mathematics - Prime to Prime: Get all first 25 Prime Numbers using up to 4 Primes

The first 25 Prime Numbers are

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

Using up to 4 prime numbers and the following mathematical operations, get all the 25 primes.

+ - x / ^ √ !

No other operators (like !!) are allowed.

Other rules

• You cannot use same Prime Number more than once.


• You can use only Prime Numbers.


• Any number that appears as a number in the equation will be counted as one of the primes out of four (e.g. 7^2 means you have used both 7 and 2).


• You do not have to use all the 4 primes in every equation.


• You must use the SAME 4 primes in every equation. If you select say 2, 13, 17, 23 then they are the only primes that to appear in every equation to get the 25 primes.

I have 1 solution. There may be more.

No programming please.
NO PARTIAL ANSWERS.

Answer

Using 2,3,7,11:

$2 = 2$

$3 = 3$

$5 = 11 + 3 - 7 -2$

$7 = 7$

$11 = 11$

$13 = 2 + 11$

$17 = 3! + 11$

$19 = 2^3 + 11$

$23 = 3 \cdot 7 + 2$

$29 = \frac{(7-2)!}{3} - 11$

$31 = 3 \cdot 11- 2$

$37 = (11-3!) \cdot 7 + 2$

$41 = 7^2 +3 - 11$

$43 = 2 \cdot 11 + 3 \cdot 7$

$47 = 3 \cdot 11 + 2 \cdot 7$

$53 = 2^ {3!} - 11$

$59 = 3! \cdot 11 - 7$

$61 = (11-2) \cdot 3! + 7$

$67 = 7 \cdot 2^3 + 11$

$71 = 2^{3!} + 7$

$73 = 3! \cdot 11 + 7$

$79 = 7 \cdot 11 + 2$

$83 = 7 \cdot 11 + 3!$

$89 = 7 \cdot 11 + 2 \cdot 3!$

$97 = (2+11) \cdot 7 + 3!$