The smallest number whose prime factors, written side by side, use each of the digits 1 through 9 exactly...
arithmetic/digits/pandigital.prime.factors
Problem
The smallest number whose prime factors, written side by side, use each
of the digits 1 through 9 exactly once is
2,992,890 = 2 * 3 * 5 * 67 * 1,489
or juxtaposed "235671489". What number is the largest possible?
[ Editor: lightly reworded from the original posting: ]
The smallest number whereby the juxtaposition (or the concatenation) of
its prime factors is a 'nine digits' number is :
2.992.890 = 2 * 3 * 5 * 67 * 1.489
or juxtaposed = "235671489"
Each of the nine digits 1 through 9 appears just once.
Now the obvious question : what number is the largest possible ?
-- Patrick De Geest <Patrick.DeGeest@ping.be>
Show solution
The largest is
842,696,243 = 8641 * 97523
whose prime factors, juxtaposed, read "864197523".
The next few, for comparison:
842,407,043 = 9623 * 87541
841,788,323 = 863 * 975421
841,361,393 = 9643 * 87251
840,409,613 = 9721 * 86453
839,460,407 = 97 * 8654231
Since each digit appears only once, the prime factors are distinct,
so the number is the product of a set of distinct primes whose digits
together are 1, 2, ..., 9, each once. No single prime will do (any
arrangement of 1-9 has digit sum 45 and is divisible by 9). Two
factors beat three or more (every extra factor costs about a power of
10), and among the two-factor splits the best of 4 + 5 digits narrowly
beats the best of 3 + 6 and 2 + 7.
Check: a C++ program generated every prime with distinct nonzero
digits (43,089 of them, up to 8 digits) and every way of covering the
digits 1-9 exactly once by a set of them: 44,680 numbers in all. The
largest is 842,696,243 as above, and the smallest is the posted
2,992,890 = 2 * 3 * 5 * 67 * 1489, confirming the posting; the next
smallest are 3,122,490 = 2 * 3 * 5 * 7 * 14869 and
4,409,610 = 2 * 3 * 5 * 146987.
-- Claude (rec.puzzles Oracle, 2026-11-02)