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

Question

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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freedom128 · Accepted Answer

(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

FINAL ANSWER IS (B)

newyork2012 · Answer

125= 5^3 has 3 0s

1000/125=8

Since 625=5^4 has 4 0s

8-1=7

CrackverbalGMAT · Answer

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!

shameekv1989 · Answer

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

Answer - D

CareerGeek · Answer

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

exc4libur · Answer

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)

unraveled · Answer

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

Answer B.

Kinshook · Answer

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

Fdambro294 · Answer

If you can understand the concept of “skipping trailing zeroes” or understand the pattern of how consecutive factorials after 5! work in regards to trailing zeroes, the problem is not that difficult.

However, most of us don’t know the concept so this is good practice at finding patterns.

Every prime factor pair of (5 *2) included within a Factorial’s prime factorization will produce 1 trailing zero.

There will always be more 2 prime factors than 5 prime factors past 5!

Thus, finding the Count of 5 prime factors in the factorial is enough to determine the number of trailing zeroes (there will always be a 2 prime to pair with the 5)

From 5! - to - 9!——-> we will only have 1 prime factor of 5 ———> only 1 trailing zero

Once we hit 10! ———> since there are now 2 prime factors of 5 ———> 2 trailing zeroes

2 trailing zeroes will continue through 14! until we hit the next multiple of 5 ——> 15! ——> which will have 3 trailing zeroes

Following this pattern:

15! - 19! ——— 3 TZ

20 - 24! ———- 4 TZ

However , when we hit a multiple of (5)^2 = 25 ———> we will “skip a trailing zero” because the added factor of 25 is now adding 2 more trailing zeroes than 24! had

25! - 29! ——- 6 TZ (not 5 TZ)

Thus, in order to “skip 2 trailing zeroes” and go from X tz ——- skip over (X + 1) and (X + 2)——- and go to (X + 3) tz

n + 1 must be a multiple of (5)^3 = 125

124! ——> will have X trailing zeroes

Now when we go to 125! ——-> a factor of (5)^3 has been added to the product chain——> there will be (X + 3) trailing zeroes

Basically, we need to find the multiples of 125 up through 1,000 with one important caveat:

n = 124! ———n + 1 = 125!

250!

375!

500!

***625!***

750!

875!

1,000!

We have 8 times that consecutive factorials will skip more than 2 trailing zeroes.

However, since 625 = (5)^4

624! ——-> will have X trailing zeroes

625! ———> will add FOUR more prime factors of 5, and thus FOUR more trailing zeroes ——-> (X + 4) trailing zeroes

Removing 625! = (n + 1)!

There are 7 numbers of N

B

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

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

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