Sum of three consecutive prime numbers is $173$

As quickly as I can add up the first few primes in trios to look them up in the OEIS to get the appropriate result. I searched for 10, 15, 23, 31, 41, which gave me http://oeis.org/A034961 sequence entry for sums of three consecutive primes.

Then I clicked on the "list" link and I spotted 173 to the right of 16. This means that the 16th, 17th, 18th primes add up to 173. The 16th prime is 53, so then I verify that 53 + 59 + 61 = 173, which means that the biggest of these numbers is 61.

If there was no OEIS, this could still be done fairly quickly, since 173 is really quite a small number. 173 divided by 3 is approximately 57.7, and since, again, 173 is a small number, the three primes we're looking for should be fairly close to 57.7.

The largest prime below 57.7 is 53, and the smallest prime above is 59, so we can start by trying 47 + 53 + 59 and 53 + 59 + 61; the former falls short, the latter is the answer. If you've got the primes from 2 to 97 and have a calculator, you can probably get the answer by this method in a couple of minutes, if that.

At least one of the primes is at most $173/3 < 58 $, and so you only have to test until $53$ to find the smallest prime in that sequence.

It so happens that $53$, $59$, $61$ is the solution.

It depends how much you know about the distribution of small primes. If $a,b$ are consecutive primes with $a<b\leq 89$ then $b-a\leq 6.$

If $p,q,r$ are consecutive primes with $p<q<r$ and $p+q+r=173$ then $q\leq 83.$ Otherwise $q\geq 89,$ but then $p+q+r>q+r>2 q\geq 178.$ So $r\leq 89.$

Therefore $q\leq p+6$ and $r\leq q+6\leq p+12.$ $$\text {Hence }\quad 173=p+q+r\leq p+(p+6)+(p+12)=3 p+18,$$ $$\text {giving }\quad p\geq 155/3.$$ $$\text {Also } \quad 173=r+q+p\geq r+(r-6)+(r-12)=3 r-18,$$ $$\text {giving }\quad r\leq 191/3.$$ There are exactly $3$ primes in the interval $[155/3,191/3],$ and they are $53,59,61.$

This is not how I solved it in my head. I just "looked around the neighborhood" of $173/3$.