Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 65241

If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
01 Jun 2020, 08:22
Question Stats:
85% (01:38) correct 15% (02:01) wrong based on 40 sessions
HideShow timer Statistics
If N is the product of the first hundred multiples of five starting from 5, what is the rightmost nonzero digit in the number N? A. 0 B. 5 C. 7 D. 8 E. 9 Are You Up For the Challenge: 700 Level Questions
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Director
Joined: 25 Jul 2018
Posts: 731

If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
01 Jun 2020, 09:29
Bunuel wrote: If N is the product of the first hundred multiples of five starting from 5, what is the rightmost nonzero digit in the number N? A. 0 B. 5 C. 7 D. 8 E. 9 Are You Up For the Challenge: 700 Level Questions\((5*1) (5*2)(5*3)(5*4).....(5*100)\)= \(5^{100} (100!)\) \([\frac{100}{2}]+ [\frac{100}{2^{2}}]+ [\frac{100}{2^{3}}] + [\frac{100}{2^{4}}]+ [\frac{100}{2^{5}}]+ [\frac{100}{2^{6}}]= 50+ 25+ 12+ 6+ 3+ 1= 97\) > (\(2^{97}\)) \([\frac{100}{5}]+ [\frac{100}{5^{2}}]...= 20 + 4 = 24\) > ( \(5^{24}\)) = (All odd number up to 99) * \(5^{24}\)*\(5^{100}\)* \(2^{97}\) = =(All odd number up to 99)* \(5^{27}\)*\(10^{97}\) > Multiplying 5 to any odd number will end with units digit 5. > Rightmost nonzero digit will be 5 Answer (B)



PS Forum Moderator
Joined: 18 Jan 2020
Posts: 1175
Location: India
GPA: 4

Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
01 Jun 2020, 11:20
lacktutor wrote: \([\frac{100}{2}]+ [\frac{100}{2^{2}}]+ [\frac{100}{2^{3}}] + [\frac{100}{2^{4}}]+ [\frac{100}{2^{5}}]+ [\frac{100}{2^{6}}]= 50+ 25+ 12+ 6+ 3+ 1= 97\) > (\(2^{97}\))
\([\frac{100}{5}]+ [\frac{100}{5^{2}}]...= 20 + 4 = 24\) > ( \(5^{24}\))
Can you brief these two points? Why did you divide them with 2&5 respectively with the powers! Posted from my mobile device



Director
Joined: 25 Jul 2018
Posts: 731

If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
01 Jun 2020, 11:28
yashikaaggarwal wrote: lacktutor wrote: \([\frac{100}{2}]+ [\frac{100}{2^{2}}]+ [\frac{100}{2^{3}}] + [\frac{100}{2^{4}}]+ [\frac{100}{2^{5}}]+ [\frac{100}{2^{6}}]= 50+ 25+ 12+ 6+ 3+ 1= 97\) > (\(2^{97}\))
\([\frac{100}{5}]+ [\frac{100}{5^{2}}]...= 20 + 4 = 24\) > ( \(5^{24}\))
Can you brief these two points? Why did you divide them with 2&5 respectively with the powers! Posted from my mobile deviceMultiplying 5 to even numbers gives the distinct rightmost non—zero digits. —> 14*5 = 70 —> 18*5 = 90 —> 12*5 = 60 But multiplying 5 to any odd numbers ends with units digit 5. That’s why, we need to remove all 2s in the product. —> there are ninty seven 2s (\( 2^{97}\) ) Hope it helps



Intern
Joined: 01 Sep 2019
Posts: 29
Concentration: Marketing, Strategy

Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
01 Jun 2020, 13:42
lacktutor wrote: Bunuel wrote: If N is the product of the first hundred multiples of five starting from 5, what is the rightmost nonzero digit in the number N? A. 0 B. 5 C. 7 D. 8 E. 9 Are You Up For the Challenge: 700 Level Questions\((5*1) (5*2)(5*3)(5*4).....(5*100)\)= \(5^{100} (100!)\) \([\frac{100}{2}]+ [\frac{100}{2^{2}}]+ [\frac{100}{2^{3}}] + [\frac{100}{2^{4}}]+ [\frac{100}{2^{5}}]+ [\frac{100}{2^{6}}]= 50+ 25+ 12+ 6+ 3+ 1= 97\) > (\(2^{97}\)) \([\frac{100}{5}]+ [\frac{100}{5^{2}}]...= 20 + 4 = 24\) > ( \(5^{24}\)) = (All odd number up to 99) * \(5^{24}\)*\(5^{100}\)* \(2^{97}\) = =(All odd number up to 99)* \(5^{27}\)*\(10^{97}\) > Multiplying 5 to any odd number will end with units digit 5. > Rightmost nonzero digit will be 5 Answer (B) Hi VeritasKarishmaCould you please advise?



Manager
Joined: 05 Jan 2020
Posts: 130

Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
01 Jun 2020, 19:39
N = 5^100 * 100! First, we need to segregate the number of 0s at the end of N. Number of times 2 occurs in 100! = 97. => N = 5^100 * 2^97 * Y = 5^3 * 10*97 * Y. Since, we have factored all the 2s from 100!, thus, Y will always be odd in nature. 5 * odd = last digit as 5.
On a side note: Here, 5^100 will use up all 2s in 2^97. For our purpose of understanding, let's assume the power of 2 is 124. After exhausting a 2 for each 5, we'll still be left with 2^24 in N. Then, we'll have to find the power of 5 in 100! which equals to 24.
Now 24 powers of 2 will account for 24 powers of 5. Three takeaways: 1) If the powers of 2 and 5 are equal in N, then the last nonzero digit will be odd, but determining the last digit will be difficult in the given time frame. 2) If the power of 5 is greater than the power of 2 in N, then the last nonzero digit of N will result in a 5. 3) If the power of 2 in greater than the power of 5 in N, then the last nonzero digit of N will be even. Determining the digit will be difficult in the given time frame.



Manager
Joined: 24 Sep 2014
Posts: 73
Concentration: General Management, Technology

Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
01 Jun 2020, 21:39
If N is the product of the first hundred multiples of five starting from 5, what is the rightmost nonzero digit in the number N?
first multiple of five = 5x1 = 5 second multiple of five = 5x2 = 10 third multiple of five = 5x3 = 15 fourth multiple of five = 5x4 = 20 . . . 50th multiple of five = 5x50 = 250 . . . 100th multiple of five = 5x100 = 500
Now, N = (5x1) x (5x2) x (5x3) x (5x4) ......... (5x50).........x (5x100)
If we observe the above series, 50 multiples have the right most number 0 (all even multiples) and 50 multiples have the right most number 5 (all odd multiples). Now, if we multiply these 50 even multiples and 50 odd multiples, then we will have the below type of multiplication: N = (xyz....0)x(pqr...5), where x, y, z, p, q, r are all digits N = abcd....50, where the last digit is always 0 and the right most nonzero digit will be 5 IMO, answer is B



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10673
Location: Pune, India

Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
03 Jun 2020, 03:56
Bunuel wrote: If N is the product of the first hundred multiples of five starting from 5, what is the rightmost nonzero digit in the number N? A. 0 B. 5 C. 7 D. 8 E. 9 Are You Up For the Challenge: 700 Level Questions\(N = 5 * 10 * 15 * 20 * 25 * ... (100 multiples) = 5^{100} * (1*2*3*4*5...) = 5^{100} * 100!\) Every term has a 5 and every alternate term is even. Most probably, there will be more 5s than 2s. Still let's find out how many 2s there are in 100!. 100/2 = 50 50/2 = 25 25/2 = 12 12/2 = 6 6/2 = 3 3/2 = 1 Total number of 2s = 97 So yes, there are more 5s than 2s. As of now, we haven't even considered the 5s in 100!. So the product will end with 97 0s. So N will look like ...............0000000... (97 0s) We will be left with a whole lot of 5s and other odd numbers. 5*Odd will end in 5. So N will look some thing like .......................50000000000000.... Answer (B) Check: https://www.veritasprep.com/blog/2011/0 ... actorials/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Joined: 16 May 2011
Posts: 262
Concentration: Finance, Real Estate
GMAT Date: 12272011
WE: Law (Law)

Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
03 Jun 2020, 04:10
5×1 is 5 5*10 is 50 5*10*15. Is 750 5*10*15* 20 is 1,500 5*10*15*20*25 is 375,000 So the pattern is always 5 followed by 000.
Posted from my mobile device



IESE School Moderator
Joined: 11 Feb 2019
Posts: 308

If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
03 Jun 2020, 09:15
IMO B Anything multiplied by 5 gives units digit as 0 or 5 Qtn asks rightmost nonzero digit i.e. first digit after 0. so it will be 5. 5*10 =50 5*10 * 15 = 750 5*10*15*20 = 1500 Am is missing some logic for this question?
_________________



Senior Manager
Joined: 16 May 2011
Posts: 262
Concentration: Finance, Real Estate
GMAT Date: 12272011
WE: Law (Law)

Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
03 Jun 2020, 11:57
NitishJain wrote: IMO B
Anything multiplied by 5 gives units digit as 0 or 5
Qtn asks rightmost nonzero digit i.e. first digit after 0. so it will be 5.
5*10 =50 5*10 * 15 = 750 5*10*15*20 = 1500
Am is missing some logic for this question? yes. it didnt ask if it ends with 5 or 0. 5*10 or 5*10*15 etc.. will allways end with 0 or 00 or 000 etc. the question was (after you know that all right digits are 0) what will be the digit b before the 0s (to the left of the 0's). so it will aLLWAYS BE 5



IESE School Moderator
Joined: 11 Feb 2019
Posts: 308

Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
Show Tags
03 Jun 2020, 12:56
dimri10 wrote: NitishJain wrote: IMO B
Anything multiplied by 5 gives units digit as 0 or 5
Qtn asks rightmost nonzero digit i.e. first digit after 0. so it will be 5.
5*10 =50 5*10 * 15 = 750 5*10*15*20 = 1500
Am is missing some logic for this question? yes. it didnt ask if it ends with 5 or 0. 5*10 or 5*10*15 etc.. will allways end with 0 or 00 or 000 etc. the question was (after you know that all right digits are 0) what will be the digit b before the 0s (to the left of the 0's). so it will aLLWAYS BE 5 Hello dimri10That is exactly my point. I read the question and picked 5 as the answer choice. But after seeing the detailed explanations by various experts, I got confused that if I am missing something.
_________________




Re: If N is the product of the first hundred multiples of five starting fr
[#permalink]
03 Jun 2020, 12:56




