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# Number theory

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Senior Manager
Joined: 01 May 2004
Posts: 335
Location: USA

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06 Jul 2004, 16:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Try to solve this one.

What is the highest power of 7 in 5000!
Senior Manager
Joined: 01 May 2004
Posts: 335
Location: USA

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06 Jul 2004, 16:45

(1) 4998 (2) 714 (3) 832 (4) 816
Senior Manager
Joined: 21 Mar 2004
Posts: 445
Location: Cary,NC

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06 Jul 2004, 16:58
the factorial sign is scaring me !

here an attempt to it.

3 and 4 are not div by 7 so they are out.

4998/7 = 714

between 714 and 4998........i would go for 4998 ......dont ask me why

- ash.
_________________

ash
________________________
I'm crossing the bridge.........

Director
Joined: 05 May 2004
Posts: 574
Location: San Jose, CA

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06 Jul 2004, 17:40
boksana wrote:
Try to solve this one.

What is the highest power of 7 in 5000!

Boksana
I am kinda confused with this Q

Do you mean .. find highest value of n such that

(7^n)*(integer)=5000!
Senior Manager
Joined: 26 Jan 2004
Posts: 400
Location: India

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07 Jul 2004, 05:49
I might be wrong as far as the answer is concerned,

8! = 1.2.3.4.5.6.7.8 = 7^1*(some integer)
Similarly for any factorial you will have atleast 1 7 from 0 to 9,
so the total number of 7's in say 15! two = 7^2*inetger,

Similarly in 5000! it will be minimum, 714 (4998/7 = 714)
Adding to this number the total of extra sevens from 0-9 number system, (in 20's we have two numbers divisible by 7, 21 and 28. So we have two sevens in this range)
I think the answer should be either 832 or 816,
Wild guess now, going for 816
Director
Joined: 05 Jul 2004
Posts: 894

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07 Jul 2004, 05:56
Ans: 816
If it is correct, then I will explain.
Manager
Joined: 16 May 2004
Posts: 64
Location: columbus

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07 Jul 2004, 18:44
so heres the explanation.
when you multiply numbers 1 to 5000
7 appears once in multiple of 7^1(7) eg 7 , 14, 21,28....
7 appears twice in multiples of 7^2(49) eg 49, 98, 147
7 appears thrice in multiples of 7^3(343) eg 343, 684
7 appears four times in multiples of 7^4(2401) eg 2401, 4802

so we have to add all these instances to get to the answer

for case 1 divide 5000 by 7 = 714 to get total number multiples of 7
for case 2 divide 5000 by 49 = 102 to get total number multiples of 49
for case 1 divide 5000 by 343 = 14 to get total number multiples of 343
for case 1 divide 5000 by 2401 = 2 to get total number multiples of 2401

hence the total number is 714 + 102 + 14 + 2 = 832
ans is 832   [#permalink] 07 Jul 2004, 18:44
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