Question 1205180
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The numbers 2,3,5, and 7 are prime since the only factors are 1 and themselves.


Make a 4x4 table listing the values 2,3,5,7 along the left and top like so
<table border=1 cellpadding=5><tbody><tr><td></td><td>2</td><td>3</td><td>5</td><td>7</td></tr><tr><td>2</td><td></td><td></td><td></td><td></td></tr><tr><td>3</td><td></td><td></td><td></td><td></td></tr><tr><td>5</td><td></td><td></td><td></td><td></td></tr><tr><td>7</td><td></td><td></td><td></td><td></td></tr></tbody></table>
Cross out the northwest main diagonal. This is because we cannot re-use the same digit twice.
<table border=1 cellpadding=5><tbody><tr><td></td><td>2</td><td>3</td><td>5</td><td>7</td></tr><tr><td>2</td><td>X</td><td></td><td></td><td></td></tr><tr><td>3</td><td></td><td>X</td><td></td><td></td></tr><tr><td>5</td><td></td><td></td><td>X</td><td></td></tr><tr><td>7</td><td></td><td></td><td></td><td>X</td></tr></tbody></table>
The remaining cells are filled in by concatenating the headers. I'll have the left header go first and then the top next.
<table border=1 cellpadding=5><tbody><tr><td></td><td>2</td><td>3</td><td>5</td><td>7</td></tr><tr><td>2</td><td>X</td><td>23</td><td>25</td><td>27</td></tr><tr><td>3</td><td>32</td><td>X</td><td>35</td><td>37</td></tr><tr><td>5</td><td>52</td><td>53</td><td>X</td><td>57</td></tr><tr><td>7</td><td>72</td><td>73</td><td>75</td><td>X</td></tr></tbody></table>
The "2" column and "5" column can be crossed out because of the divisibility by 2 and divisibility by 5 rules.
27 is composite since 27 = 3*9
57 is composite since 57 = 3*19
Or you can use the divisibility by 3 rule to check 27 and 57 are multiples of 3.


Of that table, the primes are: 23, 37, 53, 73
Refer to a list of primes. Or you can check each one by one. 


I'll let you handle the possible 3 digit primes that can be formed with 2,3,5,7.
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