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Question

What is the value of N? (1) N! ends with 28 zeroes (2) (N+2)! ends with 31 zeroes and (N-1)! ends with 28 zeroes

theCodeToGMAT · Answer

Is the Answer [spoiler]{E}[/spoiler]?

To find: value of N

Statement 1:
N! ends with 28 zeroes
N!/5 = 28
120/5 = 24, 24/5 = 4, 4/5 = 0 ==> So, 24+4=28
121/5 = 24, 24/5 = 4, 4/5 = 0 ==> So, 24+4=28
So, possible values: 120, 121, 122, 123,124
INSUFFICIENT

Statement 2:
(N+2)! ends with 31 zeroes
129/5 = 25, 25/5 = 5, 5/5 = 1 ==> So, (25+5+1=31)
Same applies to 125 to 129
So N can be 123 to 127

(N-1)! ends with 28 zeroes
120/5 = 24, 24/5 = 4, 4/5 = 0 ==> So, 24+4=28
this applies to 120 to 124
So, N can be 121 to 125

So, "N" can be 123, 124 & 125
INSUFFICIENT

Combining...
From (1) = 120, 121, 122, 123, 124
From (2) = 123, 124 & 125
So, N = 123 & 124
INSUFFICIENT

Answer [spoiler]{E}[/spoiler]

GMATGuruNY · Answer

This problem is about TRAILING 0's: the number of 0's at the end of a large product.

If N = 120, then N! = 120*119*118*....*3*2*1.

Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 120! will yield a 0 at the end of the integer representation of 120!.
The prime-factorization of 120! is composed of FAR MORE 2'S than 5's.
Thus, the number of 0's depends on the NUMBER OF 5's contained within 120!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 120.

Every multiple of 5 within 120! provides at least one 5:
120/5 = 24 --> 24 5's.
Every multiple of 5² provides a SECOND 5:
120/5² = 4 --> 4 more 5's.
Thus, the total number of 5's contained within 120! = 24+4 = 28.

Note that 121!, 122!, 123!, and 124! all contain the same number of 5's as 120!.

For more than 28 5's to be contained within N!, N must be AT LEAST 125, the next greatest multiple of 5 after 120.
Every multiple of 5 within 125! provides at least one 5:
125/5 = 25 --> 25 5's.
Every multiple of 5² provides a SECOND 5:
125/5² = 5 --> 5 more 5's.
Every multiple of 5³ provides a THIRD 5:
125/5³ = 1 --> 1 more 5.
Thus, the total number of 5's contained within 125! = 25+5+1 = 31.

Note that 126!, 127!, 128!, and 129! all contain the same number of 5's as 125!.

Onto the problem:

kop wrote:What is the value of N?

(1) N! ends with 28 zeroes

(2) (N+2)! ends with 31 zeroes and (N-1)! ends with 28 zeroes

The following cases satisfy both statements:

Case 1: N=123
(123+2)! = 125!, which has 31 5's.
(123-1)! = 122!, which has 28 5's.

Case 2: N=124
(124+2)! = 126!, which has 31 5's.
(124-1)! = 123!, which has 28 5's.

Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.

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Mathsbuddy · Answer

This looks like a really tough question, so I've tried to simplify its appearance.

x, y and z are integers which are not multiples of 10:

N! = x * 10^28
(N+2)! = y * 10^31
(N-1)! = z * 10^28

Furthermore:

(N+2)! = N!(N+1)(N+2) by definition

Substituting the given facts, makes this:

y * 10^31 = x * 10^28 * (N+1)(N+2)

so y * 10^3 = x * (N+1)(N+2)

As x cannot be a multiple of 10^3, then (N+1)(N+2)= k x 10^3 (where k is an integer)

From the given answers, we know that N = 120 + p (where 0 <= p <= 5 is an integer)

By substitution:

(p+121)(p+122) = k x 10^3

By considering the LAST DIGITS ONLY:

(p+1)(p+2) = 10t (where t is an integer)

Now try the possible values of p:

p=0 -> 1 * 2 = 2 does not end in zero
p=1 -> 2 * 3 = 6 does not end in zero
p=2 -> 3 * 4 = 12 does not end in zero
p=3 -> 4 * 5 = 20 OK
p=4 -> 5 * 6 = 30 OK
p=5 -> 6 * 7 = 42 does not end in zero

Using N = 120 + p we get

N = 123, 124

This matches solutions by other methods.

So now let's see if we can eliminate either of these 2 solutions!

N! = (N-1)! * N by definition.

So, N = N!/(N-1)! = x * 10^28/(z * 10^28) = x/z

Hence z * N = x where no variables are a factor F of 10, i.e. the last digit is not zero.

If N = 123, then it is certain (for all values of the last digit of z) that x is not 10F.

If N = 124, then z must not end in 5; so z (last digit) = 1,2,3,4,6,7,8,9.

INSUFFICIENT.

There's one more equation to test drive:

(N+2)! = (N-1)! * N(N+1)(N+2) by definition.

Lets try N = 123:

y * 10^31 = z * 10^28 * (123)(124)(125)

so y * 10^3 = z * 1906500 which is not a multiple of 10^3 -> ELIMINATE!

Now try N = 124:

y * 10^31 = z * 10^28 * (124)(125)(126)

so y * 10^3 = z * 1953000 which IS a multiple of 10^3 -> SINGLE SOLUTION

Therefore, I propose that there is just one answer: N = 124

Job done, but please somebody check through it and let me know if its correct.

Mathsbuddy · Answer

GMATGuruNY wrote:This problem is about TRAILING 0's: the number of 0's at the end of a large product.

If N = 120, then N! = 120*119*118*....*3*2*1.

Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 120! will yield a 0 at the end of the integer representation of 120!.
The prime-factorization of 120! is composed of FAR MORE 2'S than 5's.
Thus, the number of 0's depends on the NUMBER OF 5's contained within 120!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 120.

Every multiple of 5 within 120! provides at least one 5:
120/5 = 24 --> 24 5's.
Every multiple of 5² provides a SECOND 5:
120/5² = 4 --> 4 more 5's.
Thus, the total number of 5's contained within 120! = 24+4 = 28.

Note that 121!, 122!, 123!, and 124! all contain the same number of 5's as 120!.

For more than 28 5's to be contained within N!, N must be AT LEAST 125, the next greatest multiple of 5 after 120.
Every multiple of 5 within 125! provides at least one 5:
125/5 = 25 --> 25 5's.
Every multiple of 5² provides a SECOND 5:
125/5² = 5 --> 5 more 5's.
Every multiple of 5³ provides a THIRD 5:
125/5³ = 1 --> 1 more 5.
Thus, the total number of 5's contained within 125! = 25+5+1 = 31.

Note that 126!, 127!, 128!, and 129! all contain the same number of 5's as 125!.

Onto the problem:

kop wrote:What is the value of N?

(1) N! ends with 28 zeroes

(2) (N+2)! ends with 31 zeroes and (N-1)! ends with 28 zeroes
The following cases satisfy both statements:

Case 1: N=123
(123+2)! = 125!, which has 31 5's.
(123-1)! = 122!, which has 28 5's.

Case 2: N=124
(124+2)! = 126!, which has 31 5's.
(124-1)! = 123!, which has 28 5's.

Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.

Hi there GMatguruNY,

I just posted a reply that seems to prove that there is a single solution.
Here is the last part of the argument (having reduced N to N = 123, 124):
[Note: the full post explains all the terms, etc.]

(N+2)! = (N-1)! * N(N+1)(N+2) by definition.

Lets try N = 123:

y * 10^31 = z * 10^28 * (123)(124)(125)

so y * 10^3 = z * 1906500 which is not a multiple of 10^3 -> ELIMINATE!

Now try N = 124:

y * 10^31 = z * 10^28 * (124)(125)(126)

so y * 10^3 = z * 1953000 which IS a multiple of 10^3 -> SINGLE SOLUTION

Therefore, I propose that there is just one answer: N = 124

Could you please let me know if you can spot a mistake? Thanks.

GMATGuruNY · Answer

Mathsbuddy wrote:As x cannot be a multiple of 10^3, then (N+1)(N+2)= k x 10^3 (where k is an integer)

The portion in red overly constrains the value of (N+1)(N+2).
In your solution, N! = x * 10²�.
While x is not a multiple of 10³, it IS a multiple of 2³.
Thus, we can represent N! as follows:
N! = y * 2³ * 10²�.
Thus, if (N+1)(N+2) = z * 5³, then (N+2)! = y * z * 2³ * 5³ * 10²� = yz * 10³¹.
Implication:
(N+1)(N+2) = z * 5³.

Mathsbuddy · Answer

GMATGuruNY wrote:Mathsbuddy wrote:As x cannot be a multiple of 10^3, then (N+1)(N+2)= k x 10^3 (where k is an integer) The portion in red overly constrains the value of (N+1)(N+2). In your solution, N! = x * 10²�. While x is not a multiple of 10³, it IS a multiple of 2³. Thus, we can represent N! as follows: N! = y * 2³ * 10²�. Thus, if (N+1)(N+2) = z * 5³, then (N+2)! = y * z * 2³ * 5³ * 10²� = yz * 10³¹. Implication: (N+1)(N+2) = k * 5³. Thank you, I see what you mean. However x, y and z are defined to not end in zero. Therefore surely it is implicit that (N+1)(N+2)ends in three zeros?

Mathsbuddy · Answer

To eliminate 123:

to get from (n-1)! to (n+2)! we multiply by n * (n+1) * (n+2)

so if N = 123:

we multiply by 123 * 124 * 125 which ends in 2 noughts and not 3 noughts, as required by changing 28 to 31 noughts.

Therefore eliminate N = 123, or indeed any value for N that ends in a 3.

GMATGuruNY · Answer

Mathsbuddy wrote:To eliminate 123:

to get from (n-1)! to (n+2)! we multiply by n * (n+1) * (n+2)

so if N = 123:

we multiply by 123 * 124 * 125 which ends in a nought

Therefore we would have to change 28 noughts to 29 noughts, which conflicts with the question.

Therefore eliminate N = 123, or indeed any value for N that ends in a 3.

Let's dig deeper.

To count the number of 5's contained within 123!, divide increasing POWERS OF 5 into 123.
Every multiple of 5 contained within 123! provides at least one 5:
123/5 = 24 --> 24 5's.
Every multiple of 5² provides a SECOND 5:
123/5² = 4 --> 4 more 5's.
Thus, the total number of 5's contained within 123! = 24+4 = 28.

To count the number of 2's contained within 123!, divide increasing POWERS OF 2 into 123.
Every multiple of 2 contained within 123! provides at least one 2:
123/2 = 61 --> 61 2's.
Every multiple of 2² provides a SECOND 2:
123/2² = 30 --> 30 more 2's.
Every multiple of 2³ provides a THIRD 2:
123/2³ = 15 --> 15 more 2's.
Every multiple of 2� provides a FOURTH 2:
123/2� = 7 --> 7 more 2's.
Every multiple of 2� provides a FIFTH 2:
123/2� = 3 --> 3 more 2's.
Every multiple of 2� provides a SIXTH 2:
123/2� = 1 --> 1 more 2.
Thus, the total number of 2's contained within 123! = 61+30+15+7+3+1 = 117.

Thus:
123! = 2¹¹� * 5²� * k, where k is not a multiple of 2 or 5.

Every 2*5 contained within 123! will yield a trailing 0.
123! = 2�� * (2²� * 5²�) * k = 2�� * k * 10²�.
Thus, the total number of trailing 0's = 28.

Every 2*5 contained within 125! will yield a trailing 0.
125! =

= 123! * 124 * 125

= (2�� * k * 10²�) * (2² * 31) * 5³

= 2�� * k * 10²� * 31 * (2³ * 5³)

= 2�� * k * 10²� * 31 * 10³

= 2�� * 31 * k * 10³¹.
Thus, the total number of trailing 0's = 31.

I hope that the forgoing explanation clarifies why 123! will yield 28 trailing 0's, while 125! will yield 31 trailing 0's.

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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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Mathsbuddy · Answer

GMATGuruNY wrote:Mathsbuddy wrote: I hope that the forgoing explanation clarifies why 123! will yield 28 trailing 0's, while 125! will yield 31 trailing 0's. Thank you for your time and insight. I see my flaw now: To not eliminate 123! to get from (n-1)! to (n+2)! we multiply by M = n * (n+1) * (n+2) so if N = 123: we multiply (n-1)! by M = 123 * 124 * 125 = 1906500 At this point I mistakingly thought the 2 noughts at the end meant it would not increase by 3 noughts. However, as (n-1)! = 122! is even, then (n+2)! could be a multiple of 2*1906500 = 3813000 which would add three noughts to the end of (n-1)! Nonetheless, it would have to be the last digit before the noughts that needs to be even. Hence I will concede to GMATGuruNY's method and congratulate him for his patience and logic! Thank you.

Mathsbuddy · Answer

After much deliberation considering the the expert's method and my own (which contained originally a crucial oversight), I believe the quickest way to solve this in future would be to work out the multiplier of (N-1)! that makes (N+2)! with each answer option:

Using formula (N+2)!/(N-1)! = N(N+1)(N+2) = 1000/k so requires 2 or 3 zeroes

N=120 -> 120 * 121 * 122 = 1771440 = NOT ENOUGH ZEROES
N=121 -> 121 * 122 * 123 = LAST DIGIT NOT ZERO (worked out by just multiplying last digits)
N=122 -> 122 * 123 * 124 = LAST DIGIT NOT ZERO (worked out by just multiplying last digits)
N=123 -> 123 * 124 * 125 = 1906500 OK BECAUSE k is even -> 1000/k = 500 (last 3 digits)
N=124 -> 124 * 125 * 126 = 1953000 TOTALLY OK (last 3 digits are zero)
N=125 -> 125 * 126 * 127 = LAST DIGIT NOT ZERO (worked out by just multiplying last digits)

Hence N = 123, 124

Not sufficient reduce to a unique solution.

The prime factor method is excellent. This alternative could just save time in a test. Essentially it's the same logic: each zero can be 'made' by directly multiplying by 10, or by multiplying by 5 (as (N-2)! will always be even, including the last non-zero digit). Both methods assess multiplying by 2 and 5. Considering the the multiples of 10 eliminates several options very quickly. Then looking for a multiple of 5 is not too difficult for the remaining options. However, if this is not understood, then stick with prime factors! (Disclaimer: I am not an instructor, so take heed!)

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