Problems & Puzzles:
Farideh Firoozbakht sent the following puzzle:
211896 = phi(211896/2) + phi(211896/1) +
phi(211896/1) + phi(211896/8) + phi(211896/9) + phi(211896/6).
Farideh also found other numbers like this
one: 61341696, 141732864, 219483432, 1423392768, 4844814336
Q. What is the next such
Contributions came from Mr. Qu Shun Liang & W Edwin Clark.
Qu Shun Liang wrote:
I found 7 more terms!
for 1<=k<=9, phi(n/k)=phi(n)/t, there 1<=t<=9.
so, if n is a solution,
n=phi(n)*u/v, there u/v=s1*/1+s2*/2+s3*/3+…+s9/9.
so we get:
actually, The largest prime factor of u is very small.
The largest prime factor of n is very small.
take n=61341696 for example,
I have search all number that the largest prime factor is less than 256
W. E. Clark wrote:
I think, but am not absolutely sure, that the next number in
list is 16484622336.
I found 14 numbers satisfying Farideh's criterion: All of
them have the form 2^a*3^b*7^d*p^e where p is prime > 11. (This
the 6 already found by Farideh.) They are:
211896, 61341696, 141732864, 219483432, 1423392768, 4844814336,
16484622336, 23362267824, 28193299344,
But I make no claim that there aren't others in between.
Two questions: Are there any containing TWO distinct prime factors
different from 2,3,7? Is there an example containing 5 as a factor?
Fred answer to Clark's question:
No, here's why:
1)The totient(x) for every x except for 2 is an even number. The
upper bound where for n where n/d = 2 (d is some digit) is 18. There
are no solutions through 18. Therefore, there can be no odd
solutions to this puzzle in general because the sum of even numbers
can never be odd.
2) A number can only be divisible by 5 if its unit digit is 0 or 5.
3) An odd solution (ending in 5) is not possible as stated above.
4) An even solution ending in 0 is not possible for some n because
phi(n/0) is undefined.
Concerning puzzle number 500: In case you are planning on submitting
the sequence to the OEIS: I have now checked that there are no
numbers satisfying your condition between 4844814336 and
16484622336. So you can safely add 16484622336 to the list. But I
cannot speak about the lack of intermediate values for the other
numbers in my list.