(1): Consider the set of the smallest 6 odd primes: {3,5,7,11,13,17}.
What is the largest multiple of this product which has
distinct digits ?
(2) What is the maximum number of digits a square-free integer (whether even or odd) can have if its digits are all distinct?
(3) What is the largest odd square-free integer with distinct digits having exactly n prime factors for n = 1,2,3,4,5? You can extend this to larger numbers of factors if you wish.
Note: a square-free integer is one whose prime factorization has exactly one factor for each prime that appears in it.
If all digits 0-9 are present in a number with distinct digits then that number must be a multiple of 9, as the sum of digits is 45. That number cannot be squarefree. Then the number of digits a squarefree number can have with distinct digits is at most 9.
Removing one digit (not 0 or 9) will avoid the number being a multiple of 9. Its pretty easy to find 987654310 = 2 x 5 x 98765431 (prime factorization) as the largest squarefree number with distinct digits. This satisfies n=3 in Question 3.
Moving on, the number 987654302 = 2 x 347 x 661 x 2153 satisfies n=4
Then 987654301 = 2029 x 486769 satisfies n=2.
987654230 = 2 x 5 x 73 x 419 x 3229 satisfies n=5.
And 987654103 is prime (n=1).
Question 2 and 3 answers.
