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Let N = 60! + 55! + 50! The unit digit of N and a number of digits to

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Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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New post 28 Nov 2017, 18:12
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Let N = 60! + 55! + 50! The unit digit of N and a number of digits to the left of the units digits are consecutive zeros before we come to the first non-zero digit. How many such consecutive zeros are there until the first non-zero digit?

A.16
B.15
C.14
D.12
E.10

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Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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New post 28 Nov 2017, 21:12
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Gnpth wrote:
Let N = 60! + 55! + 50! The unit digit of N and a number of digits to the left of the units digits are consecutive zeros before we come to the first non-zero digit. How many such consecutive zeros are there until the first non-zero digit?

A.16
B.15
C.14
D.12
E.10


Don't do the mistake of adding all separately..
Number of 0s will depend on number of 10s, which in turn would depend on 5s ...


So \(N=60!+55!+50!=50!(51*52..60 + 51*52*..5 + 1)\)..
Don't calculate the bracket as the two terms are multiple of 5 but then 1 is added to it so entire bracket will NOT be div by 5 or 10
So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)
D

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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New post 28 Nov 2017, 22:41
chetan2u wrote:
Gnpth wrote:
Let N = 60! + 55! + 50! The unit digit of N and a number of digits to the left of the units digits are consecutive zeros before we come to the first non-zero digit. How many such consecutive zeros are there until the first non-zero digit?

A.16
B.15
C.14
D.12
E.10


Don't do the mistake of adding all separately..
Number of 0s will depend on number of 10s, which in turn would depend on 5s ...


So \(N=60!+55!+50!=50!(51*52..60 + 51*52*..5 + 1)\)..
Don't calculate the bracket as the two terms are multiple of 5 but then 1 is added to it so entire bracket will NOT be div by 5 or 10
So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)
D

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Hi, I have seen people doing this in many questions bt im still low at understanding it ....Please explain in most easiest way possible thank u

So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)
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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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New post 30 Nov 2017, 00:19
3
rocko911 wrote:
chetan2u wrote:
Gnpth wrote:
Let N = 60! + 55! + 50! The unit digit of N and a number of digits to the left of the units digits are consecutive zeros before we come to the first non-zero digit. How many such consecutive zeros are there until the first non-zero digit?

A.16
B.15
C.14
D.12
E.10




Hi, I have seen people doing this in many questions bt im still low at understanding it ....Please explain in most easiest way possible thank u

So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)


as we know to find the number of zeros we find the number of 5s in the number
so
50! has 12 zero
55! has 13 zero
60! has 14 zero

now we know that suppose we add 100,1000 and 10000 :least of the them retains the zero
10000
1000
100
11100 ; similar pattern will be seen on increasing the zero so 12 is the answer
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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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New post 30 Nov 2017, 00:21
So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)

as we know to find the number of zeros we find the number of 5s in the number
so
50! has 12 zero
55! has 13 zero
60! has 14 zero

now we know that suppose we add 100,1000 and 10000 :least of the them retains the zero
10000
1000
100
11100 ; similar pattern will be seen on increasing the zero so 12 is the answer
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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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New post 30 Nov 2017, 03:05
rocko911 wrote:
chetan2u wrote:
Gnpth wrote:
Let N = 60! + 55! + 50! The unit digit of N and a number of digits to the left of the units digits are consecutive zeros before we come to the first non-zero digit. How many such consecutive zeros are there until the first non-zero digit?

A.16
B.15
C.14
D.12
E.10


Don't do the mistake of adding all separately..
Number of 0s will depend on number of 10s, which in turn would depend on 5s ...


So \(N=60!+55!+50!=50!(51*52..60 + 51*52*..5 + 1)\)..
Don't calculate the bracket as the two terms are multiple of 5 but then 1 is added to it so entire bracket will NOT be div by 5 or 10
So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)
D

Posted from my mobile device



Hi, I have seen people doing this in many questions bt im still low at understanding it ....Please explain in most easiest way possible thank u

So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)



hi...

LOGIC is taking OUT number of 5s that are in 50!
now \(50! =1*2*3*4*5*...49*50\)
MULTIPLES of 5 are 5,10,15,20,25,30,35,40,45,50
their product will have \(5*10*15*20*25*30*35*40*45*50\) ...... single 5 from each one of them so 50/5, this is nothing but multiples of 5 till 50
but their are multiples of \(5^2\), so an EXTRA 5, that is why \(\frac{50}{5^2}=2\) , these are 25 and 50
add all of them to get TOTAL..
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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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New post 30 Nov 2017, 03:10
3
StrugglingGmat2910 wrote:
So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)

as we know to find the number of zeros we find the number of 5s in the number
so
50! has 12 zero
55! has 13 zero
60! has 14 zero

now we know that suppose we add 100,1000 and 10000 :least of the them retains the zero
10000
1000
100
11100 ; similar pattern will be seen on increasing the zero so 12 is the answer


DON'T always look for the lowest number of 5s in a SUM and answer accordingly....
It can be a TRAP too..


example 23!+24!
Both 24! and 23! have 4 * 5s, so you would be tempted to hit that 4..
BUT \(24!+23! = 23!(1+24)=23!*25\)... so you will get 4 * 5s from 23! but ANOTHER 2 from 25
ans 4+2=6
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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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New post 10 Dec 2017, 12:55
chetan2u wrote:
StrugglingGmat2910 wrote:
So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)

as we know to find the number of zeros we find the number of 5s in the number
so
50! has 12 zero
55! has 13 zero
60! has 14 zero

now we know that suppose we add 100,1000 and 10000 :least of the them retains the zero
10000
1000
100
11100 ; similar pattern will be seen on increasing the zero so 12 is the answer


DON'T always look for the lowest number of 5s in a SUM and answer accordingly....
It can be a TRAP too..


example 23!+24!
Both 24! and 23! have 4 * 5s, so you would be tempted to hit that 4..
BUT \(24!+23! = 23!(1+24)=23!*25\)... so you will get 4 * 5s from 23! but ANOTHER 2 from 25
ans 4+2=6

Thank you sir for the concept while study the topic i have memorized the numbers where we get an extra zero


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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to   [#permalink] 24 Jan 2019, 08:02
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