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If z is a positive integer and r is the remainder when z^2 + 2z + 4 is

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If z is a positive integer and r is the remainder when z^2 + 2z + 4 is  [#permalink]

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New post 31 Aug 2017, 23:14
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If z is a positive integer and r is the remainder when z^2 + 2z + 4 is divided by 8, what is the value of r?[/b]

(1) When (z−3)^2 is divided by 8, the remainder is 4.

(2) When 2z is divided by 8, the remainder is 2.

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Re: If z is a positive integer and r is the remainder when z^2 + 2z + 4 is  [#permalink]

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New post 03 Sep 2017, 07:09
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If z is a positive integer and r is the remainder when z^2 + 2z + 4 is divided by 8, what is the value of r?[/b]
(1) When (z−3)^2 is divided by 8, the remainder is 4.
(2) When 2z is divided by 8, the remainder is 2.
Solution -
(1) (z-3)^2 = 8k+4
z^2 - 6z + 9 = 8k + 4
z^2 -6z + 8z + 9 = 8k + 8z + 4 (Adding "8z" on both sides)
z^2 +2z + 4 = 8(k+z) - 1
Therefore, r = "-1" i.e 7. SUFFICIENT

(2) 2z = 8p + 2
z = 4p + 1
z^2 = (4p + 1)^2 = 16p^2 + 8p + 1
So, z^2 + 2z + 4 = (16p^2 + 8p + 1) + (8p + 2) + (4) = 16p^2 + 16p + 7 = 16p(p+1) + 7
Therefore, r = "7". SUFFICIENT

Ans - D
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If z is a positive integer and r is the remainder when z^2 + 2z + 4 is  [#permalink]

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New post 31 Aug 2017, 23:48
Bunuel wrote:
If z is a positive integer and r is the remainder when z^2 + 2z + 4 is divided by 8, what is the value of r?

(1) When (z−3)^2 is divided by 8, the remainder is 4.

(2) When 2z is divided by 8, the remainder is 2.


A positive number \(x\) has the remainder \(\{0, \pm 1, \pm 2, \pm 3, \pm 4\}\) when divided by 8, hence \(x^2\) has the remainder \(\{0, 1, 4, 9, 16\}\) or \(\{0, 1, 4, 1, 0\}\) or \(\{0, 1, 4\}\) when divided by 8.

We have \(z^2+2z+4=(z+1)^2+3\).

Since \((z+1)^2\) has the remainder \(\{0, 1, 4\}\) when divided by 8,
\((z+1)^2+3\) has the remainder \(\{3, 4, 7\}\) when divided by 8, or r could be \(\{3, 4, 7\}\).

(1) If \((z-3)^2\) has the remainder 4 when divided by 8, then \(z-3\) has the remainder \(\pm 2\) or \(z\) has the remainder 1 or 5.

If \(z\) has the remainder 1, we have \(r=4\).
If \(z\) has the remainder 5, we have \(r=1\).

Insufficient.

(2) If \(2z\) has the remainder 2, or 10, then \(z\) has the remainder 1 or 5.

If \(z\) has the remainder 1, we have \(r=4\).
If \(z\) has the remainder 5, we have \(r=1\).

Insufficient.

Combine (1) and (2).

We still have 2 cases: \(z\) has the remainder 1 or 5. Insufficient. Answer D.
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Re: If z is a positive integer and r is the remainder when z^2 + 2z + 4 is  [#permalink]

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New post 01 Sep 2017, 03:12
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Bunuel wrote:
If z is a positive integer and r is the remainder when z^2 + 2z + 4 is divided by 8, what is the value of r?[/b]

(1) When (z−3)^2 is divided by 8, the remainder is 4.

(2) When 2z is divided by 8, the remainder is 2.


(1)

This is not sufficient as we don't know what remainder we would get if z is divided by 8

(2)

From this we get -
2z = 8k + 2 (k is a constant)
z = 4k + 1

z^2 = (4k + 1)^2 = 16(k^2) + 8k + 1 -> as 16 and 8 are both divisible by 8 -> z^2 will give a remainder of 1 when divided by 8
2z as given gives a remainder of 2 when divided by 8
And constant 4 gives a remainder of 4 when divided by 8

Hence -
remainder((z^2 + 2z + 4)/8) = remainder(z^2/8) + remainder(2z/8) + remainder(4/8)
=1+2+4
=7

Hence (2) alone is sufficient but (1) is not

Hence (B) is correct
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If z is a positive integer and r is the remainder when z^2 + 2z + 4 is  [#permalink]

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New post 27 Apr 2018, 17:23
Giving this a shot...

\(z^2+2z+4=(z+2)^2=8q+r\)

Need to find r.

If z is odd, remainder will be 1. I tried some numbers and saw a pattern. Turns out it is always true that \(\frac{odd^2}{8}\) has a remainder of 1.
If z is even, remainder will be 0 or 4.

Statement 1
\((z-3)^2=8a+4=4(2a+1)\)
This implies that the expression \((z-3)^2\) is even, which further implies that z is odd. By analysis above, if z is odd, remainder will be 1.
Sufficient

Statement 2
\(2z=8b+2=2(4b+1)\)
\(z=4b+1\)
z is odd. By analysis above, if z is odd, remainder will be 1.
Sufficient

Answer: D

P.S. I agree with karankhurana04 analysis but dont understand how his solution gives r=7, while mine gives r=1. Did one of us make a mistake?
If z is a positive integer and r is the remainder when z^2 + 2z + 4 is &nbs [#permalink] 27 Apr 2018, 17:23
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