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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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04 Dec 2017, 05:22
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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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04 Dec 2017, 08:43
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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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05 Dec 2017, 09:01
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 [ 51.39 KiB  Viewed 801 times ]
Answer is A 28 trailing zeros Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app



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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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05 Dec 2017, 09:51
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 Sent from my ONEPLUS A3003 using GMAT Club Forum mobile appHi lesner19The 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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05 Dec 2017, 12:02
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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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05 Dec 2017, 14:42
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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05 Dec 2017, 17:47
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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06 Dec 2017, 02:38
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. Sent from my ONEPLUS A5000 using GMAT Club Forum mobile app



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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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07 Dec 2017, 13:07
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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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07 Dec 2017, 13:48
Thanks a lot !! It really helped . Sent from my ONEPLUS A5000 using GMAT Club Forum mobile app



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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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12 Jan 2018, 06:14
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 5and2 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 5and2 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 5and2 pairs and, thus, 31 trailing zeros. Answer: D
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