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

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 amFollowed by: 2 members

by Gmat_mission » Fri Aug 20, 2021 1:01 pm

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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

Answer: C

Source: GMAT Club Tests

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Source: Beat The GMAT — Problem Solving | Fri Aug 20, 2021 1:01 pm

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

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

by regor60 » Wed Aug 25, 2021 5:20 am

Rearrange the statement to read

N! = 2^k-8 = 2^k- 2^3. Factor out 2^3.

N! = 2^3[2^(k-3) - 1]. So

N!/2^3 = 2^(k-3) -1

Since the right side must be odd, so must the left. Let's expand the left with a hypothetical series:

(9x8x7x6x5x4x3x2x1)/8

In order for this to be odd all multiples of 2 have to divide out. Any sequence greater than 5 will have at least one multiple of 2 remaining after the division by 8.

So, n=5 works and so does n= 4. The corresponding k's are:
5x4x3x2=120+8=128= 2^7. One ordered pair is 5,7

4x3x2=24+8=32=2^5.
The other ordered pair is 4,5
C,2

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