Hardest question in the world

9514 1404 393

  3

Step-by-step explanation:

The center square must be the middle digit of two 3-digit squares. Those squares must begin with an odd number (the units digit of a 2-digit prime).

The possible 3-digit squares that have a middle digit shared with another 3-digit square are ...

  121, 324, 529, 729  and  169, 361, 961

The squares cannot end in 9, because there would be no 2-digit numbers satisfying 4 Down and 5 Across in those cases. The squares sharing middle digit 6 don't work because the only square starting with 1 is even (16), meaning there is no matching prime starting with 1.

After these eliminations, we're down to ...

So, the choices for square 1 are the leading digits of 2-digit primes such that x1 and x3 are both primes. This is the case for x=1, 4, and 7.

That is, there are exactly 3 solutions. The above list, together with ...

answer: i wpould go with c

step-by-step explanation:

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