Problems & Puzzles:
2d(n) = n + E
Joseph L. Pe sent the following
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
Contributions came from: Farideh Firoozbakht, Shyam Sunder Gupta,
Jacques Tramu, Dan Dima & J. C. Rosa:
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
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
the equation, d(2^(y - 1) -19) = y/2 is y = 128 so the earliest
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
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.
Question 2 : I have found the following
If E=2^k then n=E=2^k is a
solution of the
I have only searched a solution
for E=2*p (p is an odd
prime ) . I have found a solution for
For p=19 ; E=38 ,
but for p=31 I don't find any solution