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If a, b, c, and z are nonnegative integers, what is the remainder when

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If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 30 Apr 2016, 03:37
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Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 01 May 2016, 03:39
Bunuel wrote:
If a, b, c, and z are nonnegative integers, what is the remainder when 3^(az)×3^(bz)×3^(cz) is divided by 4?

(1) z is even.
(2) The sum of a, b, and c is odd.


Hi,
3^(az)×3^(bz)×3^(cz) =\(3^{az+bz+cz} = 3^{(a+b+c)z}\)..

Two methods --



I. check for pattern -

3^1 leaves a remainder 3
and 3^2 leaves a remainder 1
3^3 leaves a remainder 3
and 3^4 leaves a remainder 1..
so ODD power leaves remainder 3 and EVEN power leaves a remainder 1..
SO if we know whether POWER is ODD or EVEN, we can answer..

POWER is product of a+b+c and z

lets see the statements--

(1) z is even.
Hence POWER is EVEN irrespective of what a+b+c is..
remainder is 1
Suff

(2) The sum of a, b, and c is odd.
we do not know about z..
If z is ODD.. power is ODD and ans is 3..
But if z is Even .. power is EVEN and ans is 1
different answers possible
Insuff

ans A


Algebrically through expansion..
(1) z is even.

so \(3^{(a+b+c)z} = 3^{(a+b+c)*2*y}\) as z= 2y where y is an integer, nonnegative
\((3^2)^{(a+b+c)*y} = 9^{(a+b+c)*y} = (8+1)^{(a+b+c)*y}\)..
so all terms will be div by 8 and therefore by 4 except \(1^{((a+b+c)*y}\)
so remainder = 1..
Suff

(2) The sum of a, b, and c is odd.
we cannot work further on it..
Insuff
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Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 01 May 2016, 03:42
1
Bunuel wrote:
If a, b, c, and z are nonnegative integers, what is the remainder when 3^(az)×3^(bz)×3^(cz) is divided by 4?

(1) z is even.
(2) The sum of a, b, and c is odd.


Hi Bunuel,
there is a typo in the topic name..
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Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 01 May 2016, 03:53
3^(az)×3^(bz)×3^(cz) --> 3^z(a + b + c)

St1: z is even --> z = 2k

3^z(a + b + c) = 3^even --> 3^2K = 9^K
9/4 --> Remainder = 1 --> (1)^K is always 1.
Remainder is always 1
Sufficient

St2: a + b + c is odd
If z = 0 --> 3^0/4 --> Remainder = 1
If z = 1 and a + b + c = 3 --> 3^3/4 --> Remainder = 3
Not Sufficient

Answer: A
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Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 03 May 2016, 01:54
Sol:
3^(az)×3^(bz)×3^(cz) = 3^(az+bz+cz)= (4-1)^(az+bz+cz)

4^(az+bz+cz) is always divisible by 4, or remainder = 0.
What matters is the remainder when (-1)^(az+bz+cz) is divided by 4

(1) z is even => (az+bz+cz) is even =>(-1)^(az+bz+cz) = 1 => remainder when divided by 4 is 1
Sufficient

(2) (a+b+c) is odd => we don't know whether z(a+b+c) is odd or even without info regarding z
Insufficient

Answer (A)
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If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 02 Sep 2018, 10:23
Hi, what about the case when a=b=c=0. I'm guessing if they are non negative, their value could be zero ?
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Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 22 Sep 2019, 12:01
chetan2u wrote:
Bunuel wrote:
If a, b, c, and z are nonnegative integers, what is the remainder when 3^(az)×3^(bz)×3^(cz) is divided by 4?

(1) z is even.
(2) The sum of a, b, and c is odd.


Hi,
3^(az)×3^(bz)×3^(cz) =\(3^{az+bz+cz} = 3^{(a+b+c)z}\)..

Two methods --



I. check for pattern -

3^1 leaves a remainder 3
and 3^2 leaves a remainder 1
3^3 leaves a remainder 3
and 3^4 leaves a remainder 1..
so ODD power leaves remainder 3 and EVEN power leaves a remainder 1..
SO if we know whether POWER is ODD or EVEN, we can answer..

POWER is product of a+b+c and z

lets see the statements--

(1) z is even.
Hence POWER is EVEN irrespective of what a+b+c is..
remainder is 1
Suff

(2) The sum of a, b, and c is odd.
we do not know about z..
If z is ODD.. power is ODD and ans is 3..
But if z is Even .. power is EVEN and ans is 1
different answers possible
Insuff

ans A


Algebrically through expansion..
(1) z is even.

so \(3^{(a+b+c)z} = 3^{(a+b+c)*2*y}\) as z= 2y where y is an integer, nonnegative
\((3^2)^{(a+b+c)*y} = 9^{(a+b+c)*y} = (8+1)^{(a+b+c)*y}\)..
so all terms will be div by 8 and therefore by 4 except \(1^{((a+b+c)*y}\)
so remainder = 1..
Suff

(2) The sum of a, b, and c is odd.
we cannot work further on it..
Insuff



Hi,
Which method do you think is more efficient for most of the problems?
Best regards
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Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 23 Sep 2019, 08:14
chetan2u wrote:
Bunuel wrote:
If a, b, c, and z are nonnegative integers, what is the remainder when 3^(az)×3^(bz)×3^(cz) is divided by 4?

(1) z is even.
(2) The sum of a, b, and c is odd.


Hi,
3^(az)×3^(bz)×3^(cz) =\(3^{az+bz+cz} = 3^{(a+b+c)z}\)..

Two methods --



I. check for pattern -

3^1 leaves a remainder 3
and 3^2 leaves a remainder 1
3^3 leaves a remainder 3
and 3^4 leaves a remainder 1..
so ODD power leaves remainder 3 and EVEN power leaves a remainder 1..
SO if we know whether POWER is ODD or EVEN, we can answer..

POWER is product of a+b+c and z

lets see the statements--

(1) z is even.
Hence POWER is EVEN irrespective of what a+b+c is..
remainder is 1
Suff

(2) The sum of a, b, and c is odd.
we do not know about z..
If z is ODD.. power is ODD and ans is 3..
But if z is Even .. power is EVEN and ans is 1
different answers possible
Insuff

ans A


Algebrically through expansion..
(1) z is even.

so \(3^{(a+b+c)z} = 3^{(a+b+c)*2*y}\) as z= 2y where y is an integer, nonnegative
\((3^2)^{(a+b+c)*y} = 9^{(a+b+c)*y} = (8+1)^{(a+b+c)*y}\)..
so all terms will be div by 8 and therefore by 4 except \(1^{((a+b+c)*y}\)
so remainder = 1..
Suff

(2) The sum of a, b, and c is odd.
we cannot work further on it..


In GMAT, isnt 0 considered as an Even number?

Insuff
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If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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New post 23 Sep 2019, 14:08
Bunuel wrote:
If a, b, c, and z are nonnegative integers, what is the remainder when 3^(az)×3^(bz)×3^(cz) is divided by 4?

(1) z is even.
(2) The sum of a, b, and c is odd.


We want to find the remainder of \(3^{az + bz + cz} = 3^{z(a+b+c)}\) divided by 4. We can do some remainder arithmetic to check how the remainder acts as the power of 3 increases. \(\frac{{3^1}}{4}\) => 3. The remainder is 3. For the next power of 3 divided by 4, \(\frac{{3^2}}{4}\), we can multiply the remainder from \(\frac{{3^1}}{4}\) itself by 3. We can treat this as if we had a remainder of 3*3 = 9 for \(\frac{{3^2}}{4}\) which reduces to a remainder of 1 since we are in the div by 4 world. Repeat this process again to see the next remainder is 1*3 = 3, therefore the remainders will keep repeating from now on. This is called remainder arithmetic and we just proved that the remainders cycle through 1 and 3. Hence, we just need to figure out whether z(a+b+c) is even or odd to answer this question.

(1) Sufficient.
(2) Insufficient.

Answer: A
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If a, b, c, and z are nonnegative integers, what is the remainder when   [#permalink] 23 Sep 2019, 14:08
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