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

Question

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.

karankhurana04 · Accepted Answer

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. 
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

GMATGuruNY · Answer

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:
(z-3)² = 8b + 4
(z-3)² = 4(2b + 1)
(z-3)² = 2²odd

Since (z-3)² is a perfect square, the odd integer on the right side must be an odd perfect square:
(z-3)² = 2²odd²
(z-3)² = (2 * odd)²
z-3 = 2*odd

An odd integer can be represented as 2c+1.
Substituting odd = 2c+1 into z-3 = 2*odd, we get:
z-3 = (2)(2c+1)
z-3 = 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.

D

broall · Answer

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.

sindhya1781 · Answer

(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

T1101 · Answer

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{(z-3)^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

Mo2men · Answer

Dear  GMATGuruNY

Can you share your thoughts in solving such a question without plugging numbers?

Thanks

rahul16singh28 · Answer

I did the following way -

Statemnt I :

As the remainder is 4 when divided by 8,  (z-3)^2  has to be 4,12, 20, 28, 36, 44 etc.
But as Z is a positive integer  (z-3)^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.

MsInvBanker · Answer

Bunuel  ,
Can we have your explanation for this question? I have been trying to understand this question for two days now.

dabaobao · Answer

z^2 + 2z + 4 = 8a + r
r = ?

Stmt 1: (z-3)^2 = z^2 - 6z + 9 gives r=4 => z^2 - 6z + 9 = 8a + 4
Adding 8z on both sides
z^2 - 6z + 9 + 8z = 8a + 4 + 8z
z^2 + 2z + 9 = 8(a+z) + 4
z^2 + 2z + 9 = 8(a) + 4

If z^2 + 2z + 9, gives r=4
Then z^2 + 2z + 4, gives r=4-5 = -1 => r=7 => Sufficient

Stmt 2: 2z = 8a + 2
z = 4a + 1
(4a + 1)^2 + 2(4a+1)+4 = 16a^2 + 8a + 1 + 8a + 2 + 4 => r=7 => Sufficient

Kritisood · Answer

Hi, Bunuel has discussed this question here  https://gmatclub.com/forum/divisibility-and-remainders-on-the-gmat-207418.html

sreddy07 · Answer

Bunuel   VeritasKarishma

I have a quick question here.

St 1) (z-3)^2 when divided by 8, R=4 Therefore it follows that when z-3 is divided by 8, R=2 and thus, when z/8 R=5.

Using this info, R when (z^2+2z+4) /8 R=7

St 2) 2z/8 R=2 therefore, R=1 when z/8
Using this info too R when (z^2+2z+4) /8 R=7.

What I'm caught up in is that st1 and st2 should actually give the same value for z which means value of R when z/8 should be same in both statements as GMAT does not give contradictory/different information in ST1 and 2 for DS questions. So is something wrong with my approach here or am I missing something obvious?

yashikaaggarwal · Answer

I have a slight different approach, if one can crack it.

Given: If z is a positive integer and r is the remainder when z^2 + 2z + 4 is divided by 8, 
To find: what is the value of r?

(1) When (z−3)^2 is divided by 8, the remainder is 4.
The no. Leaving 4 as remainder when divided by 8 = 4,12,20,28,36.......

So, Let (Z-3)^2 = 4
(Z-3)^2 = 2^2 
Z-3 = 2
Z = 5

Putting value of Z = 5 in equation z^2 + 2z + 4 = 5^2+2*5+4 = 39,which when divided by 8 leaves remainder 7

2) Let (Z-3)^2 = 12
(Z-3)^2 = (2√2)^2 
Z-3 = 2√2
Z = 2√2+3 (not a perfect integer) (Crossed out)

3) Let (Z-3)^2 = 36
(Z-3)^2 = 6^2 
Z-3 = 6
Z = 9

Putting value of Z = 9 in equation z^2 + 2z + 4 = 9^2+2*9+4 = 81+18+4 = 103,which when divided by 8 leaves remainder 7

So for every value of z which is integer z^2 + 2z + 4 when divided by 8 will leave remainder 7
(Sufficient)

(2) When 2z is divided by 8, the remainder is 2.
No. Leaving remainder 2 when divided by 8 = 2,10,18,26,34,........

1) Put 2Z = 2
Z = 1

Putting value of Z = 1 in equation z^2 + 2z + 4 = 1^2+2*1+4 = 1+2+4 = 7 ,which when divided by 8 leaves remainder 7

2) Put 2Z = 10
Z = 5

Putting value of Z = 5 in equation z^2 + 2z + 4 = 5^2+2*5+4 = 39 ,which when divided by 8 leaves remainder 7

3) Put 2Z = 18
Z = 9

Putting value of Z = 9 in equation z^2 + 2z + 4 = 9^2+2*9+4 = 103 ,which when divided by 8 leaves remainder 7

So for every value of 2z which leave remainder 2 when divided by 8, leaves remainder 7 when z^2 + 2z + 4 is divided by 8.
(Sufficient)

Answer is D

Posted from my mobile device

Archit3110 · Answer

#1
Valid for z =1,5 remainder value the. For given expression would be 7
Sufficient
#2
2z divided by 8 remainder 2 possible z 1,5,13..
Remainder is 7 
Sufficient
Option D

Posted from my mobile device

KarishmaB · Answer

How do you figure this?

3 divided by 8, remainder 3
9 divided by 8, remainder 1

If we know that z = 8a + 2 (remainder 2)
Then we know that z^2 = 64a^2 + 32a + 4 (remainder 4)

But if we know that z^2 = 8a + 4, how can we say that z must be 8b + 2?
Say a = 4, z^2 = 36, z = 6 (remainder 6, not 2)
Say a = 12, z^2 = 100, z = 10 (remainder 2)

ManifestDreamMBA · Answer

Tried a different approach: Accounting for R of individual terms

Statement 1:
(z−3)^2 = R4
(z−3) = R2
z = R5

z^2 + 2z + 4 = R25 +R10 + R4 = R39 = R7
Sufficient

Statement 2:
2z = R2
z = R1

z^2 + 2z + 4 = R1 +R2 + R4 = R7
Sufficient

Answer D

Bunuel   KarishmaB  can you please verify this approach?

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

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