gmat prep 2

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gmat prep 2

by jainrahul1985 » Fri Sep 02, 2011 4:13 am
If t is a positive integer and r is the remainder when t2 + 5t + 6 is divided by 7, what is the value of r ?

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

(2) When t2 is divided by 7, the remainder is 1

OA A

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by Anurag@Gurome » Fri Sep 02, 2011 4:46 am
jainrahul1985 wrote:If t is a positive integer and r is the remainder when t2 + 5t + 6 is divided by 7, what is the value of r ?
(1) When t is divided by 7, the remainder is 6.
(2) When t2 is divided by 7, the remainder is 1
(t² + 5t + 6) = (t + 2)(t + 3)

Statement 1: When t is divided by 7, the remainder is 6. Hence, t is of the form (7k + 6), where k is an integer.

Hence, (t + 2)(t + 3) = (7k + 8)(7k + 9) = 49k² + 119k + 72 = 7*(7k² + 17k + 10) + 2
Hence, the remainder is 2.

Sufficient

Statement 2: When t² is divided by 7, the remainder is 1. But from this we cannot conclude the remainder when t is divided by 7. For example, t can be 6 or 9.
  • For t = 6, the required remainder = (t² + 5t + 6)|7 = 72|7 = 2
    For t = 9, the required remainder = (t² + 5t + 6)|7 = 132|7 = 6
Not sufficient

The correct answer is A.
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by GMATGuruNY » Fri Sep 02, 2011 6:38 am
M
jainrahul1985 wrote:If t is a positive integer and r is the remainder when t2 + 5t + 6 is divided by 7, what is the value of r ?

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

(2) When t2 is divided by 7, the remainder is 1

OA A
When positive integer x is divided by the divisor D, the remainder is R.

Given the information above:

-- Make a list of possible values for x.
-- The smallest possible value of x = the remainder R.
-- To determine the other possible values of x, just keep adding multiples of the divisor D.

The question at hand rephrased: What is the remainder when (t+2)(t+3) is divided by 7?

Statement 1: When t is divided by 7, the remainder is 6.
Smallest value = R = 6.
Now add multiples of the divisor 7:
t = 6, 13, 20, 27...

If t=6, then (t+2)(t+3) = 8*9 = 72.
72/7 = 10 R2.

If t=13, then (t+2)(t+3) = 15*16 = 240.
240/7 = 34 R2.

Just to be safe:
If t=27, then (t+2)(t+3) = 29*30 = 870.
870/7 = 124 R2.

Since in each case R=2, sufficient.

Statement 2: When t² is divided by 7, the remainder is 1.
Smallest value = R = 1.
Now add multiples of the divisor 7:
t² = 1, 8, 15, 22, 29, 36...

The list above includes two perfect squares: 1 and 36.
Thus, integer values for t = 1, 6...

If t=1, then (t+2)(t+3) = 3*4 = 12.
12/7 = 1 R5.

If t=6, then (t+2)(t+3) = 8*9 = 72.
72/7 = 10 R2.

Since R can be different values, insufficient.

The correct answer is A.
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by prateek_guy2004 » Fri Sep 02, 2011 1:43 pm
t2 + 5t + 6 is divided by 7, what is the value of r ?

For the value of t- Lets do FOIL method

(t+3) (t+2)

When t is divided by 7, the remainder is 6.

when t is divided by 2 we get the remainder of 6

sufficient

When t2 is divided by 7, the remainder is 1

Whether t is 2or 3 remainder is not 1.

Insufficient.

Hence A
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by samyukta » Fri Sep 02, 2011 4:58 pm
Anurag@Gurome wrote: Statement 2: When t² is divided by 7, the remainder is 1. But from this we cannot conclude the remainder when t is divided by 7. For example, t can be 6 or 9.
  • For t = 6, the required remainder = (t² + 5t + 6)|7 = 72|7 = 2
    For t = 9, the required remainder = (t² + 5t + 6)|7 = 132|7 = 6
Not sufficient

The correct answer is A.
@Anurag Ji,

Why u took t = 9?? t^2 = 81 and t^2/7 gives 4 as remainder but we need 1 as remainder... Is nt it??