What is the largest value of non negative integer N for which 10N is a

Hi,
The reasoning for these type of Qs is...

What does ! Or factorial mean ..
It means product of all positive integers till that number..
So 50! Means 1*2*3*4*5....*49*50
Now when we multiply 2 and 5 we get a 10..
So 2s will be more than 5s and that is why number of 10s will be equal to number of 5s...

Now how do we find number of 5s in 50!..
When you divide by 5 you get all multiples of 5 till 50..
50/5=10..
These 10 numbers are 5,10,15,20,25,30,35,.40,45,50..
When we divide by 25 we get numbers which had 5^2 in it..
50/25 =2..
And these are 25 Aand 50..
So you have 12 5s or 10s..
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To determine the largest number of N, we need to determine how many times 10 divides 50!. Since 10 breaks into primes of 5 and 2, and since there are there are fewer 5s in 50! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 50!, we can use the following shortcut in which we divide 50 by 5, then divide the quotient of 50/5 by 5 and continue this process until we no longer get a nonzero quotient.

50/5 = 10

10/5 = 2

Since 2/5 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 50!.

Thus, there are 10 + 2 = 12 factors of 5 and hence 12 factors of 10 within 50!

Answer: C
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Should the question read \(10^N\) rather than 10N? That's what you'd need in order for the answers to make sense (otherwise, 10N would be a factor for all of these answer choices.)
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Considering the highlighted part to be 10^n

50! has as many 10's as it have 5's
calculating 5's
50!/5=10
50!/25=2
sum=12

Ans C

Thanks for your response!

Could you elaborate a bit more on the reasoning?

why do we have to find the no of 5s and how exactly did it come to 12?

Is there a faster way than counting all the 5s and the 2s?

So I think the following method is helpful and short too. Finding non-prime factors in factorials (based on a different thread I read).

50/(5^1)+50/(5^2)
=>10+2
Therefore, there are 12 5's in 50.
We know every alternate number is a multiple of 2.

Given the maximim number of 2's and 5's to create a ten can be 12 of each (10^N = 2^12 * 5^12)
The answer is C - 12.

Extra:
Say we calculate the number of 2's in 50 regardless
50/(2^1)+50/(2^2)+50/(2^3)+50/(2^4)
=>25+12+8+3+1
Therefor there are 49 2's in 50.

Hi Chetan,

Can you please explain from here onwards "So 2s will be more than 5s and that is why number of 10s will be equal to number of 5s..."

1) why are 2s more than 5s?
2) why do we need to calculate how many 5s are there?

Thanks

You need to identify how many 10s are there in 50!
For this first ask, what is 10 made of?
---> it's made of 5*2 ... ... ...(5*2=10)

So we understand that to create a 10 you will need one 5 and one 2.

Now ask, in 50! how many 5s are there and how many 2s are there?
(side note: once you identify the number of 5s, you don't actually need to find the number of 2s; because the number of 2s> the number of 5s in 50!)

So there are ten 5s in 50! because there are ten multiples of 5 till 50. And, there is one additional 5 in 25 (5*5=25) and one additional in 50 (5*5*2=50)

Thus there is a total of twelve 5s in 50!

So with twelve 5s, we can make twelve 10s in 50!
Thus ans. is C

This is basically a question of finding the number of trailing zeroes in a given factorial.

Using the following formula, The formula is from the GMAT CLUB MATH GUIDE.

(50/5) + (50/25) = 10 + 2 = 12 trailing zeroes.

Therefore, in 10^N, N = 12.

You're welcome.

avigutman sir I cannot understand how the answer is 12
I chose 5 as an answer as there are 5 factors of 10 = 2^n * 5^n inside 50! i.e. 10,20,30,40,50

Successively dividing 50 by 5

=>50/5=>10/5=>2

Hence largest N = 10+2 = 12

Answer C

NOTE :- 10 = 5*2, we could have checked for 2 as well but for all 10 we need 5 also so the largest power of 5 will dominate here

pdfff You're missing numbers that contribute just a '2' or just a '5'.
An analogy about pairs of shoes might help. Imagine that factors of 2 are right shoes, and factors of 5 are left shoes, and we want to know how many pairs of shoes we can get out of all those numbers. So, for example, 9 has no shoes, but 12 has two right shoes and 15 has one left shoe and 20 has two right shoes and one left shoe.
There are obviously going to be more right shoes in the end, so our limiting factor (pun intended) is the left shoes (right shoes without matching left shoes are useless to us).
So, if I can figure out how many left shoes there are, I'm done.