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

Question

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.

chetan2u · Answer

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

Kurtosis · Answer

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

zennith512 · Answer

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)

sheen22 · Answer

Hi, what about the case when a=b=c=0. I'm guessing if they are non negative, their value could be zero ?

auradediligodo · Answer

Hi,
Which method do you think is more efficient for most of the problems?
Best regards

kritiminocha · Answer

TestPrepUnlimited · Answer

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

dylanl1218 · Answer

This tricked me up a bit also when I first went through this question.

However, it's still true that if a+b+c=0, OR if z = 0 that the remainder would be equal to 1, because:
(1) 3^0 = 1
(2) 1/4 = remainder of 1

So then statement one must be true because any even exponent, including 0, will result in a remainder of 1.

Basshead · Answer

If a, b, c, and z are nonnegative integers, what is the remainder when  3^{az}3^{bz}3^{cz}  is divided by 4?

3^{az}3^{bz}3^{cz}  =  3^{z(a+b+c)}

(1) z is even.

If z is even, then  z(a+b+c)  = even.

Lets test some values:
 3^2  = 9/4= remainder 1
 3^4  = 81/4 = remainder 1

The remainder will always be 1 when z is even. Sufficient.

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

We don't know the what z is; insufficient.

Answer is A.

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

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