If a is divisible by 5!, then a/4 must be

Question

I. an odd integer
II. a multiple of 3
III. a multiple of 10

A. I only
B. II only
C. III only
D. I and III only
E. II and III only

Vinayak Shenoy · Answer

IMO E
If a is divisible by 5! then a must be of the form 120*k when k is an integer 1, 2, 3
then a can be 120, 240, 360, 480....
a/4= 30, 60, 90, 120.....
from this we can say a/4 satisfies E.

stonecold · Answer

Nice Question.
This is a must be true question.
Statement 1 => a/4 must be odd
Hmm Let a = 5! => clearly a/4 is even>
Hence 1 must not be true.
Statement 2 => of course it will be true as a must contain a 3 => a/4 must be a multiple of 3
statement 3 => Officourse the least value of a is 5! and 5!/4 is clearly divisibly by 10
Hence E

Luckisnoexcuse · Answer

IMO it is (E)

a is divisible by 5*4*3*2
a/4 is divisible by 5*3*2

I is not odd integer hence No
II a multiple of 3 Yes
III a multiple of 10 Yes

smartarvinz · Answer

I getting E.

a/2*3*4*5*4 = a/ 240

so a  must be  divisible by 3 and 10 but need not necessarily be an odd number

herbgatherer · Answer

If a is divisible by 5!, then a/4 must be

a = 5*4*3*2*1*K (K = integer)

a/4 = 5*3*2*1*K

I. an odd integer 
a/4 contains a 2. This cannot be odd. Eliminate.

II. a multiple of 3 
a/4 contains a 3. Is a multiple of 3. Retain.

III. a multiple of 10 
a/4 contains 5 & 2. Retain

Answer = E. II and III only

Abhishek009 · Answer

5! = 120

Now Check the options -

I. a/120 can be ODD/EVEN

a = 240
a/120 = 2

a = 360
a/120 =3

II. Since 5! = 120 a multiple of 3 , any number " a "will be a multiple of 3

III. Since 5! = 120 a multiple of 10 , any number " a " will be a multiple of 10

Thus only option (E) is true...

deenesh2309 · Answer

If a is divisible by 5! ie a is divisible by 2,3,4,5.
So a/4 is divisible by 2,3,5.

I) Is it odd integer? No, as it is still divisible by 2
II) Is it multiple of 3? Yes, as it is divisible by 3
III) Is it multiple of 10? Yes, as it is divisible by 2 and 5

So answer is Only II and III ie E

mvictor · Answer

first of all, 5! = 1*2*3*4*5

a/4 = 1*2*3*5 - an even number. I is not true. A and D are out.
II - we have a factor of 3, so yes, IT IS a multiple of 3. C is out.
III - we have a factor of 2 and a factor of 5 - so yes, IT IS a multiple of 10. B is out.

the answer is E.

ScottTargetTestPrep · Answer

Since 5! = 5 x 4 x 3 x 2 x 1 = 120, and a is divisible by 5!, we know that a is divisible by some multiple of 120. However since we are determining what MUST be true, we can set a equal to the smallest positive multiple of 120, which is 120.

Thus a/4 = 120/4 = 30.

Now we can analyze each Roman numeral.

I. a/4 is an odd integer

Since 120/4 = 30, a/4 does not have to result in an odd integer. Roman numeral one is not correct.

II. a/4 is a multiple of 3

Since 120/4 = 30, a/4 will always be a multiple of 3, Roman numeral II is correct.

III. a/4 is a multiple of 10

Since 120/4 = 30, a/4 will always be a multiple of 10, Roman numeral III is correct.

Answer: E

GMATinsight · Answer

a is divisible by 5! i.e. a is a multiple of 120

i.e. then a/4 must be a multiple of 120/4 = multiple of 30
Which is Even, a multiple of 3 and a multiple of 10 as well

Hence
Answer: Option E

Bunuel · Answer

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If a is divisible by 5!, then a/4 must be

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