If x = a!, which of the following values of a is the least possible va

Question

If x = a!, which of the following values of a is the least possible value for which the last 8 digits of the integer x will all be zero?
A) 21
B) 29
C) 38
D) 40
E) 100

syahasa2 · Accepted Answer

(A) 21/5 + 21/5^2= 4+0=4 5s, therefore 4 0s.
(B) 29/5 + 29/5^2= 5+1=6 5s, therefore 6 0s.
(C) 38/5 + 38/5^2= 7+1=8 5s, so 8 0s.  Correct

abhimahna · Answer

We need to find out the No. of 5s in a! to find out the number of zeros at the end.

Option A = 21/5 = 4. Hence, max 5s are 4. Hence, max zero would be 4.

Option B = 29/5 = 5/5 = 1. Hence, max 5s are 6 Hence, max zero would be 6.

Option C = 38/5 = 7/5 = 1. Hence, max 5s are 8 Hence, max zero would be 8. Hence, the answer C

Abhishek009 · Answer

a! must have 8 trailing zeroes...

We know that in order to have 8 trailing zero's we must decide a by 5 and continue the process till be get answer 8 or More than 8

a/5 = 8

So, a ~ 40

Or, a must be close to 40...

Check options close to 40 , Only option (C) and (D) in this case...

Check option (C)

38/5 = 7
7/5 = 1
  
So, Option (C) is correct...   
  
PS: I just used approximation to arrive at the answer quickly

Krasniqi123 · Answer

Why should we find the No. of 5s in a!. I don't understand. 
And what do you mean by 5s? For example, number 21 has 5,10,15,20 - so, 4 times 5s. Is this what you mean by the number of 5s?

abhimahna · Answer

One zero will come when we multiply a number with 10.

Now, 10 = 2 *5

Out of the 2 and 5, we will always have the number of 5s less than the number of 2's.

Hence, if we find the number of 5's we will atleast have that many 2's but the vice viersa is not true.

hence, we are find the number of 5's

Think it like this :

in 10!, we will have more 2's than 5's.

So, I can say 10/5 = 2  5's  are present in 10!

if we find the number of 2's, I will get 10/2 = 5 / 2= 2 or 5+2 = 7 2's. But since we have only 2 5's we have only two zero's at the end of 10!.

I hope the concept is clear now.

Abhishek009 · Answer

Go through the concept on Factorials in our GMAT Quantz Book Here - math-number-theory-88376.html#p666609 Hope this will surely help you..

mvictor · Answer

i think this question is testing our knowledge of trailing zeroes...
to have 8 digits of zero in the end, we need to have 8 factors of two, and 8 factors of 5.
let's see.
since we can very fast get to 8 factors of two, let's see how we can get to 8 factors of 5...
5 (1),... 10 (2), ... 15 (3), ... 20 (4), ... 25 (6), ... 30 (7), ... 35 (8).
A, B are right away out.
since we need the least possible value, 38 works...
C is the answer.

vitaliyGMAT · Answer

We have infinite series which sum approximately equals to  8

\frac{a}{5} + \frac{a}{5^2} + \frac{a}{5^3} + … ≈ 8

\frac{a}{5} / (1 - \frac{1}{5}) ≈ 8

a ≈ 32

+   = 7

From 30 to 34 to 34 there are 7 trailing zeros.

From 35 to 39 there are 8 trailing zeros.

Answer C.

JeffTargetTestPrep · Answer

We are given that x = a! and need to determine the least value for a that will create 8 trailing zeros in integer x. We must remember that a trailing zero is created with each 5-and-2 pair within the prime factorization of a!. Furthermore, since we know there are fewer 5s in a! (or any factorial) than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

Let’s test answer choice A:

To determine the number of 5s within 21!, we can use the following shortcut in which we divide 21 by 5, and then divide the quotient of 21/5 by 5 and continue this process until we no longer get a nonzero quotient:

21/5 = 4 (we can ignore the remainder)

Since 4/5 does not produce a nonzero quotient, we can stop.

21! contains four 5-and-2 pairs, and thus 21! does not contain 8 trailing zeros.

Next we can test answer choice B:

29/5 = 5 (we can ignore the remainder)

5/5 =1 (we can ignore the remainder)

Since 1/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 29!. Thus, there are 5 + 1 = 6 factors of 5 within 29!

29! contains six 5-and-2 pairs, and thus 29! does not contain 8 trailing zeros.

Next we can test answer choice C:

38/5 = 7 (we can ignore the remainder)

7/5 =1 (we can ignore the remainder)

Since 1/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 38!. Thus, there are 7 + 1 = 8 factors of 5 within 38!

38! contains eight 5-and-2 pairs, and thus 38! does contain 8 trailing zeros.

Answer: C

cavana · Answer

1) We need to find the first 8 5's in the sequence of consecutive integers: 5*10*15*20*25(5*5)*30*35
2) 35 is the smallest value of a, but from the given choices, it is 38

The correct answer is C

exc4libur · Answer

num of zeros in a factorial  n!  is the sum of the quotients of  n  divided by  5^1,5^2,5^3,5^4… 
num of zeros in  40!  is  40/5+40/25=8+1=9 
least possible value for which the last 8 digits are zeros must be a  35≤factorial<40

Answer (C).

orangebicycle5 · Answer

We need 8 pairs of 5*2. Limited by the primes of 5.

Pick a number in the middle of answer choices and test.

38/5 = 7, 38/25 = 1. This would = 8 trailing zeroes and is smaller than 40 and 100.

DanTheGMATMan · Answer

https://www.youtube.com/watch?v=9qIOg06DQPk

bumpbot · Answer

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If x = a!, which of the following values of a is the least possible va

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If x = a!, which of the following values of a is the least possible va

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