#math trivia #231 solution

There are six day-numbers consisting of the digits 1, 2, and 3, and they have the following factorizations:

  • 123 = 3 * 41
  • 132 = 2^2 * 3 * 11
  • 213 = 3 * 71
  • 231 = 3 * 7 * 11
  • 312 = 2^3 * 3 * 13
  • 321 = 3 * 107

The day-number 231 is the only one that has 7 as a prime factor.

Answer:  7.

Notes:

  • For obvious reasons, the two day-numbers in this series that end in 2 have 2 as a prime factor.
  • For perhaps not as obvious reasons, all six day-numbers have 3 as a prime factor.  But recall that one of the ways to test that a number is divisible by 3 is to add its digits and test whether their sum is divisible by 3.  The sum of the digits of each of these day-numbers is the same, 1 + 2 + 3 = 6, hence the divisibility by 3.
  • The condition “consisting of the digits 1, 2 and 3” might be interpreted as “consisting of no digits other than 1, 2 and 3”  rather than “consisting of all of the digits 1, 2 and 3”.  However, it’s clear that the latter meaning is intended, because otherwise 231 would not be the fourth in the series, there would be many others preceding it, starting with 1, 2, 3, 11, 12, etc.

Related question:  What relationship between 132 and 231 makes it obvious that if one of the two numbers is divisible by 11, then so is the other, as is the case here?

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