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48! + 49!)/10^n is an integer. What is the maximum value of

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48! + 49!)/10^n is an integer. What is the maximum value of [#permalink] New post 14 Dec 2005, 06:18
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(48! + 49!)/10^n is an integer.
What is the maximum value of n?

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 [#permalink] New post 14 Dec 2005, 07:10
(48!+49!)/10^n => 48!(50)/10^n=>largest power of 10^n that divides 48! is 10 and largest power of 50 is 1...hence 11 !
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 [#permalink] New post 14 Dec 2005, 07:13
christoph wrote:
(48!+49!)/10^n => 48!(50)/10^n=>largest power of 10^n that divides 48! is 10 and largest power of 50 is 1...hence 11 !


Isn't the largest power of 50 is 2?
50 is 5 x 10, and there are plenty of 2s in 48!.
So I guess we can make two 10s from 50.

What do you think?

(However, I do not have an OA for this question)
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 [#permalink] New post 14 Dec 2005, 07:16
uhmm, n max = 12 right?
48!+49!= 48! * 50
5*even = 10
15*even = 10
25*even = 10^2
35*even = 10
45*even = 10
and 10, 20, 30, 40
50*even=10^2
Therefore, n max = 12
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 [#permalink] New post 14 Dec 2005, 08:04
gamjatang wrote:
christoph wrote:
(48!+49!)/10^n => 48!(50)/10^n=>largest power of 10^n that divides 48! is 10 and largest power of 50 is 1...hence 11 !


Isn't the largest power of 50 is 2?
50 is 5 x 10, and there are plenty of 2s in 48!.
So I guess we can make two 10s from 50.

What do you think?

(However, I do not have an OA for this question)


yes ur right...i lost my concentration somewhere... :(
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 [#permalink] New post 14 Dec 2005, 08:53
OK let me take a knock at this.....I am a bit rusty...

(48!+49!)/10^n

well lets first break 10^n into its prime factors 2^a * 5^b....well obviously the power of 5 will be the limiting case here, and so N cannot be greater than b.....

OK now lets re-write 48! +49! into something more easy...48!*50

oK so lets see how many 5s are there in 50,
50/5=10, 50/25=2, so there are 12 5s from 50...

now 48!, has a 5 from 5, 10, 15, 20, 30, 35, 40, 45 plus 2 5s from 25... so 48! gives 10 5s ....so in all as we can see....the power of N=12+10....22!
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 [#permalink] New post 14 Dec 2005, 09:33
fresinha12 wrote:
OK let me take a knock at this.....I am a bit rusty...

(48!+49!)/10^n

well lets first break 10^n into its prime factors 2^a * 5^b....well obviously the power of 5 will be the limiting case here, and so N cannot be greater than b.....

OK now lets re-write 48! +49! into something more easy...48!*50

oK so lets see how many 5s are there in 50,
50/5=10, 50/25=2, so there are 12 5s from 50...


now 48!, has a 5 from 5, 10, 15, 20, 30, 35, 40, 45 plus 2 5s from 25... so 48! gives 10 5s ....so in all as we can see....the power of N=12+10....22!


fresinha12, i guess you thought its 50! not 50.
50 has 2 5's
total: 10 + 2 = 12
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 [#permalink] New post 14 Dec 2005, 09:40
you are correct...man...I am so out of it...

50=5*10...so 10=5*2...therefore you get 2 5s...

:oops:

duttsit wrote:
fresinha12 wrote:
OK let me take a knock at this.....I am a bit rusty...

(48!+49!)/10^n

well lets first break 10^n into its prime factors 2^a * 5^b....well obviously the power of 5 will be the limiting case here, and so N cannot be greater than b.....

OK now lets re-write 48! +49! into something more easy...48!*50

oK so lets see how many 5s are there in 50,
50/5=10, 50/25=2, so there are 12 5s from 50...


now 48!, has a 5 from 5, 10, 15, 20, 30, 35, 40, 45 plus 2 5s from 25... so 48! gives 10 5s ....so in all as we can see....the power of N=12+10....22!


fresinha12, i guess you thought its 50! not 50.
50 has 2 5's
total: 10 + 2 = 12
  [#permalink] 14 Dec 2005, 09:40
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