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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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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
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06 Sep 2018, 15:55
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. Statement 2: In other words, 2z is 2 more than a multiple of 8: 2z = 8a + 2 z = 4a + 1 Substituting z = 4a+1 into z² + 2z + 4, we get: (4a+1)² + 2(4a+1) + 4 16a² + 8a + 1 + 8a + 2 + 4 16a² + 16a + 7 (multiple of 8) + (multiple of 8) + 7 (multiple of 8) + 7 Since the resulting expression is 7 more than a multiple of 8, dividing z² + 2z + 4 by 8 will yield a remainder of 7. SUFFICIENT. Statement 1: In other words, (z−3)² is 4 more than a multiple of 8: (z3)² = 8b + 4 (z3)² = 4(2b + 1) (z3)² = 2²odd Since (z3)² is a perfect square, the odd integer on the right side must be an odd perfect square: (z3)² = 2²odd² (z3)² = (2 * odd)² z3 = 2*odd An odd integer can be represented as 2c+1. Substituting odd = 2c+1 into z3 = 2*odd, we get: z3 = (2)(2c+1) z3 = 4c + 2 z = 4c + 5 z = 4c + 4 + 1 z = 4(c+1) + 1 z = (multiple of 4) + 1 z = 4a + 1 As shown under Statement 2, z = 4a+1 implies that dividing z² + 2z + 4 by 8 will yield a remainder of 7. SUFFICIENT.
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Re: If z is a positive integer and r is the remainder when z^2 + 2z + 4 is
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03 Sep 2017, 07:09
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) (z3)^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
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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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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 \((z3)^2\) has the remainder 4 when divided by 8, then \(z3\) 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
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01 Sep 2017, 03:12
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
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06 Sep 2018, 07:34
karankhurana04 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. Solution  (1) (z3)^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 I have taken another approach (plugging in numbers), which is a little bit more time consuming than yours.... Given: \(\frac{z^2+2z+4}{8}\) S(1) \(\frac{(z3)^2}{8}\) R=4 > Possible values for z: 5,9,13,17,21... If you plug these values of z in the original equation you get: z=5 \(\frac{5^2+2*5+4}{8}\) > \(\frac{25+10+4}{8}\) >\(\frac{39}{8}\) r=7 z=9 \(\frac{9^2+2*9+4}{8}\) > \(\frac{81+18+4}{8}\) >\(\frac{103}{8}\) r=7 Therefore sufficient S(2) \(\frac{2z}{8}\) R=2 > possible values for z: 1,5,9,13... If you plug these values of z in the original equation you get: z=1 \(\frac{1^2+2*1+4}{8}\) > \(\frac{1+2+4}{8}\) >\(\frac{7}{8}\) r=7 z=5 \(\frac{5^2+2*5+4}{8}\) > \(\frac{25+10+4}{8}\) >\(\frac{39}{8}\) r=7 z=9 \(\frac{9^2+2*9+4}{8}\) > \(\frac{81+18+4}{8}\) >\(\frac{103}{8}\) r=7 Therefore sufficient Hence, D
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Re: If z is a positive integer and r is the remainder when z^2 + 2z + 4 is
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06 Sep 2018, 13:04
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. Dear GMATGuruNYCan you share your thoughts in solving such a question without plugging numbers? Thanks



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Re: If z is a positive integer and r is the remainder when z^2 + 2z + 4 is
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06 Sep 2018, 19:39
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. I did the following way  Statemnt I : As the remainder is 4 when divided by 8, \((z3)^2\) has to be 4,12, 20, 28, 36, 44 etc. But as Z is a positive integer \((z3)^2 = 4,36\) etc. So, Z = 1,5,9,13.... etc. Now, \((z+1)^2 = (1=1)^2, (5+1)^2, (9+1)^2\).... All these number will leave a remainder of 4. Hence, Sufficient. Statmnt II: Now for remainder to be 2, \(2z = 2,10,18,26\) etc. So, z will be 1,5,9,13.. etc. \((z+1)^2\) will leave a remainder of 4 when divided by 8. Hence, Sufficient. So, D.
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