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What are the number of zeroes in the end of integer x, if x=123!+124!?

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What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 04 Dec 2017, 06:22
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What are the number of zeroes in the end of integer x, if x = 123!+124! ?
(1) 28
(2) 29
(3) 30
(4) 31
(5) 56


self made - tricky

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What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 04 Dec 2017, 09:43
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chetan2u wrote:
What are the number of zeroes in the end of integer x, if x = 123!+124! ?
(1) 28
(2) 29
(3) 30
(4) 31
(5) 56

self made - tricky


In order to find the number of zeroes, we first need to simplify x = 123!+124!

x = \(123!+124! = 123!+124*123! = 123!(1+124) = 123!*125\)

123! has \(\frac{123}{5} + \frac{123}{5^2} = 24 + 4 = 28\) zeroes

123! has 8, which when multiplied by 125, will give 3 zeroes, making the total number of zeroes, 31(Option D)
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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 05 Dec 2017, 10:01
chetan2u wrote:
What are the number of zeroes in the end of integer x, if x = 123!+124! ?
(1) 28
(2) 29
(3) 30
(4) 31
(5) 56


self made - tricky

Attachment:
IMG_20171205_222937.jpg
IMG_20171205_222937.jpg [ 51.39 KiB | Viewed 1494 times ]


Answer is A
28 trailing zeros

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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 05 Dec 2017, 10:51
lesner19 wrote:
chetan2u wrote:
What are the number of zeroes in the end of integer x, if x = 123!+124! ?
(1) 28
(2) 29
(3) 30
(4) 31
(5) 56


self made - tricky

Attachment:
IMG_20171205_222937.jpg


Answer is A
28 trailing zeros

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Hi lesner19

The 123! when taken out will give 1+124 = 125, which when multiplied by 8 gives 1000(which is 3 zeroes)
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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 05 Dec 2017, 13:02
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123! + 124! should be first converted into the form of A!*B

123!(1+124) = 123! * 125

To find the number of zeroes is similar to finding the highest power of 10 in given factorial

10 has 2 and 5 as its prime factors. 5 will have the lesser power and that will be considered as the highest power of 10

123! --> 123/5= 24-->24/5 = 4
total 24+4 = 28

Now we have to find the number of 5s in 125
125 = 5*5*5 = 5^3

total = 28+3 = 31 zeroes.
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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 05 Dec 2017, 15:42
pushpitkc wrote:
chetan2u wrote:
What are the number of zeroes in the end of integer x, if x = 123!+124! ?
(1) 28
(2) 29
(3) 30
(4) 31
(5) 56

self made - tricky


In order to find the number of zeroes, we first need to simplify x = 123!+124!

x = \(123!+124! = 123!+124*123! = 123!(1+124) = 123!*125\)

123! has \(\frac{123}{5} + \frac{123}{5^2} = 24 + 4 = 28\) zeroes

123! has 8, which when multiplied by 125, will give 3 zeroes, making the total number of zeroes, 31(Option D)



Hi. Please can you explain why 123!+124*123!= 123!(1+124)? When you expand the last one you get 123!*124 right?
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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 05 Dec 2017, 18:47
HarryAxel wrote:
pushpitkc wrote:
chetan2u wrote:
What are the number of zeroes in the end of integer x, if x = 123!+124! ?
(1) 28
(2) 29
(3) 30
(4) 31
(5) 56

self made - tricky


In order to find the number of zeroes, we first need to simplify x = 123!+124!

x = \(123!+124! = 123!+124*123! = 123!(1+124) = 123!*125\)

123! has \(\frac{123}{5} + \frac{123}{5^2} = 24 + 4 = 28\) zeroes

123! has 8, which when multiplied by 125, will give 3 zeroes, making the total number of zeroes, 31(Option D)



Hi. Please can you explain why 123!+124*123!= 123!(1+124)? When you expand the last one you get 123!*124 right?


When you say 123! it means 1*2*3*4*.......122*123= 123! (123 factorial), same with 124! it has= 1*2*3*4*.......122*123*124= 124!
So basically the question is asking x = 123!+124! or x= 1*2*3*4*.......122*123 + 1*2*3*4*.......122*123*124

Now if you take 123! (123 factorial) common, you will have x = 123!+124! = 123! (1+124)
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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 06 Dec 2017, 03:38
Pratyaksh2791 wrote:
123! + 124! should be first converted into the form of A!*B

123!(1+124) = 123! * 125

To find the number of zeroes is similar to finding the highest power of 10 in given factorial

10 has 2 and 5 as its prime factors. 5 will have the lesser power and that will be considered as the highest power of 10

123! --> 123/5= 24-->24/5 = 4
total 24+4 = 28

Now we have to find the number of 5s in 125
125 = 5*5*5 = 5^3

total = 28+3 = 31 zeroes.

Can u explain the part as to why we are dividing by 5 ? I did not understand it exactly.

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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 07 Dec 2017, 14:07
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Well it's a formula

Trying to explain. Lets try trailing zeroes for 26!

According to our formula, the number of trailing zeroes is

26/5 + 26/5^2 = 5+1 = 6 zeroes

Rationale behind formula

26! Is nothing but 26*25*24..... 1

25 has 2 5's and 20,15,10,5 each has a 5.
Similarly we have 2, 4 with 2 2's, and 8 with 3 2's. In total we have 6 2's and 5's giving us 1000000. This number when multiplied by any number will have 6 zeroes in the end.

Similarly we need to extrapolate this formula and get the number of trailing zeroes for 123!

The extra 3 zeroes come because when you multiply 125 with 8, we get 1000(giving us the total of 31 zeroes)

Hope that clears your confusion.
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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 07 Dec 2017, 14:48
Thanks a lot !! It really helped .

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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 12 Jan 2018, 07:14
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chetan2u wrote:
What are the number of zeroes in the end of integer x, if x = 123!+124! ?
(1) 28
(2) 29
(3) 30
(4) 31
(5) 56


To determine the number of trailing zeros in a number, we need to determine the number of 5-and-2 pairs within the prime factorization of that number. Let’s start by simplifying 123!+124!.

123!+124! = 123!(1 + 124) = 123!(125)

Since we know there are fewer 5s in 123!(125) 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 123!, we can use the following shortcut in which we divide 123 by 5, then divide the quotient of 123/5 by 5 and continue this process until we no longer get a nonzero quotient.

123/5 = 24

24/5 = 4 (we can ignore the remainder)

Since 4/5 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 123!.

Thus, there are 24 +4 = 28 factors of 5 within 123!

Finally, we see that there are three factors of 5 within 125, since 125 = 5^3.

Since there are a total of 31 factors of 5 within 123!(125), there are 31 5-and-2 pairs and, thus, 31 trailing zeros.

Answer: D
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Re: What are the number of zeroes in the end of integer x, if x=123!+124!?  [#permalink]

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New post 26 May 2018, 07:06
chetan2u wrote:
What are the number of zeroes in the end of integer x, if x = 123!+124! ?
(1) 28
(2) 29
(3) 30
(4) 31
(5) 56


self made - tricky
Yes I got it right here is my solution. I am extremely sorry for the dirty handwriting though !!

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Re: What are the number of zeroes in the end of integer x, if x=123!+124!? &nbs [#permalink] 26 May 2018, 07:06
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