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prime factors ps [#permalink] New post 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 ?
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 [#permalink] New post 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] New post 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.


:good
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 [#permalink] New post 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.
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 [#permalink] New post 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
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 [#permalink] New post 09 Jan 2006, 00:22
13...as above. took the numbers as r, r+1, r+2.
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 [#permalink] New post 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
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 [#permalink] New post 22 Aug 2006, 15:27
Can someone just clarify how 3^n(13) = 13 (3^n)????

My brain is hurting!
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 [#permalink] New post 22 Aug 2006, 19:57
13...
Used the same solution as above... 3^n x 13.
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 [#permalink] New post 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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 [#permalink] New post 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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