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Math Expert V
Joined: 02 Sep 2009
Posts: 58464
If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 57% (02:01) correct 43% (02:20) wrong based on 130 sessions

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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.

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Math Expert V
Joined: 02 Aug 2009
Posts: 8023
Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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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
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
_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 8023
Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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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..
_________________
Marshall & McDonough Moderator D
Joined: 13 Apr 2015
Posts: 1683
Location: India
Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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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

Intern  Joined: 12 Dec 2014
Posts: 1
Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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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

Intern  B
Joined: 04 Jun 2018
Posts: 15
Location: India
GMAT 1: 660 Q48 V31 If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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Hi, what about the case when a=b=c=0. I'm guessing if they are non negative, their value could be zero ?
Manager  G
Joined: 31 Jan 2019
Posts: 172
Location: Switzerland
Concentration: General Management
GPA: 3.9
Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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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
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
Intern  B
Joined: 20 Jun 2019
Posts: 6
Re: If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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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
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
GMAT Tutor S
Joined: 17 Sep 2014
Posts: 208
Location: United States
GMAT 1: 780 Q51 V45 GRE 1: Q170 V167 If a, b, c, and z are nonnegative integers, what is the remainder when  [#permalink]

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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.

_________________
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