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Hi all, Again a value DS question. 1). Statement 1 is insufficient. To find the number of zeros all we need is find number of five’s. For N values 120,121,122,123,124… the number of zeros(Number of five's) is 28 So insufficient 2). Statement 2 is insufficient. N values again can be 123,124,125… Together also insufficient, Because N can be 123 or 124. So answer is E
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Hi all, Again a value DS question. 1). Statement 1 is insufficient. To find the number of zeros all we need is find number of five’s. For N values 120,121,122,123,124… the number of zeros(Number of five's) is 28 So insufficient 2). Statement 2 is insufficient. N values again can be 123,124,125… Together also insufficient, Because N can be 123 or 124. So answer is E

hi N cannot be 125.... because it is given N! has 28 zeroes in end..
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(2) (N+2)! ends with 31 zeroes and (N-1)! ends with 28 zeroes

Kudos for a correct solution.

ans E..

firstly the value of zeroes depends on power of 5 as power of 2 will always be greater than power of 5... as we see (N+2)! increases the zeroes by 3 digits...it means between N and N+2, there is 5^3 multiple..125,250... here 125 fits in ..

1) statement one gives us values 120 to 124....as 119 will give 27 zeroes and 125 will give 31 zeroes.. insufficient.. 2) statement two gives us values 123 to 127 as N+2 < 130... insufficient..

combined two values still remain... 123 and 124.. insufficient
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In any N!, the number of trailing zeroes = number of factors of 2 and 5. For example, for 6! (which is 720), the ending 0 is created by taking the numbers 2 and 5 in (6)(5)(4)(3)(2)(1) and multiplying them together. It's also important to realize that since every second number is divisible by 2 but every fifth number is a multiple of 5, the number of occurrences of 5 is less than 2, hence the limiting factor is 5.

Sine we know we're working with giant numbers, we can start getting an idea of what we're looking at by seeing how many zeroes 100! ends with

Every multiple of 5 between that 1 and 100 provides one factor of 5, and at 25, 50, 75, and 100, you get an extra 5 (for example, 75 = 3 * 5 * 5, so 75 provides two factors of 5). So the number of factors in 100! is 24 as a starting point for us.

Now since N! ends with 28 zeroes, we need to four more 5’s in 100!, so N should be 120. But one thing is to be noticed here is 120!, 121!, 122!, 123! and 124! also have 28 zeroes as from 120! to 124! there is no more extra 5 factor.

Hence Statement 1 is not sufficient.

As per Statement 2 the (N+2)! has 31 zeroes.

So let’s check which numbers can have 30 zeroes. We know from the above that 120!-124! should have 28 zeroes.

But when it comes to 125!, we have 5^3 factor so 125!-129! have 31 zeroes.

So N+2 can be any number between 125 to 129, but the statement also says N-1 has 28 zeroes that means N-1 can be any number from 120! to 124!.

So N can be 125 or 124 or 123 so again Statement 2 is not sufficient.

Let’s combine Statements 1 and 2: in this case, N could be 124 or it could be 123, so E is the correct answer.
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In any N!, the number of trailing zeroes = number of factors of 2 and 5. For example, for 6! (which is 720), the ending 0 is created by taking the numbers 2 and 5 in (6)(5)(4)(3)(2)(1) and multiplying them together. It's also important to realize that since every second number is divisible by 2 but every fifth number is a multiple of 5, the number of occurrences of 5 is less than 2, hence the limiting factor is 5.

Sine we know we're working with giant numbers, we can start getting an idea of what we're looking at by seeing how many zeroes 100! ends with

Every multiple of 5 between that 1 and 100 provides one factor of 5, and at 25, 50, 75, and 100, you get an extra 5 (for example, 75 = 3 * 5 * 5, so 75 provides two factors of 5). So the number of factors in 100! is 24 as a starting point for us.

Now since N! ends with 28 zeroes, we need to four more 5’s in 100!, so N should be 120. But one thing is to be noticed here is 120!, 121!, 122!, 123! and 124! also have 28 zeroes as from 120! to 124! there is no more extra 5 factor.

Hence Statement 1 is not sufficient.

As per Statement 2 the (N+2)! has 31 zeroes.

So let’s check which numbers can have 30 zeroes. We know from the above that 120!-124! should have 28 zeroes.

But when it comes to 125!, we have 5^3 factor so 125!-129! have 31 zeroes.

So N+2 can be any number between 125 to 129, but the statement also says N-1 has 28 zeroes that means N-1 can be any number from 120! to 124!.

So N can be 125 or 124 or 123 so again Statement 2 is not sufficient.

Let’s combine Statements 1 and 2: in this case, N could be 124 or it could be 123, so E is the correct answer.

I dont understand why has everyone picked 124, 125 as the starting point for N

Not clear - Can somebody pls explain

hi

firstly the value of zeroes depends on power of 5 as power of 2 will always be greater than power of 5... as we see (N+2)! increases the zeroes by 3 digits...it means between N and N+2, there is 5^3 multiple..125,250... here 125 fits in .. because no of zeroes=125/5+125/5^2+125/5^3=25+5+1=31...
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In any N!, the number of trailing zeroes = number of factors of 2 and 5. For example, for 6! (which is 720), the ending 0 is created by taking the numbers 2 and 5 in (6)(5)(4)(3)(2)(1) and multiplying them together. It's also important to realize that since every second number is divisible by 2 but every fifth number is a multiple of 5, the number of occurrences of 5 is less than 2, hence the limiting factor is 5.

Sine we know we're working with giant numbers, we can start getting an idea of what we're looking at by seeing how many zeroes 100! ends with

Every multiple of 5 between that 1 and 100 provides one factor of 5, and at 25, 50, 75, and 100, you get an extra 5 (for example, 75 = 3 * 5 * 5, so 75 provides two factors of 5). So the number of factors in 100! is 24 as a starting point for us.

Now since N! ends with 28 zeroes, we need to four more 5’s in 100!, so N should be 120. But one thing is to be noticed here is 120!, 121!, 122!, 123! and 124! also have 28 zeroes as from 120! to 124! there is no more extra 5 factor.

Hence Statement 1 is not sufficient.

As per Statement 2 the (N+2)! has 31 zeroes.

So let’s check which numbers can have 30 zeroes. We know from the above that 120!-124! should have 28 zeroes.

But when it comes to 125!, we have 5^3 factor so 125!-129! have 31 zeroes.

So N+2 can be any number between 125 to 129, but the statement also says N-1 has 28 zeroes that means N-1 can be any number from 120! to 124!.

So N can be 125 or 124 or 123 so again Statement 2 is not sufficient.

Let’s combine Statements 1 and 2: in this case, N could be 124 or it could be 123, so E is the correct answer.

I got a little doubt

Zeros in 120! to 124! is 29

Correct me i assumed it like: dividing 120! with 5,25&75 i will get 24+4+1 number of 5's i.e 29

while taking 100!= 24 and then counting 5's in 105,110,115,120 will take to 28 . whats missing here?

Correct me i assumed it like: dividing 120! with 5,25&75 i will get 24+4+1 number of 5's i.e 29

while taking 100!= 24 and then counting 5's in 105,110,115,120 will take to 28 . whats missing here?

hi, to find power of 5 or zeroes in any factorial, we have to divide the number with increasing power of 5.... 5^1,5^2,5^3 and so on.... you have correctly divided by 5,25 but 75 is wrong as next number would be 5^3=125.. and you cannot divide 120 by 125... so answer is 24+4=28 and not 29... hope it is clear...
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Correct me i assumed it like: dividing 120! with 5,25&75 i will get 24+4+1 number of 5's i.e 29

while taking 100!= 24 and then counting 5's in 105,110,115,120 will take to 28 . whats missing here?

hi, to find power of 5 or zeroes in any factorial, we have to divide the number with increasing power of 5.... 5^1,5^2,5^3 and so on.... you have correctly divided by 5,25 but 75 is wrong as next number would be 5^3=125.. and you cannot divide 120 by 125... so answer is 24+4=28 and not 29...

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