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.
Conjecture: that the proposition is true for all odd numbers, not just powers of 5.
Computer testing every odd integer up to 10^4 shows that the multiplier to achieve a totally odd number is lowest for odd numbers ending in 5 and 7; about 20% more than that for odds ending in 3 and 9; and most for odds ending in 1.
But for every odd number up to 10^4, the conjecture held.
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Posted by Larry
on 2024-10-06 09:13:18 |
also no proof, but a conjecture
