3
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$Given$

$Devil$ goes by $666$

Express him as

1). Sum of all Primes in 3 Different ways using all the digits 0 to 9.

Use only + sign.

2). Sum of all Squares of Primes. Use only + and ^ signs.

3). Sum of Squares of Prime Products

   $ (P1*P2)^2 + (P1*P3)^2$


    where P1,P2,P3 are primes.
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  • $\begingroup$ Edit made..to include all digits for 1) $\endgroup$ – Uvc May 24 at 4:31
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  1. $2 + 29 + 37 + 53 + 61+ 83 + 401 = 666$
    $2 + 11 + 53 + 61 + 83 + 149 + 307 = 666$
    $3 + 5 + 17 + 23 + 59 + 67 + 83 + 409 = 666$
    Strategy: Start with a 3 digit prime with different digits that contains a 0 since it's going to be hard to use a zero after that. Picked 401 then 307 then 409. Subtract that from 666 and then try to find primes with the remaining digits and keep subtracting. For even digits I tried to find a 2 digit prime that starts with that digit. Not worring much about the 2 because that's a prime and can be added at the end. after a few tries found some combinations.

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  1. $17^2+13^2+11^2+7^2+5^2+3^2+2^2=666$
    strategy: Basically the same thing that phenomist did. Start with the largets possible prime that squares to something below 666 subtract from 666 and repeat the process.

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  1. $(3 \times 7)^2 + (3 \times 5)^2 = 666$
    Strategy. $25^2 < 666 < 26^2$.
    so start down from 25 and look for the numbers composed of 2 prime factors (25, 22, 21, 15, 14, 10, 9, 6, 4).
    take them 1 by one, square them and subtract the result from 666. See if what's left matches the criteria.
    25: $666-25^2 = 41$. Not a square
    22: $666 - 22^2 = 182$. Not s square
    21: $666 - 21^2 = 225 = 15^2 = (3 \times 5)^2$. There you go.
    15: $666 - 15^2 = 441 = 21^2 = (3 \times 7)^2$ It's the reverse of 21. We get the same valid result.
    14: $666 - 14^2 = 470$. Not a square
    10: $666 - 10^2 = 566$. Not a square
    9: $666 - 9^2 = 585$. Not a square
    6: $666 - 6^2 = 630$. Not a square
    4: $666 - 4^2 = 650$. Not a square.

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  • $\begingroup$ Well explained strategy towards final solution $\endgroup$ – Uvc May 24 at 15:22
5
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1.

$89+97+101+103+107+109 = 606$, so let's just give three ways to sum to $60$. $7+53$, $13+47$, and $17+43$. Great, we're done.

Revised 1.

Cover up the digits, and hope that Goldbach works out in the end. For example, $2+23+41+53+67+89+101+103+127=606$. Then proceed as before.

2.

Of course, we can add $3^2$ 74 times, but where's the fun in that? Can we do it using distinct squares? We can do a sort of a recursive search: $666 - 23^2 = 137$, $137 - 11^2 = 26$, $26 - 5^2 = 1$, bad. But $2^2+3^2 = 13 < 26$, so this is entirely bad. But $7^2+5^2+3^2+2^2 = 87 < 137$, so we can conclude that $23^2$ is not part of the sum. $666 - 19^2 = 305$. $305 - 17^2 = 16$, not coverable by $2^2 + 3^2$. $305 - 13^2 = 136$, $136 - 11^2 = 15$, not coverable by $2^2+3^2$. $11^2+7^2+5^2+3^2+2^2 = 208 < 305$ so this branch dies out. $666-17^2 = 377$, $377-13^2=208$, oh interesting. Turns out $17^2+13^2+11^2+7^2+5^2+3^2+2^2 = 666$ (the seven smallest prime squares).

3.

$666 = 9 \cdot 74$. We can use the sum of two squares theorem and the Brahmagupta–Fibonacci identity to help us. Namely, $74 = 2 \cdot 37 = (1^2 + 1^2)(1^2+6^2) = (1-6)^2+(1+6)^2 = 5^2+7^2$. So $666 = (3\cdot5)^2+(3\cdot7)^2$.

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