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If N is the product of the first hundred multiples of five starting fr

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If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 01 Jun 2020, 08:22
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If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 01 Jun 2020, 09:29
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Bunuel wrote:
If N is the product of the first hundred multiples of five starting from 5, what is the rightmost non-zero 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)
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Re: If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 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!

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If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 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!

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Multiplying 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
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Re: If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 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 non-zero 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 VeritasKarishma

Could you please advise?
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Re: If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 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 non-zero 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 non-zero 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 non-zero digit of N will be even. Determining the digit will be difficult in the given time frame.
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Re: If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 01 Jun 2020, 21:39
1
If N is the product of the first hundred multiples of five starting from 5, what is the rightmost non-zero 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 non-zero digit will be 5
IMO, answer is B
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Re: If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 03 Jun 2020, 03:56
1
Bunuel wrote:
If N is the product of the first hundred multiples of five starting from 5, what is the rightmost non-zero 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/
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Re: If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 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.

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If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 03 Jun 2020, 09:15
IMO B

Anything multiplied by 5 gives units digit as 0 or 5

Qtn asks rightmost non-zero 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?
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Re: If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 03 Jun 2020, 11:57
NitishJain wrote:
IMO B

Anything multiplied by 5 gives units digit as 0 or 5

Qtn asks rightmost non-zero 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
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Re: If N is the product of the first hundred multiples of five starting fr  [#permalink]

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New post 03 Jun 2020, 12:56
1
dimri10 wrote:
NitishJain wrote:
IMO B

Anything multiplied by 5 gives units digit as 0 or 5

Qtn asks rightmost non-zero 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 dimri10

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

If N is the product of the first hundred multiples of five starting fr

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