For any positive integer n greater than 1, n! denotes the product of a

Question

For any positive integer n greater than 1, n! denotes the product of all the integers from 1 to n, inclusive.
If A is a positive integer such that the greatest number that divides both  A^3  and 13! is 448, which of the following can be the value of A?

A. 14
B. 56
C. 140
D. 196
E. 448

sujoykrdatta · Accepted Answer

The greatest common divisor (GCD) of  A^3  and  13!  is 448

448 = 4 * 112 = 4 * 4 * 28 = 4 * 4 * 4 * 7 = 2^6 * 7

Observe that  13!  has only  7^1  as factor
Also, highest power of  2  in  13! =   +   +   = 6 + 3 + 1 = 10  (where     denotes the  greatest integer  less than or equal to  n )

Since  2^6 * 7  is the GCD of  A^3  and  13! , we can conclude that   2^6 * 7  must be a factor of  A^3

Thus, we have:

1.  A^3  must NOT have the power of 2 greater than 6 (since 13! has power of 2 as 10, if  A^3  had a power of 2 greater than 6, the GCD would also have a power of 2 greater than 6) => Highest power of 2 in A must be 2 i.e. A is a multiple of  2^2  (not more than that) ... (i)

2.  A^3  must have  7^3  as a factor as well (it is the cube of a number), implying  A  is a multiple of  4 * 7  i.e.  28  ... (ii)

Working with options:

A.  A = 14  ---- not possible since A should be a multiple of 28 ---> from (ii)

B.  A = 56 = 2^3 * 7  ---- Violates (i)

C.  A = 140 = 2^2 * 7 * 5  ---- If A is a multiple of 5, and since 13! is also a multiple of 5, the GCD would have 5 as a factor ---- Violates given info

D.  A = 196 = 2^2 * 7^2  ---- Satisfies both (i) and (ii). Note: even though A has a factor  7^2 ,  13!  has a factor only  7^1 , hence GCD would still have  7^1  as factor

E.  A = 448 = 2^6 * 7  ---- Violates (i)

Answer D

shameekv1989 · Answer

Is that A3 =  A^3  or  A*3 ??

unraveled · Answer

You are right.
Thanks for pointing the error. Corrected it.

shameekv1989 · Answer

If  A^3  is divisible by 448 (which is  2^6 *7) then A must have atleast one 7 ->  A^3  must have 7^3 = 343

The only value that is greater than 343 is 448 - Answer - E

unraveled · Answer

My friend you were right till halfway. Revisit, i am sure you would find why.

Apt0810 · Answer

Answer would be I think 196 since 448= 2^6 *7
And 196^3 = 7^6*2^6
And it would be divisible by 448

Posted from my mobile device

shameekv1989 · Answer

You are right. That's A^3 not A; Anyways :-

Minimum that A has is one 7 and two 2's. Two 2's is fixed but 7 is flexible. Thus it can be 7^2 * 4 = 196

Answer - D

freedom128 · Answer

Both A and 13! must have common factors of minimum one 7 and exactly two 2s. No other factors is allowed.

Eliminate choices A, B, E since the number doesn't have exact factors of 2^2.
A. 14=2*7 
B. 56=2^3*7
E. 448=2^6*7

Eliminate choice C because it has additional factor 5. If so, A^3 and 13! must have been divisible by 448*5^2
C. 140=2^2*7*5

FINAL ANSWER IS (D)
Both A and 13! must have common factors of minimum one 7 and exactly two 2s.
D. 196=2^2*7^2

Posted from my mobile device

unraveled · Answer

Good that you corrected the mistake made earlier, a mistake that i also made under timed condition. Anyway great learning

rocky620 · Answer

448 =  2^6 *7

Highest power of 2 in 13! = 9
Highest power of 7 in 13! = 1

1.  Since, the HFC of  A^3  & 13! is 448. The highest and the only power of 2 in  A^3  has to be 6, otherwise the HCF will increase or decrease.
 2.  The power of 7 in  A^3  has to be minimum 3. It can even be more than 3, but that will not affect the HCF.
 3.  Since there are 3 more prime numbers in 13! (apart from 2 & &),  A^3  cannot have any of these numbers or their multiple as its factor. If so, the HCF will increase again.
 4.   A^3  can have any other prime number greater than 13 as its factor, as it will not affect the HCF.

So, from 1,2,3, & 4 we can deduce that, A will be of the form =  2^2  *  7^p  *  X^q

Where, P=<1, X=any prime number greater than 13, and q=any integer

Only option D satisfies the above conditions.

14 = 2*7 (only one power of 2)
56 = 8*7 (3 power of 2)
140 = 4*7*5 (We do not need 3/5/11)
 196 = 4*49 (2 powers of 2) 
448= 64*7 (6 powers of 2)

Ulfat · Answer

For any positive integer n greater than 1, n! denotes the product of all the integers from 1 to n, inclusive. If A is a positive integer such that the greatest number that divides both  A^3  and 13! is 448, which of the following can be the value of A?
A. 14 
B. 56 
C. 140 
D. 196 
E. 448

SOLUTION

Option  
  A  
  A  
  A^3  
  13!  
  448  
  Actual Common Factors between  A^3  and 13!  
 
  A 
 14 
  2^1 * 7^1  
  2^3 * 7^3  
 (2^10)* 7^1 *K 
  2^6 * 7^1  
   7^1  is fine, but,  2^3  is less than  2^6 .  
 
  B 
 56 
  2^3 * 7^1  
  2^9 * 7^3  
 (2^10)* 7^1 *K 
  2^6 * 7^1  
   7^1  is fine, but,  2^9  is greater than  2^6 .  
 
  C 
 140 
  2^2 * 5^1 * 7^1  
  2^6 * 5^3 * 7^3  
 (2^10)* 5^2 * 7^1 *P 
  2^6 * 7^1  
   7^1  and  2^6  are fine, but,  5^2  is an extra common factor.  
 
  D 
 196 
  2^2 * 7^2  
  2^6 * 7^6  
 (2^10)* 7^1 *K 
  2^6 * 7^1  
   2^6  and  7^1  are the Highest common factors, which is same as 448  
 
  E 
 448 
  2^6 * 7^1  
 (2^18)* 7^3  
 (2^10)* 7^1 *K 
  2^6 * 7^1  
   7^1  is fine, but, (2^10) is greater than  2^6 .

Hence Choice (D) is correct.

Option	A	A	A^3	13!	448	Actual Common Factors between \(A^3\) and 13!
A	14	\(2^1\)*\(7^1\)	\(2^3\)*\(7^3\)	(2^10)\(7^1\)K	\(2^6\)*\(7^1\)	\(7^1\) is fine, but, \(2^3\) is less than \(2^6\).
B	56	\(2^3\)*\(7^1\)	\(2^9\)*\(7^3\)	(2^10)\(7^1\)K	\(2^6\)*\(7^1\)	\(7^1\) is fine, but, \(2^9\) is greater than \(2^6\).
C	140	\(2^2\)\(5^1\)\(7^1\)	\(2^6\)\(5^3\)\(7^3\)	(2^10)\(5^2\)\(7^1\)*P	\(2^6\)*\(7^1\)	\(7^1\) and \(2^6\) are fine, but, \(5^2\) is an extra common factor.
D	196	\(2^2\)*\(7^2\)	\(2^6\)*\(7^6\)	(2^10)\(7^1\)K	\(2^6\)*\(7^1\)	\(2^6\) and \(7^1\) are the Highest common factors, which is same as 448
E	448	\(2^6\)*\(7^1\)	(2^18)*\(7^3\)	(2^10)\(7^1\)K	\(2^6\)*\(7^1\)	\(7^1\) is fine, but, (2^10) is greater than \(2^6\).

For any positive integer n greater than 1, n! denotes the product of a

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