# The Prime Directive

I don’t intend for this blog to turn into commentary on YouTube videos, but here’s another one.

If you like math puzzles, the Mind Your Decisions channel from Presh Talwalkar is worth looking into. Unlike many math YouTubers, Presh doesn’t stretch his videos out by going through each step in painful detail. He presents the puzzle, works through a solution that respects your intelligence, and finishes up in about five minutes.

In this recent video, the puzzle concerns prime numbers.

The question is:

There are 168 prime numbers less than 1000. What is their sum?

You’re not expected to work out the sum. It’s a multiple choice problem with four possible answers:

1. 11,569
2. 76,127
3. 57,298
4. 81,744

I started the solution the same way Presh did, by noticing that two of the answers were even and two were odd. Since every prime number other than 2 is odd, that means the answer is the sum of 167 odd numbers plus 2. Since the sum of an odd number of odd numbers is an odd number (yes, I enjoyed writing that), the sum we’re looking for must be odd. That eliminates (c) and (d).

At this point, Presh and I took different paths, and when I watched his solution, I felt he had cheated a bit. His reasoning was based on a simple truth: the sum of the first 168 primes must be greater than the sum of the first 168 positive integers, i.e.,

$∑n=1168pn>∑n=1168n=168(168+1)2=14,196\sum_{n=1}^{168} p_n > \sum_{n=1}^{168} n = \frac{168 \,(168+1)}{2} = 14,196$

where I’m using ${p}_{n}$ to represent the nth prime. This eliminates 11,569 as a possible answer, leaving 76,127.

I don’t object to Presh using Gauss’s trick to simplify the sum of the first 168 integers, but I thought an unwritten rule of this problem was that it was to be done entirely in one’s head. I know there are people who can multiply 168 by 169 more or less instantly, but most of us can’t. So my first instinct was that this was cheating.

Later, I realized the same logic could be used by slipping in a simpler calculation:

$∑n=1168pn>∑n=1168n=168·1692>160·1602=25,6002=12,800\sum_{n=1}^{168} p_n > \sum_{n=1}^{168} n = \frac{168 \cdot 169}{2} > \frac{160 \cdot 160}{2} = \frac{25,600}{2} = 12,800$

We can all do this in our heads because working with computers has gotten us familiar with powers of 2. This also eliminates answer (a).

Anyway, that’s not how I finished the problem. I figured the primes less than 1000 would be nearly uniformly distributed (more on this later) and therefore the average would be about 500. Since

$168·500=168,0002=84,000168 \cdot 500 = \frac{168,000}{2} = 84,000$

the answer should be close to this. Also, since primes tend to get sparser as they get larger, I would expect the sum to be less than 84,000. That made me feel good about choosing 76,127.

I should have also realized that if 11,569 were the answer, the average of the primes under 1000 would have to be less than 100, which is ridiculous. But I didn’t think of that argument until I started writing this post.

After making my guess and watching Presh’s solution, I opened Mathematica and started playing around. Mathematica has a function, Prime[n], that returns the nth prime number. So

Table[Prime[n], {n, 1, 168}]


gave me the list of the first 168 primes,

  2      3      5      7     11     13     17     19
23     29     31     37     41     43     47     53
59     61     67     71     73     79     83     89
97    101    103    107    109    113    127    131
137    139    149    151    157    163    167    173
179    181    191    193    197    199    211    223
227    229    233    239    241    251    257    263
269    271    277    281    283    293    307    311
313    317    331    337    347    349    353    359
367    373    379    383    389    397    401    409
419    421    431    433    439    443    449    457
461    463    467    479    487    491    499    503
509    521    523    541    547    557    563    569
571    577    587    593    599    601    607    613
617    619    631    641    643    647    653    659
661    673    677    683    691    701    709    719
727    733    739    743    751    757    761    769
773    787    797    809    811    821    823    827
829    839    853    857    859    863    877    881
883    887    907    911    919    929    937    941
947    953    967    971    977    983    991    997


and

Sum[Prime[n], {n, 1, 168}]


gave me the answer of 76,127.

As for my assumption that the primes were nearly uniformly distributed, I checked their histogram,

Histogram[Table[Prime[n], {n, 1, 168}], 10]


which gave me this:

Pretty uniform, except for those under 200. Close enough to uniform for the kind of rough calculation I was doing.