarly, for n = 999 the number of 0's are
Number of 0's in 999! = $$\frac{999}{5^1} + \frac{999}{5^2} + \frac{999}{5^3} + \frac{999}{5^4}$$
= 199 + 39 + 7 + 1 = 246
&
Number of 0's in 1000! = $$\frac{1000}{5^1} + \frac{1000}{5^2} + \frac{1000}{5^3} + \frac{1000}{5^4}$$
= 200 + 40 + 8 + 1 = 249

Checking for n = 624 the number of 0's are
Number of 0's in 624! = $$\frac{624}{5^1} + \frac{624}{5^2} + \frac{624}{5^3} + \frac{624}{5^4}$$
= 124 + 24 + 4 + 0 = 152
&
Number of 0's in 625! = $$\frac{625}{5^1} + \frac{625}{5^2} + \frac{625}{5^3} + \frac{625}{5^4}$$
= 125 + 25 + 5 + 1 = 156
So there is a difference of 4 0's so this is not an applicable case here.

So, possible number of values that n can take = 8 - 1 = 7

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Re: n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  [#permalink]

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Bunuel wrote:

Competition Mode Question

n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes at the end. If n is a positive integer and 1 ≤ n ≤ 1000. How many values of n are possible ?

A. 8
B. 7
C. 6
D. 4
E. 1

Are You Up For the Challenge: 700 Level Questions

Since by adding 1 to n; number of zeros at end increase by 3.
n+1 should have 5^3 has a factor
n+1 = 5^3k ; where k is an integer but not a multiple of 5

n+1 = {125;250;375;500;750;875;900}

IMO B
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Email: kinshook.chaturvedi@gmail.com Re: n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes   [#permalink] 28 Mar 2020, 04:08

# n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes  