Really prime

$\begin{aligned}p+q+r+s+t+u&=253173\hspace{4mm}\color{#3D99F6}\small (Given)\\ p+q+r+s+t+u\pmod{2}&\equiv1\pmod{2}\\\\ \text{If all of p,q,r,s,t,u were odd, then}\\\\ p+q+r+s+t+u\pmod{2}&\equiv1+1+1+1+1+1\pmod {2}\\ &\equiv0\pmod {2}\\ \implies \text{1,3, or 5 among p,q,r,s,t,u are }&\text{even }\\ \text{Since p,q,r,s,t,u are primes and 2 is the } &\text{only even prime,exactly one of them is equal to } 2. \\ \text{Since p<q<r<s<t<u and since 2 is the } &\text{smallest prime,}\\ \implies p&=2\\ p^p&=2^2=\color{#EC7300}\boxed{\color{#333333}4}\end{aligned}$

using Mathematica
Firstly I found all the possible sums of the first 10 primes
Union[Total /@ Union[Sort /@ Permutations[Prime[Range@10], {5}]]]

which gives {28, 30, 34, 36, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68,69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85,86, 87, 89, 90, 91, 93, 95, 99, 101}

Then I found the first 10 differences of 253173 and the 10 largest possible primes
Table[253173 - Prime[PrimePi@253173 - i], {i, 10}]

which gives {16, 20, 40, 64, 70, 92, 110, 124, 160, 170}

As we can see the key number is 40 and we need to use the third greatest prime (which is 253133) plus 5 Primes whose Sum is 40 (like {2,3,5,7,23},{2,3,5,11,19},{2,3,5,13,17},{2,3,7,11,17})

So, a possible solution would be 2+3+5+7+23+253133=253173
and the prime we are searching for is **2

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