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If r,s,t are consecutive integers, what is the greatest

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If r,s,t are consecutive integers, what is the greatest  [#permalink]

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New post Updated on: 19 Jun 2008, 23:35
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If r,s,t are consecutive integers, what is the greatest prime factor of 3^r+3^s+3^t
5
7
11
13
17

Guys can u please tell me how do i solve this

many thanks

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Originally posted by vd on 19 Jun 2008, 04:53.
Last edited by vd on 19 Jun 2008, 23:35, edited 1 time in total.
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Re: PS question  [#permalink]

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New post 19 Jun 2008, 05:55
2
I believe the question is 3^r + 3^s + 3^t.
(the last term's exponent is t and not 7)

then since, r,s,t are consecutive number, we can put the above equation as:

--> 3^r * ( 1 + 3 ^1 + 3^2)
--> 3^r * 13

thus 13 is the greatest prime factor. The other factor is 3.

Ans is D
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Re: PS question  [#permalink]

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New post 19 Jun 2008, 23:37
gmatcrook wrote:
I believe the question is 3^r + 3^s + 3^t.
(the last term's exponent is t and not 7)

then since, r,s,t are consecutive number, we can put the above equation as:

--> 3^r * ( 1 + 3 ^1 + 3^2)
--> 3^r * 13

thus 13 is the greatest prime factor. The other factor is 3.

Ans is D


Good explanation gmatcrook

Thanks and +1 for you
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Re: PS question  [#permalink]

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New post 20 Jun 2008, 07:25
gmatcrook wrote:
I believe the question is 3^r + 3^s + 3^t.
(the last term's exponent is t and not 7)

then since, r,s,t are consecutive number, we can put the above equation as:

--> 3^r * ( 1 + 3 ^1 + 3^2)
--> 3^r * 13

thus 13 is the greatest prime factor. The other factor is 3.

Ans is D



I got D as well. Since it says consecutive numbers I replaced the letters r,s,t with 1,2,3 and solved. 3^1 + 3^2 + 3^3 = 39 13 is a factor of 39 hence D.

Gmatcrook, I don't understand your explanation, can you please elaborate. Thanks
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Re: PS question  [#permalink]

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New post 20 Jun 2008, 08:06
gmatcrook wrote:
I believe the question is 3^r + 3^s + 3^t.
(the last term's exponent is t and not 7)

then since, r,s,t are consecutive number, we can put the above equation as:

--> 3^r * ( 1 + 3 ^1 + 3^2)
--> 3^r * 13

thus 13 is the greatest prime factor. The other factor is 3.

Ans is D


The logic jump that he made is saying 3^r * ( 1 + 3 ^1 + 3^2). He is saying this b/c 3^r is the lowest number in a group on consecutive numbers. For example.

Say R = 5. => 3^5 + 3^6 + 3^7 = 3^5 * (1 + 3^1 + 3^2).

Any number for R works like that because the other two are 1 & 2 greater than R respectively. That was a brilliant answer gmatcrook.

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Re: If r,s,t are consecutive integers, what is the greatest  [#permalink]

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New post 07 Mar 2017, 11:33
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If r,s,t are consecutive integers, what is the greatest &nbs [#permalink] 07 Mar 2017, 11:33
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