Who wants to be a Millionaire? Primes between $1$ and $1000$.

Question

I played the quiz game mentioned in the title (in an app) and got to the last question:

In what number do the most primes between $1$ and $1000$ end?

• 1
• 3
• 7
• 9

Up to $100$ we have $7$ primes ending in $3$, $6$ ending in $7$ and $5$ in $1$ and $5$ in $9$. I guess you can figure that out in the short time of a quiz show. Nevertheless, the correct answer is $7$ and that was my gut feeling so I picked it.

I am now, virtually, a millionare, but not a bit smarter about the matter. Is there any heuristic calculation, reasoning or idea to help one get to $7$ in a short amount of time?

I thought about factorizations. Because you need multiples of $3$ for a number ending in $7$, but for $1$ you need the $7$ so its also not a lot of ways. Any ideas?

Answer 1

Guessing 3 or 7 is more likely to be correct than guessing 1 or 9.

This is related to a topic in “comparative prime number theory” called “prime number races”. Andrew Granville and I have written an accessible introduction to this topic.

Answer 2

I happen to have a pretty large list of primes and did some counting:

                                            1         3         7         9
            0 < p <             10:         0         1         1         0
           10 < p <            100:         5         6         5         5
          100 < p <          1,000:        35        35        40        33
        1,000 < p <         10,000:       266       268       262       265
       10,000 < p <        100,000:      2081      2092      2103      2087
      100,000 < p <      1,000,000:     17230     17263     17210     17203
    1,000,000 < p <     10,000,000:    146487    146565    146590    146439
   10,000,000 < p <    100,000,000:   1274194   1274244   1274284   1274154
  100,000,000 < p <  1,000,000,000:  11271088  11272025  11271819  11271147
1,000,000,000 < p < 10,000,000,000: 101050133 101053126 101051725 101049993

As MathUnicorn pointed out, there's no real "winner"

To answer Levent's comment, if I count the ending digits of primes for all primes between 0 and 1,000,000, then between 1,000,000 and 2,000,000, continuing up to 100,000,000, I yield the following winners:

1 : 21  
3 : 23  
7 : 28  
9 : 28

In those 100 ranges the 1 was over-represented in 21 of them, 3 in 23, and so on. And if I change the size from 1 million to 10 million, the picture looks a bit different:

1 : 21  
3 : 27  
7 : 21  
9 : 31

Hence the explanation is (in my humble opinion): 3 and 7 were lucky.

Answer 3

Since I happen to be sitting in front of a computer, I can take the brute-force approach of just finding all the primes under 1000 (there are 168 of them) and counting the frequency of the last digits.

• 1: 40
• 2: 1
• 3: 42
• 5: 1
• 7: 46
• 9: 38

So, the answer is indeed 7. But is there a way to find that without actually listing all the primes?

Obviously, 2 and 5 are unique special cases, and aren't even answer choices, so we can ignore them. So, let's look at the base-ten multiplication table for how many ways there are to obtain composite numbers ending in those digits.

• 1: 4 ways (1×1, 3×7, 7×3, 9×9)
• 3: 4 ways (1×3, 3×1, 7×9, 9×7)
• 7: 4 ways (1×7, 3×9, 7×1, 9×3)
• 9: 4 ways (1×9, 3×3, 7×7, 9×1)

So, a four-way tie. Not helpful. I guess that's the reason for the form of Dirichlet's theorem saying “Equivalently, the primes are evenly distributed (asymptotically) among the congruence classes modulo d [i.e., the last digit in base d] containing a's coprime to d.”

But that's a long-run trend. Obviously it doesn't apply to the finite set of primes under 1000, because there is a clear winner of 7. Let's try other finite sets of primes, similar to @Roland's answer.

Domain	1	3	7	9	Mode	Disuniformity
$p < 10$	0	1	1	0	tie: 3, 7	$0.25$
$p < 100$	5	7	6	5	3	$0.00519849$
$p < 1000$	40	42	46	38	7	$0.00127014$
$p < 10^4$	306	310	308	303	3	$1.77678 \times 10^{-5}$
$p < 10^5$	2387	2402	2411	2390	7	$4.01226 \times 10^{-6}$
$p < 10^6$	19617	19665	19621	19593	3	$4.38196 \times 10^{-7}$
$p < 10^7$	166104	166230	166211	166032	3	$5.89336 \times 10^{-8}$
$p < 10^8$	1440298	1440474	1440495	1440186	7	$1.96717 \times 10^{-9}$
$p < 10^9$	12711386	12712499	12712314	12711333	3	$4.3115 \times 10^{-10}$
$p < 10^{10}$	113761519	113765625	113764039	113761326	3	$6.23018 \times 10^{-11}$
$p < 10^{11}$	1029517130	1029509448	1029518337	1029509896	7	$3.88107 \times 10^{-12}$
$p < 10^{12}$	9401960980	9401979904	9401997000	9401974132	7	$4.73196 \times 10^{-13}$
$p < 10^{13}$	86516370000	86516427946	86516367790	86516371101	3	$2.13443 \times 10^{-14}$

In this chart, “disuniformity” is defined as the sum of squares of the difference between the relative frequency of each digit and the expected relative frequency (i.e. 1/4). Note that the distribution becomes consistently more uniform as the upper bound for p increases.

AFAIK, the “winner” is always either 3 or 7, not 1 or 9. Maybe there's a counterexample for a higher order of magnitude, but I'd need a much faster prime-listing program to check. The first few rows have an interesting alternating 3-7 pattern, but it breaks down for $10^7$.

Well, I have to admit that I just don't know how to answer the question without brute-force computation or sheer luck. Maybe that's a lack of education, since I only have an M.S. in mathematics instead of a Ph.D. But then, @GregMartin has co-written an academic paper on the exact subject of “prime number races” (linked in his answer), and can only conclude that “3 or 7 is more likely to be correct”, so I don't know how an “average” person could reasonably be expected to know the answer.