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thankont · 2007 January 19th, 08:00 PM

What is the highest power of 7 in 5000! (five thousand factorial)?
answer: 832

Swiss_boy · 2007 January 19th, 10:25 PM

5000/7 = 714
714/7 = 102
102/7 = 14
14 / 7 = 2
2 / 7 = 0

714+102+14+2+0 = 832

Don't ask me how. I just know that to identify power of a number in a factorial, u need to sum the successive quotients.

aashu_79 · 2007 January 20th, 01:50 AM

thats a great approach..

i think the logic goes like this.. consider highest power of 2 that devides 10!

10! = 1*2*3*4*5*6*7*8*9*10

we know that every 2nd(2^1) number is even no, hence divisible by 2
numbers of such factors of 2 in 10! = 10/2 = 5 (so 2^5 divides 10!)

But there are more numbers, every 4th (2^2) number has an extra factor of 2(apart from what we have already counted).

numbers of such additional factors of 2 in 10! = 10/4 = 2 (integral part of 10/4)

similarly every 8th (2^3) number has one additional factor of 2 that is not counted yet.

numbers of such additional factors of 2 in 10! = 10/8 = 1 (integral part of 10/8)

so, in general highest power of a number x that devides n! is given by

= n/x + n/x^2 + n/x^3.... +n/x^k (where k is such that x^k <= n)
each term here denotes the integral part of the division.

thankont · 2007 January 20th, 06:55 AM

great work from both of you v_spirada and of course aashu.
logic of v_spirada is that first we see how many divisors of 7 are there
in numbers 1...5000. There are 5000/7 = 714. Now that we are done with 7 look how many divisors
of 7^2 are there in 5000 so 5000/7^2 = 714/7= 102 etc.
so results are equivalent since 5000/7 + 714/7+... = 5000/7+5000/7^2+...

natalie82 · 2007 January 21st, 12:38 PM

thnx for posting this q. I would have surely missed this one if it were on my exam!!

MikeJung · 2007 January 22nd, 09:24 PM

thanks for posting.

asdada1 · 2007 January 23rd, 02:13 AM

Quote:

Originally Posted by Swiss_boy

5000/7 = 714
714/7 = 102
102/7 = 14
14 / 7 = 2
2 / 7 = 0

714+102+14+2+0 = 832

Don't ask me how. I just know that to identify power of a number in a factorial, u need to sum the successive quotients.

great stuff

Divin · 2007 January 23rd, 10:51 AM

great !

target_g · 2007 April 30th, 05:43 AM

Is it not simply sum of - 5000/7, 5000/49, 5000/7^3, ... 5000/7^5

pach2212 · 2007 April 30th, 02:42 PM

target_g.

You are correct. the method illustrated above is just another way of presenting the same.

2007 January 19th, 08:00 PM	#1 (permalink)
thankont TestMagic Guru-in-Training Join Date: Dec 2006 Location: Athens Posts: 656	Number theory What is the highest power of 7 in 5000! (five thousand factorial)? answer: 832

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2007 January 19th, 10:25 PM	#2 (permalink)
Swiss_boy Within my grasp! Join Date: Dec 2006 Location: Zurich Posts: 424	5000/7 = 714 714/7 = 102 102/7 = 14 14 / 7 = 2 2 / 7 = 0 714+102+14+2+0 = 832 Don't ask me how. I just know that to identify power of a number in a factorial, u need to sum the successive quotients.

2007 January 20th, 01:50 AM	#3 (permalink)
aashu_79 Within my grasp! Join Date: Dec 2006 Posts: 166	thats a great approach.. i think the logic goes like this.. consider highest power of 2 that devides 10! 10! = 12345678910 we know that every 2nd(2^1)* number is even no, hence divisible by 2 numbers of such factors of 2 in 10! = 10/2 = 5 (so 2^5 divides 10!) But there are more numbers, every 4th (2^2) number has an extra factor of 2(apart from what we have already counted). numbers of such additional factors of 2 in 10! = 10/4 = 2 (integral part of 10/4) similarly every 8th (2^3) number has one additional factor of 2 that is not counted yet. numbers of such additional factors of 2 in 10! = 10/8 = 1 (integral part of 10/8) so, in general highest power of a number x that devides n! is given by = n/x + n/x^2 + n/x^3.... +n/x^k (where k is such that x^k <= n) each term here denotes the integral part of the division.

2007 January 20th, 06:55 AM	#4 (permalink)
thankont TestMagic Guru-in-Training Join Date: Dec 2006 Location: Athens Posts: 656	great work from both of you v_spirada and of course aashu. logic of v_spirada is that first we see how many divisors of 7 are there in numbers 1...5000. There are 5000/7 = 714. Now that we are done with 7 look how many divisors of 7^2 are there in 5000 so 5000/7^2 = 714/7= 102 etc. so results are equivalent since 5000/7 + 714/7+... = 5000/7+5000/7^2+...

2007 January 21st, 12:38 PM	#5 (permalink)
natalie82 I JUST got here. Join Date: Jan 2007 Posts: 24	thnx for posting this q. I would have surely missed this one if it were on my exam!!

2007 January 22nd, 09:24 PM	#6 (permalink)
MikeJung TestMagic Guru Join Date: May 2006 Location: Los Angeles Posts: 1,339	thanks for posting.

2007 January 23rd, 10:51 AM	#8 (permalink)
Divin Within my grasp! Join Date: May 2006 Posts: 154	great !

2007 April 30th, 05:43 AM	#9 (permalink)
target_g I JUST got here. Join Date: Apr 2007 Posts: 21	Is it not simply sum of - 5000/7, 5000/49, 5000/7^3, ... 5000/7^5

2007 April 30th, 02:42 PM	#10 (permalink)
pach2212 Within my grasp! Join Date: Jan 2005 Posts: 313	target_g. You are correct. the method illustrated above is just another way of presenting the same.

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