If a is divisible by 5!, then a/2 must be

Question

I. a factor of 10
II. a multiple of 8
III. a multiple of 10

A. None
B. I only
C. III only
D. I and III only
E. I, II, and III

vitaliyGMAT · Answer

\frac{a}{5!} = integer

in other words  a  is a multiple of  5!

a = 120x = 2^3*3*5*x

\frac{a}{2} = 2^2*3*5*x

I  In order to be a factor of 10  a  must divide 10 evenly - Impossible --->  \frac{2*5}{2^2*3*5} ≠ integer

II  In order to be a multiple of 8  a  must have 3 factors of 2, but a is only a multiple of 4 Negative.

III   a  is definitely a multiple of 10 because it has both 2 and 5 as factors.

We don't know value of x, but question asks "what must be true"

Hence answer C

Abhishek009 · Answer

a = k*5*4*3*2

So, a/2 = k*5*4*3

I. a is a not a factor of 10, but a multiple of 10

II. a is not a multiple of 8

III. a is a multiple of 10

Hence, correct answer must be (C) III only

RR88 · Answer

Option C

a = 5! * X = 120 * X = 2^3 * 3^1 * 5^1 * X

a/2 = 2^2 * 3^1 * 5^1 * X

I:  \frac{a}{2}  - is not a divisor but a multiple of 10 - so NO

II:  \frac{a}{2}  - can be multiple of 8 or can't be, depends on value of X - so NO

III:  \frac{a}{2}  - can be written as  2^1 * 3^1 * 10^1 * X  - is definitely a multiple of 10 - so YES

Bunuel · Answer

MHIKER · Answer

5!=5*4*3*2*1
a=5*4*3*2*1/2=5*4*3
is only multiple of 10
Ans. C

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