If Z is a positive integer and 192^5 is a multiple of 8^Z, what is the

Question

If Z is a positive integer and 192^5 is a multiple of 8^Z, what is the largest possible value of Z?

A. 5.
B. 7.
C. 8.
D. 10.
E. 12.

amarsinha · Answer

192 = (2^6)*3

192^5 = (2^30)*3^5
Now 192^5 = x*8^z
 = x*(2^3z)
So (2^30)*3^5 = x*(2^3z)

for largest value of z, the power of 2 on both side should be equal

So 30 = 3z
=> z = 10

omkartadsare · Answer

192^5 =((2^6)*3)^5 and 8^Z is 2^3z
In order to maximize Z, we will have to equate it with (2^6)^5.
Hence 2^30 = 2^3z....3Z= 30
z=10
IMO (D)

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Archit3110 · Answer

192^5 = 2^30* 3^5
so 8^z= 2^3z 
z has to be 10
IMO D

rkoberoi · Answer

192^5 = ((2^6)*3)^5 = (2^30)*3^5

Given - 192^5 = x*8^z
= x*(2^3z)

So (2^30)*3^5 = x*(2^3z)

To get the largest value of z, the power of 2 on both sides should be equal

So 30 = 3z
 z = 10

IMD - D

NuttyIan · Answer

Realized that 192= 64*3, thus 8^2 is a factor 192. Given 192 is brought to the 5th power, we can multiply 5*2 to get 10 8s in the factor.

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