Problems & Puzzles: Puzzles

 Puzzle 349. 2d(n) = n + E Joseph L. Pe sent the following puzzle:  I would like to propose the following puzzle: 1. Let d(n) = number of divisors of n. For a fixed even number E, does the equation 2d(n) = n + E always have at least one solution for n? (I have checked that there are solutions for E = 2, 4, ..., 28.) 2. An easier question: Find an n such that 2^d(n) = n + 92.

Contributions came from: Farideh Firoozbakht, Shyam Sunder Gupta, Jacques Tramu, Dan Dima & J. C. Rosa:

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Farideh wrote:

Answer to easier  question: The first three solutions  for the equation  2^d(x) = x + 92  are: 16777124,  281474976710564  & 79228162514264337593543950244
If E is of the form 2^m where m>0 then it's obvious that x =  E  is
the only solution for the equation  2^d(x) = x + E.

And  if  E = 2^s * t  where  s & t are natural numbers and t is odd
&  t>2 then for the following two equations (*) & (**)  we can say (*)
has solution iff (**) has solution and y is a solution for (**) iff  2^y - E
is a solution for (*).

(*):    2^d(x) = x + E                (**):     d(2^(y - s) - t) = y/(s + 1)

For example if E=38  then s = 1 & t = 19  and the earliest solution for
the equation,  d(2^(y - 1) -19) = y/2  is y = 128 so the earliest solution
for the equation   2^d(x) = x + 38   is
x = 2^128 - 38=340282366920938463463374607431768211418.

Also the smallest solution for the equation  2^d(x) = x + 30  is greater
than 2^210.

Shyam wrote:

Part 1 of the puzzle seems to be very difficult. But it appears that: For a fixed even number E,  the equation 2d(n) = n + E may not always have solution for n?

2. An easier question: Find an n such that 2^d(n) = n + 92. 16777124 is the smallest n such that 2^d(n) = n + 92.

Jacques Tramu wrote:

Q2: E,n, and d(n)

92 16777124 24
92 281474976710564 48
92 79228162514264337593543950244 96
92 6277101735386680763835789423207666416102355444464034512804 192

Dan Dima wrote:

The smallest integer n such that 2^d(n) = n + 92 is:
n = 16777124 = 2^2 * 7 * 13 * 46091
d(16777124) = 24, 2^24 = 16777216.
Unfortunately, right now I don't have a rigorous proof for the 1st part - just some facts - but I believe this is true.

Rosa wrote:

Question 2 : I have found the following solution:
n=16777124=(2^2).7.13.46091; d(n)=24; 2^24=n+92

Question 1:
If E=2^k then n=E=2^k is a solution of the
equation 2^d(n)=n+E

I have only searched a solution for E=2*p (p is an odd
prime ) . I have found a solution for p=3,5,7,11,13,17,19,23,29,37,41,43,47
For p=19 ; E=38 , n=2.103.137.2137.10739.800549.656289375387864848117
but  for p=31 I don't find any solution for n=<2^180

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