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prime factors ps

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prime factors ps [#permalink]  08 Jan 2006, 21:50
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If r, s, t are consecutive integers, what is the greatest prime factor of 3^r +3^s + 3^t ?
VP
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[#permalink]  08 Jan 2006, 22:42
Is it 13?

Let r = k,
s = k+1
t = k +2

N = 3^k + 3^(k+1) + 3^(k+2)
= 3^k (1 + 3 + 9)
= 3^k * 13

The only prime factors of N are 3 and 13.
So 13 is the highest prime.
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[#permalink]  08 Jan 2006, 22:51
giddi77 wrote:
Is it 13?

Let r = k,
s = k+1
t = k +2

N = 3^k + 3^(k+1) + 3^(k+2)
= 3^k (1 + 3 + 9)
= 3^k * 13

The only prime factors of N are 3 and 13.
So 13 is the highest prime.

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Director
Joined: 17 Oct 2005
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[#permalink]  08 Jan 2006, 23:05
oa is 13, can you explain this part? thanks in advance.

giddi77 wrote:
Is it 13?

Let r = k,
s = k+1
t = k +2

N = 3^k + 3^(k+1) + 3^(k+2)
= 3^k (1 + 3 + 9)

= 3^k * 13

The only prime factors of N are 3 and 13.
So 13 is the highest prime.
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
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[#permalink]  08 Jan 2006, 23:18
joemama142000 wrote:
oa is 13, can you explain this part? thanks in advance.

giddi77 wrote:
Is it 13?

Let r = k,
s = k+1
t = k +2

N = 3^k + 3^(k+1) + 3^(k+2)
= 3^k (1 + 3 + 9)

= 3^k * 13

The only prime factors of N are 3 and 13.
So 13 is the highest prime.

3^(k+1) = 3*3^k = 3*3^k
3^(k+2) = 3^2*3^k = 9*3^k
HTH
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

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[#permalink]  09 Jan 2006, 00:22
13...as above. took the numbers as r, r+1, r+2.
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[#permalink]  09 Jan 2006, 00:28
If r = n, s = n+1, t = n+2

3^r +3^s + 3^t = 3^n + 3^(n+1) + 3^(n+2) = 3^n (1 + 3 + 9) = 13(3^n)

Largest prime factor = 13
Senior Manager
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[#permalink]  22 Aug 2006, 15:27
Can someone just clarify how 3^n(13) = 13 (3^n)????

My brain is hurting!
VP
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[#permalink]  22 Aug 2006, 19:57
13...
Used the same solution as above... 3^n x 13.
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Joined: 20 Nov 2005
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[#permalink]  22 Aug 2006, 22:50
MBAlad wrote:
Can someone just clarify how 3^n(13) = 13 (3^n)????

My brain is hurting!

Because r,s,t are consecutive

Let r = x
s = x+1
t = x+2

Then 3^r + 3^s + 3^t becomes
3^k + 3^(k+1) + 3^(k+2)
= 3^k + 3^k * 3^1 + 3^k * 3^2
= (3^k) * (1+3+9)
= (3^k) * (13)
Hence largest prime factor = 13
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Senior Manager
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[#permalink]  23 Aug 2006, 00:11
ps_dahiya,

I just didn't understand the step "3^n(13) = 13 (3^n)" in ywilfred's post.

Does this mean a^b(c) = c(a^n)

I think I may be confusing the terms....does 3^n(13) mean 3 to the power of 13n OR 3 to the power of n, multiplied by 13?

Enlighten me!
[#permalink] 23 Aug 2006, 00:11
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