Exponent Products

Directions: Using the digits 1 to 9 at most one time each, place a digit in each box to make a correct equation.

Hint

How might three powers with different bases satisfy this equation?

Answer

Some of the possible answers: 2^7 x 4^1 = 8^3, 2^9 x 4^3 = 8^5, 1^7 x 3^4 = 9^2, 1^4 x 2^9 = 8^3. Clearly other powers of 1 work too.

Source: Stephen Cox

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Greatest Common Factor

Directions: Using the digits 0 to 9 at most one time each, place a digit …

4 comments

  1. Rudolf Österreicher

    60 solutions (120 if you swap factors). 48 of them contain powers of one. Here are the other 12:
    2^1 * 4^7 = 8^5
    2^1 * 8^3 = 4^5
    2^3 * 4^6 = 8^5
    2^3 * 4^9 = 8^7
    2^3 * 8^5 = 4^9
    2^5 * 8^3 = 4^7
    2^7 * 4^1 = 8^3
    2^7 * 8^1 = 4^5
    2^9 * 4^3 = 8^5
    2^9 * 4^6 = 8^7
    2^9 * 8^1 = 4^6
    4^3 * 8^1 = 2^9

    • Rudolf Österreicher

      And here the 48 solutions with powers of 1:
      1^2 * 3^8 = 9^4
      1^2 * 4^9 = 8^6
      1^2 * 8^6 = 4^9
      1^2 * 9^4 = 3^8
      1^3 * 4^9 = 8^6
      1^3 * 8^6 = 4^9
      1^4 * 2^9 = 8^3
      1^4 * 8^3 = 2^9
      1^5 * 2^6 = 4^3
      1^5 * 2^9 = 8^3
      1^5 * 3^4 = 9^2
      1^5 * 3^8 = 9^4
      1^5 * 4^3 = 2^6
      1^5 * 4^3 = 8^2
      1^5 * 4^9 = 8^6
      1^5 * 8^2 = 4^3
      1^5 * 8^3 = 2^9
      1^5 * 8^6 = 4^9
      1^5 * 9^2 = 3^4
      1^5 * 9^4 = 3^8

    • Rudolf Österreicher

      1^6 * 2^9 = 8^3
      1^6 * 3^4 = 9^2
      1^6 * 3^8 = 9^4
      1^6 * 4^3 = 8^2
      1^6 * 8^2 = 4^3
      1^6 * 8^3 = 2^9
      1^6 * 9^2 = 3^4
      1^6 * 9^4 = 3^8
      1^7 * 2^6 = 4^3
      1^7 * 2^9 = 8^3
      1^7 * 3^4 = 9^2
      1^7 * 3^8 = 9^4
      1^7 * 4^3 = 2^6
      1^7 * 4^3 = 8^2
      1^7 * 4^9 = 8^6
      1^7 * 8^2 = 4^3
      1^7 * 8^3 = 2^9

    • Rudolf Österreicher

      1^7 * 8^6 = 4^9
      1^7 * 9^2 = 3^4
      1^7 * 9^4 = 3^8
      1^8 * 2^6 = 4^3
      1^8 * 3^4 = 9^2
      1^8 * 4^3 = 2^6
      1^8 * 9^2 = 3^4
      1^9 * 2^6 = 4^3
      1^9 * 4^3 = 2^6
      1^9 * 4^3 = 8^2
      1^9 * 8^2 = 4^3

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