If m and n are positive integers, is n a multiple of 24?

Question

(1) n=  \frac{(m+7)!}{(m+3)!} 
(2) n is a multiple of (m+4)

G770 · Answer

Consider statement 1), Resolve the fraction and you get the product of 4 consecutive integers. It is always divisible by 4! Which is 24. 
Hence n is a multiple of 24.
Consider statememt 2), no information about m. So cannot ascertain if m+4 is a multiple of 24.

Hence , IMO A)

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ash124 · Answer

If m and n are positive integers, is n a multiple of 24?

Inference  
 m and n > 0
n should have 2^3 * 7 (24) inside it

n = (m+7)(m+6)(m+5)(m+4)(m+3)! / (m+3)!
n = (m+7)(m+6)(m+5)(m+4)

for any value of m,n will also be multiple of 24

(2) n is a multiple of (m+4)
 For some values of m, n is the multiple of 24
m = 20
For some other values of m, n is not the multiple of 24
m = 12
Insufficient.

Answer: A

sj296 · Answer

m & n are positive integers, is n a multiple of 24

If a number is a multiple of 24, then it should be divisible by 24 i.e. should be divisible by 2, 3, 4,and 8

Statement 1

n =  \frac{(m+7)!}{(m+3)!}  = (m+7)(m+6)(m+5)(m+4)

n is a multiple of 4 consecutive positive integers

n would be divisible by 2, 3, 4, and 8

Therefore, n would be divisible by 24.

Sufficient

Statement 2

n is a multiple of (m+4)

n can be (m+4) or 2(m+4) or 3(m+4) ......so on so forth

n depends on m. We don't know the value of m. m can be 1, 2, 3 ....etc

Hence, n can be divisible by 24 or can not be divisible 24.

Statement 2 is insufficient

Hence, IMO OA should be A

TarunKumar1234 · Answer

If m and n are positive integers, is n a multiple of 24?

Inference
m and n > 0
n should have 2^3 * 7 (24) inside it

n = (m+7)(m+6)(m+5)(m+4)(m+3)! / (m+3)!
n = (m+7)(m+6)(m+5)(m+4)

for any value of m,n will also be multiple of 24

(2) n is a multiple of (m+4)
For some values of m, n is the multiple of 24
m = 20
For some other values of m, n is not the multiple of 24
m = 12
Insufficient.

Answer: A

prachisaraf · Answer

What is the property used here to say that the product of all 4 must be divisible by 24 ?

Bunuel · Answer

The product of x consecutive integers is always divisible by x!. As an example, let's consider x = 4 consecutive integers: {3, 4, 5, 6}. The product of these integers is 3*4*5*6 = 360, which is divisible by 4! = 24.

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If m and n are positive integers, is n a multiple of 24?

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If m and n are positive integers, is n a multiple of 24?

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