Call a positive integer totally odd if all of its decimal digits are odd.
Show that for any positive integer k there is a k-digit totally odd integer which is a multiple of 5k.
*** Adapted from a problem which appeared at the USA Mathematical Olympiad in 2003.
In addition to 5, 3 works. I.e.,
For any positive integer k there is a k-digit totally odd integer which is a multiple of 3^k.
k = 1,2,3,4 results are shown:
k coefficient 3^k Totally odd
-------------------------------------------------
1 1 3 3
1 3 3 9
2 11 9 99
3 5 27 135
3 13 27 351
3 19 27 513
3 37 27 999
4 17 81 1377
4 19 81 1539
4 39 81 3159
4 71 81 5751
4 73 81 5913
4 91 81 7371
4 93 81 7533
4 113 81 9153
4 115 81 9315
I note that these are _not_ in the form of the 5^k results.
In all the 5^k examples, an odd digit is appended on the left
as the new most significant digit to make a k-length odd number.
This is very nicely motivated by Brian Smith's inductive proof.
The "append method" does not work for the 3^k case, however.
For example, there is no "complimentary residual" to be found from
10, 30, 50, 70, or 90 mod 9 to advance from N=3 (or N=9)
to go from k=1 to k=2.
So, a different proof would be needed for the 3^k case.
It is interesting to note that the appended odd numbers in the
5^k case are the _only_ solutions, whereas for the 3^k case,
there are multiple solutions, as shown above.
Past 5^k, there are no solutions adhering to the k-digit requirement
(that Larry's conjecture ignores) E.g., there are no 2 digit totally odd
numbers that are a multiple of 7^2. (49, 98 being the only available
possibilities)
Edited on October 17, 2024, 11:52 pm
limited solutions
