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

Question

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

pushpitkc · Accepted Answer

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)

lesner19 · Answer

IMG_20171205_222937.jpg Answer is A 28 trailing zeros Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

pushpitkc · Answer

Hi  lesner19

The 123! when taken out will give 1+124 = 125, which when multiplied by 8 gives 1000(which is 3 zeroes)

Pratyaksh2791 · Answer

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.

HarryAxel · Answer

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

Turkish · Answer

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)

kartik22 · Answer

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

pushpitkc · Answer

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.

kartik22 · Answer

Thanks a lot !! It really helped .

Sent from my ONEPLUS A5000 using  GMAT Club Forum mobile app

ScottTargetTestPrep · Answer

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

SonalSinha803 · Answer

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

Sent from my Lenovo K53a48 using  GMAT Club Forum mobile app

Kinshook · Answer

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

x = 123!+124! = 123! (1 + 124) = 123!*125 = 5^3*123!

Highest power of 5 in x = 5^3 *123! = 3 + 24 + 4 = 31
Highest power of 2 in x = 5^3*123! = 61 + 30 + 15 + 7 + 3 + 1 > 31

Number of zeros in the end of integer x = 31

IMO D

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards