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For an added challenge, time yourself to 4 minutes tops to [#permalink] New post 04 Oct 2007, 17:56
For an added challenge, time yourself to 4 minutes tops to do these questions.

Does anyone have a fast way to do these? Brutal
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 [#permalink] New post 04 Oct 2007, 18:23
For 1st Question, I am getting D.

ST1 gives r = 2
ST2 gives r = 5

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 [#permalink] New post 04 Oct 2007, 18:28
For Question2, I am getting C.

You need to quickly draw combination of lines.
Concept is "One Line should go upward and other Line should go downward" then the product of their slope will be negative.

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 [#permalink] New post 04 Oct 2007, 19:57
1. D?
2. B? (B because x value of l and k we know from the question itself - the value is x=4 for both l and k - am not sure if this logic is correct).
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 [#permalink] New post 05 Oct 2007, 05:20
For the first question, it is helpful to factor, as it is for most questions involving number properties
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 [#permalink] New post 05 Oct 2007, 10:11
First question

ST. 1
7n+6= t
let n=2
t= 20
then t^2+5t+6= 506 which yields remainder of 2 when divided by 7

try with n=3, remainder will be 2 again
SUFF

ST. 2
7n+1= t^2
let n= 5
then t= 6
and t^2+5t+6= 72 which leaves remainder of 2 when divided by 7
SUFF

D
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 [#permalink] New post 05 Oct 2007, 13:25
Wrong.

I said D as well.

OA is A.





Like I said, Hard Quant Q's :-)
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 [#permalink] New post 05 Oct 2007, 14:16
First question:

I. Factor t^2+5t+6 into (t +3)(t + 2). If t/7 has remainder 6, then t = k*7+6, where k is an integer. So t+3 = k*7 + 9 = (k+1)*7 + 2.
t+2 = (k+1)*7 + 1. Multiplying them gives you:

(k*7)^2*49 + (k+1)*21 + 2. Whatever t is, the remainder when t^2+5t + 6 is divided by 7 will be 2. SUFF.

II. t^2/7 has a remainder of 1. Then t^2+6 is divisible by 7. So the value of r is the remainder when 5t is divided by 7. We don't have enough information to know. Since t^2/7 has a remainder of 1, t could be 6 (36-1 = 35), in which case r = 2. (30/7). Or t could be 8 (64- 1 = 63), in which case r = 5 (40/7).

So it's A.
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 [#permalink] New post 05 Oct 2007, 16:15
John, fantastic explanation. I understand how to do the question now. Thanks :-) I have to brush up on my remainder arithmetic.
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 [#permalink] New post 14 Oct 2007, 14:28
I think the explanation provided by r019h was correct for first statement.

But for the second statement, the remainder is not 2, when some other number is considered, for e.g. when n = 9, in statement 2, the remainder is not 2. So the answer is not D, but it is A.

Honestly, I think the approach taken by r019h was much more straightforward and can be quick.
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 [#permalink] New post 15 Oct 2007, 07:54
I dont think this is that hard...



asks what is r if (t+3)(t+2) is divided by 7!

1) says t=7N+6

so (t+3)=7N+9 => 7(N+1)+2
(t+2) => 7N+8 => 7(N+1)+1

so you will notice that r will always be 2... for any value of N...

sufficient

2) t^2=7N+1

t=6 then in that case (9)(8)/7 gives remainder 2...

t can also be 8... in that case (10)(11)/7 gives remainder 5..insuff

A it is..
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 [#permalink] New post 15 Oct 2007, 08:25
fresinha12 wrote:
I dont think this is that hard...



asks what is r if (t+3)(t+2) is divided by 7!

1) says t=7N+6

so (t+3)=7N+9 => 7(N+1)+2
(t+2) => 7N+8 => 7(N+1)+1

so you will notice that r will always be 2... for any value of N...

sufficient

2) t^2=7N+1

t=6 then in that case (9)(8)/7 gives remainder 2...

t can also be 8... in that case (10)(11)/7 gives remainder 5..insuff

A it is..



excellent approach..

agree with A.

can be done this way as well:

1: since t has 6 reminder if t is divided by 7, t = 7k +6.

= t^2 + 5t + 6
= (7k+6)(7k+6) + 5 (7k+6) + 6
= 49k^2 + 42k + 42k + 36 + 35k + 30 + 6
= 49k^2 + 119k + 72

in the above expression, 49k^2 and 119k are evenly divided by 7. so remains 72 which as 2 as reminder when it is divided by 7.

so suff...

2: t^2 has 1 reminder if t^2 is divided by 7.

t could be 6 or 8 or 15 each has 1 as reminder so the reminder of the expression t^2 + 5t + 6 is different with t values. nsf.

A.
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 [#permalink] New post 15 Oct 2007, 10:31
All these are great explanations guys!!! Greatly appreciated.

If i may add my thoughts...i thought the best way to solve it, since i am nowhere near as advanced as all of you are is to plug in nuymbers.

I tried different ones, but only 13 diveded by 7 gave the reminder 6. All other numbers will divide and give reminders but not 6. Since t is 13 if we do the brutal multiplications and divisions we could find the reminder of the main quadratic equation. Therefore SUFF.

The second one states that t*2 divided by 2, reminder is 1. Now off the top of my head i know that t can be either positive or negative, there may be differentmunbers giving the same remainder 1, thefore i ignore it and call INSUFF.
Thereofre i would go with A

Please let me know what you guys think about my reasoning.
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 [#permalink] New post 15 Oct 2007, 13:48
Fistail wrote:
fresinha12 wrote:
I dont think this is that hard...



asks what is r if (t+3)(t+2) is divided by 7!

1) says t=7N+6

so (t+3)=7N+9 => 7(N+1)+2
(t+2) => 7N+8 => 7(N+1)+1

so you will notice that r will always be 2... for any value of N...

sufficient

2) t^2=7N+1

t=6 then in that case (9)(8)/7 gives remainder 2...

t can also be 8... in that case (10)(11)/7 gives remainder 5..insuff

A it is..



excellent approach..

agree with A.

can be done this way as well:

1: since t has 6 reminder if t is divided by 7, t = 7k +6.

= t^2 + 5t + 6
= (7k+6)(7k+6) + 5 (7k+6) + 6
= 49k^2 + 42k + 42k + 36 + 35k + 30 + 6
= 49k^2 + 119k + 72

in the above expression, 49k^2 and 119k are evenly divided by 7. so remains 72 which as 2 as reminder when it is divided by 7.

so suff...

2: t^2 has 1 reminder if t^2 is divided by 7.

t could be 6 or 8 or 15 each has 1 as reminder so the reminder of the expression t^2 + 5t + 6 is different with t values. nsf.

A.


I think these approaches are too time consuming. I just finished my GMAT today and I saw a problem VERY VERY similar to this question. The general idea/concept was the same, but the quadratic and divisor was different. Here's my take on how to approach (1):

The quadratic factors to: (t + 2)(t +3)

hmm...what pattern have we consistently seen involving divisors and quadratics? I'm thinking consecutive integers.

(1) Provides info on t so add to this consecutive integer string: t, t + 1, t + 2, t + 3

we know t has a remainder 6 when divided by 7 so t + 1 MUST be divisible by 7 which means t + 2 will have a remainder of 1 when divided by 7 and t + 3 will have a remainder of 2 when divided by 7. How do I know this? Simple: remainders increase from 0 to x - 1 as you iterate over consecutive integers and reset back to 0 once you reach a number that divides evenly. 7%7 = 0, 8%7 = 1, 9%7 = 2 etc etc where % is the modulus function (gives the remainder).

So now pick two values for t + 2 and t + 3 which have remainders 1 and 2, respectively.
8 and 9: 8*9 = 72; 72%7 = 2
15 and 16: 15*16 = 240%7 = 2
etc etc

SUFF. This method is simple, easy to understand and just involves picking numbers - no long winded quadratic calculations necessary.

(2) I'm not going to discuss because I just wanted to mention how I solved for (1)
  [#permalink] 15 Oct 2007, 13:48
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