## #math trivia #47 solution

The 10s, 40s and 70s each have at least three primes:

• 10s:  11, 13, 17,  19
• 40s:  41, 43, 47
• 70s:  71, 73, 79

The reason?  Any number less than 100 that’s not a prime must be divisible by 2, 3, 5 or 7.  (If a number is not a prime, and not divisible by 2, 3, 5, or 7,  then its smallest prime factor must be at least 11 — meaning that the number must be at least 121, which is greater than 100.)

The primes 2 and 5 take out six of the numbers in any given decade:  the ones that end in 0, 2, 4, 5, 6, or 8.  This leaves just the ones that end in 1, 3, 7, and 9.

The 10s, 40s and 70s have many primes because 3 doesn’t take out any additional numbers:  In each of these decades, the number ending in 5 is divisible by both 3 and 5, so the 3 takes out just the numbers that end in 2, 5, and 8, nothing new.

That means that it only matters in these decades which number the 7 takes out — and 7 can take at most one odd number in a decade.  In the 10s, the 7 takes out 14, which is even, leaving all four possibilities.  In the 40s, it takes out 49, leaving three; and in the 70s, it takes out 77, leaving three.

In all the other decades up to the 90s, the 3 takes out two additional odd numbers (1 and 7, or 3 and 9), which leaves at most two possibilities.  (Recall the other decades’ primes to see the pattern:  23 and 29; 31 and 37; 53 and 59; 61 and 67; 83 and 89; and just 97 — the 91 being taken out by 7.)

The 100s, by the way, are also popular.  Because 105 is the product of 3, 5 and 7, all four possibilities again remain:  101, 103, 107 and 109.

Number theory note:  A decade with four primes contains an instance of a prime quadruplet:  four primes spaced as p, p+2, p+6, p+8.  It has not yet been proven whether there are infinitely many prime quadruplets.  If the number of quadruplets is finite, then the number of four-prime decades must also be finite.  If the number of quadruplets is infinite, the number of four-prime decades presumably also is.  It’s possible in principle that the quadruplets starting with a number ending in 1 might be finite, while the full set is infinite, but it would seem an unusual result (though my number theory isn’t deep enough to be sure).