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If t is a positive integer and r is the remainder when

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If t is a positive integer and r is the remainder when [#permalink]

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If t is a positive integer and r is the remainder when t^2+5t+6 is divided by 7, what is the value of r?

(1) When t is dividd by 7, the remainder is 6.
(2) When t^2 is divided by 7, the remainder is 1.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-t-is-a-positive-integer-and-r-is-the-remainder-when-t-128031.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jul 2012, 05:25, edited 1 time in total.
Edited the question and added the OA.
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Re: DS: Remainder [#permalink]

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New post 15 Feb 2008, 15:03
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netcaesar wrote:
If t is a positive integer and r is the remainder when t^2+5t+6 is divided by 7, what is the value of r?

1) When t is dividd by 7, the remainder is 6.
2) When t^2 is divided by 7, the remainder is 1.


1) sufficient
let's denote a(7) - reminder of a when divided by 7

(t^2 + 5*t + 6) (7) = 36(7) + 5*6(30) + 6(7) -> enough to calculate r (actually it's equal to 2)

2) insufficient
t^2(7) = 1 -> t(7) = 1 or t(7) = -1 (or 6)

t(7) = 1 -> (t^2 + 5*t + 6) (7) = (1 + 5 + 6)(7) = 5
t(7) = 6 -> from previous example r = 2

A is the answer
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Re: DS: Remainder [#permalink]

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New post 15 Feb 2008, 19:11
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i just picked numbers and saw what happened ... i know its risky sometimes, but boy does it make life easy.
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Re: DS: Remainder [#permalink]

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New post 15 Feb 2008, 23:24
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maratikus wrote:
netcaesar wrote:
If t is a positive integer and r is the remainder when t^2+5t+6 is divided by 7, what is the value of r?

1) When t is dividd by 7, the remainder is 6.
2) When t^2 is divided by 7, the remainder is 1.


1) sufficient
let's denote a(7) - reminder of a when divided by 7

(t^2 + 5*t + 6) (7) = 36(7) + 5*6(30) + 6(7) -> enough to calculate r (actually it's equal to 2)

2) insufficient
t^2(7) = 1 -> t(7) = 1 or t(7) = -1 (or 6)

t(7) = 1 -> (t^2 + 5*t + 6) (7) = (1 + 5 + 6)(7) = 5
t(7) = 6 -> from previous example r = 2

A is the answer


how is r = 2 in above highlighted place?
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Re: DS: Remainder [#permalink]

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New post 16 Feb 2008, 01:36
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t=7x+6...
so t can hold values 13...20...27....so on

now the equation is (t+3)(t+2)/7....try for couple f values....the remainder is always 2....
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Re: DS: Remainder [#permalink]

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New post 16 Feb 2008, 02:21
GMAT TIGER wrote:
how is r = 2 in above highlighted place?


t=7k+6

\(t^2 + 5*t + 6=(7k+6)^2+5*(7k+6)+6=7*[7k^2+12k]+36+7*[5k]+30+6=7*[7k^2+17k]+72=7*[7k^2+17k+10]+2\)
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Re: DS: Remainder [#permalink]

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New post 28 Feb 2008, 09:21
1)t = 7 (y) + 6
+2 +2
(t+2) = 7y + 8
(t+3) = 7y + 9
(t+2)(t+3) = (7y + 8) (7y + 9)
7 (bunch of stuff) + 72 remainder = 2 suff.

2) (t^2+5t+6)/7
t^2/7 with remainder 1 (added to rest of previous numerator)
(5t+6 + 1)/7 = 5t/7 + 7/7 so 5t/7 r = 3,1, 6, etc. insuff.
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Re: DS: Remainder [#permalink]

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New post 28 Feb 2008, 12:08
You are right.

OA is A
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Re: DS: Remainder [#permalink]

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New post 29 Feb 2008, 00:58
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lets factor the stem as (t+3)(t+2) we are asked if (t+3)(t+2)=7N+r what is r?

1) says t=7N+6 i.e t can be 13

ok lets plug ig (13+3)(13+2) => (15)(16)

(15)(16)/7 gives remainder r=2

ok lets try t=27 gives remainder r=2

sufficient..
Re: DS: Remainder   [#permalink] 29 Feb 2008, 00:58
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