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amanvermagmat
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N is the product of all positive multiples of 5 from 1 to 11 Aug 2018, 23:29
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95% (hard)

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50% (02:31) correct 50% (02:22) wrong based on 236 sessions

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N is the product of all positive multiples of 5 from 1 to 100 inclusive. And x is a positive integer such that N is divisible by 10^x.

What is the highest possible value of x?

A. 16
B. 18
C. 20
D. 24
E. 26

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B
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N is the product of all positive multiples of 5 from 1 to 12 Aug 2018, 00:12
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amanvermagmat
N is the product of all positive multiples of 5 from 1 to 100 inclusive. And x is a positive integer such that N is divisible by 10^x.

What is the highest possible value of x?

A. 16
B. 18
C. 20
D. 24
E. 26

Posted from my mobile device
Show more

Given N=5*10*15*.....*90*95*100
Or, N=\(5^{20}*(1*2*3*....*19*20\)=\(5^{20}*20!\)
We have to find out power of 2 and power of 5 in 20!
\(\frac{20}{5}\)=4
\(\frac{20}{2}+\frac{20}{2^2}+\frac{20}{2^3}+\frac{20}{2^4}\)=10+5+2+1=18
So, N=\(5^{20}*5^4*2^18\)=\((5*2)^{18}*5^6\)=\(10^{18}*5^6\)

So the maximum value of x for which N is divisible by \(10^x\) is 18.
Ans. (B)
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N is the product of all positive multiples of 5 from 1 to Updated on: 11 Aug 2018, 23:43
Originally posted by Chethan92 on 11 Aug 2018, 23:42.
Last edited by Chethan92 on 11 Aug 2018, 23:43, edited 1 time in total.
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Given N is a product of all multiples of 5 from 1 to 100.
N = 5x10x15x20x....x100
Writing N in terms of 2s and 5s we have N = \(2^{20}\)*\(5^{24}\)
10 can be written as 2*5. So the maximum number of 10^x can be 20.
Hence C is the ans.
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N is the product of all positive multiples of 5 from 1 to Updated on: 12 Aug 2018, 00:33
Originally posted by push12345 on 11 Aug 2018, 23:43.
Last edited by push12345 on 12 Aug 2018, 00:33, edited 1 time in total.
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And has to be 18

5 has power of 24
20 multiples of 5
Plus 4 for 25,50 and 75

But 2 is limiting factor here
It has factor of 18

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N is the product of all positive multiples of 5 from 1 to 13 Aug 2018, 09:24
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Afc0892
Given N is a product of all multiples of 5 from 1 to 100.
N = 5x10x15x20x....x100
Writing N in terms of 2s and 5s we have N = \(2^{20}\)*\(5^{24}\)
10 can be written as 2*5. So the maximum number of 10^x can be 20.
Hence C is the ans.
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how did you calculate 2s and 5s in N?
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N is the product of all positive multiples of 5 from 1 to 13 Aug 2018, 21:23
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amanvermagmat
N is the product of all positive multiples of 5 from 1 to 100 inclusive. And x is a positive integer such that N is divisible by 10^x.

What is the highest possible value of x?

A. 16
B. 18
C. 20
D. 24
E. 26

Posted from my mobile device
Show more

Given N=5*10*15*.....*90*95*100
Or, N=\(5^{20}*(1*2*3*....*19*20\)=\(5^{20}*20!\)

Hi,

Given N=5*10*15*.....*90*95*100
This should be 5 (1*2*3*4.....*19*20) i.e. 5 * 20!
How did you arrive at 5^20 * 20! Kindly explain

Thanks,
Chirag
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N is the product of all positive multiples of 5 from 1 to 13 Aug 2018, 21:36
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chiragmaniar
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amanvermagmat
N is the product of all positive multiples of 5 from 1 to 100 inclusive. And x is a positive integer such that N is divisible by 10^x.

What is the highest possible value of x?

A. 16
B. 18
C. 20
D. 24
E. 26

Posted from my mobile device

Given N=5*10*15*.....*90*95*100
Or, N=\(5^{20}*(1*2*3*....*19*20)\)=\(5^{20}*20!\)
Hi,
Given N=5*10*15*.....*90*95*100
This should be 5 (1*2*3*4.....*19*20) i.e. 5 * 20!
How did you arrive at 5^20 * 20! Kindly explain

Thanks,
Chirag
Show more

Hi chiragmaniar,
5=5*1
10=5*2
15=5*3
.
.
.
100=5*20
x(multiply)
------------------
N=(5*1)*(5*2)*(5*3)*......................*(5*20)
Or, N={5*5*5*....(20 times)}* (1*2*3*.....*20)
Or, \(N=(5^{20})* (20!)\)

Hope it clarifies.
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N is the product of all positive multiples of 5 from 1 to Updated on: 30 Nov 2023, 06:31
Originally posted by KarishmaB on 13 Aug 2018, 22:55.
Last edited by KarishmaB on 30 Nov 2023, 06:31, edited 1 time in total.
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hasnain3047
Afc0892
Given N is a product of all multiples of 5 from 1 to 100.
N = 5x10x15x20x....x100
Writing N in terms of 2s and 5s we have N = \(2^{20}\)*\(5^{24}\)
10 can be written as 2*5. So the maximum number of 10^x can be 20.
Hence C is the ans.
how did you calculate 2s and 5s in N?
Show more

Check this post for a detailed discussion: https://anaprep.com/number-properties-h ... actorials/
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N is the product of all positive multiples of 5 from 1 to 01 Jul 2024, 23:28
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amanvermagmat
N is the product of all positive multiples of 5 from 1 to 100 inclusive. And x is a positive integer such that N is divisible by 10^x.

What is the highest possible value of x?

A. 16
B. 18
C. 20
D. 24
E. 26

Posted from my mobile device
Given N=5*10*15*.....*90*95*100
Or, N=\(5^{20}*(1*2*3*....*19*20\)=\(5^{20}*20!\)
We have to find out power of 2 and power of 5 in 20!
\(\frac{20}{5}\)=4
\(\frac{20}{2}+\frac{20}{2^2}+\frac{20}{2^3}+\frac{20}{2^4}\)=10+5+2+1=18
So, N=\(5^{20}*5^4*2^18\)=\((5*2)^{18}*5^6\)=\(10^{18}*5^6\)

So the maximum value of x for which N is divisible by \(10^x\) is 18.
Ans. (B)
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­Hi,

Great answer!, i was trying to understand how you were finding out the occurances of powers of 2 and 5 in 20!. I saw that you divide it repeatedly to find it but you were dividing 20 instead of 20!, I was wondering if it would be same. I know for a regular number (not factorial) we use it but is the method also applicable even if it is factorial of that number?
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N is the product of all positive multiples of 5 from 1 to 27 Sep 2025, 23:26
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