If n and k are positive integers, then how many ordered pairs (n, k)

Question

If n and k are positive integers, then how many ordered pairs (n, k) satisfy  n! + 8 = 2^k  ?

A. 0
B. 1
C. 2
D. 3
E. Infinitely many

M37-79

Bunuel · Accepted Answer

Official Solution:

If  n  and  k  are positive integers, then how many ordered pairs  (n, k)  satisfy  n! + 8 = 2^k  ?

A.  0 
B.  1 
C.  2 
D.  3 
E. Infinitely many

n! + 8 = 2^k ;
 
 n! = 2^k-8 ;
 
 n! = 2^k-2^3 ;
 
 n! = 2^3(2^{k-3}-1)= 2^3*odd .
 
 n! = 2^3*odd  means that  n! \geq 8 , thus  n \geq 4 .
 
 n! = 2^3*odd  also means that  2^3  is the highest power of 2 that can divide  n! . This on the other hand means that  n \leq 5  (if  n > 5 , then  n!  will have more than three 2's).
 
So, we get that  4 \leq n \leq 5 . 
 
If  n = 4 , then  k=5 .
 
If  n = 5 , then  k=7 .
 
Thus,  (n, k)  can only be  (4, 5)  and  (5, 7)

Answer: C

minustark · Answer

n!= 2^k -8 or, n!= 2^3 (2^(k-3) -1) = 8 (even - 1) = 8 * odd
Since n! follows the formula of either odd*even*odd*even.......or even *odd*even*odd....so we can limit the test to the expression = 2*4 *odd*odd
Two cases possible n!= 4!(here we take two odd integers 3 and 1) or, n! = 5!(two odd integers are 3 and 5).
Lets take, n!=4!= 24, so 2^k = 32, k =5.
When n! = 5!, 2^k = 128, k = 7.
C is the answer.

neshantz · Answer

Would k=3 and n=0 be an answer also?

0! + 8 = 2^3

Or is 0! not a valid answer.

Bunuel · Answer

First of all, we are told that n and k are positive integers and 0 is not a positive integer. Next, 0! = 1, so 0! + 8 = 9, not 8.

neshantz · Answer

Ok thank you! I missed that and did not know 0! is 1.

So 0!=1!?

Posted from my mobile device

Bunuel · Answer

Yes.

Zero Factorial 
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sujoykrdatta · Answer

n! + 8 = 2^k  (clearly 8 is smaller than  2^k )

=>  n! = 2^k - 2^3 = 2^3 * (2^{(k-3)} - 1)  = 8 * Odd

We need to make n! a multiple of 8 => n = 4, 5, 6, 7, ...

But, we also need to make n! equal to 8 * odd, i.e. an odd multiple of 8
Clearly, 6! is a multiple of 16, i.e. an even multiple of 8. Hence, all n = 6, 7, ... will result in n! to be multiple of 16 - hence can be ignored

Thus, n = 4 or 5:

n = 4: 4! + 8 = 24 + 8 = 32 = 2^5 (satisfies)
n = 5: 5! + 8 = 120 + 8 = 128 = 2^7 (satisfies)

Thus, there are 2 solutions

Option C

Archit3110 · Answer

target to get ordered pairs of (n.k) for the given expression  n! + 8 = 2^k 
value of 2^k = 2,4,8,16,32,64,128,256,512,1024
now
 n!+8 we see that least value would be 9 at n=0,1 
the observed pattern at LHS
n!+8 
n=4 ; 4*3*2*1 +8 
=> 8*4 where both values are factors of 2^k at k=3,2
similarly at n=5
5*4*3*2*1+8 
=> 8 * (16) ; the values are also factors of 2^k at 3,4
further at n=6 onwards we see that
6*5*4*3*2*1+8
=> 8* ( 90) ; and n=7 ; 8*630 so on the values are not factors of 2^k

so the min value equal to 2^k would be at n=4 ; 32 i.e k=5 and n= 5 where k = 7
only 2 possible values
option C

bumpbot · Answer

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If n and k are positive integers, then how many ordered pairs (n, k)

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