# Devil in Different Disguises but in Prime Shape…Can you Figure out his Diverse Expressions?

$$Given$$

$$Devil$$ 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.

• Edit made..to include all digits for 1) – Uvc May 24 at 4:31

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.

.

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.

.

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.

• Well explained strategy towards final solution – Uvc May 24 at 15:22

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 $$666 = (3\cdot5)^2+(3\cdot7)^2$$.