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Determine all pairs (m, n) of a two-digit natural number m and a single-digit natural number n satisfying the following conditions.
1) If the number n is given as the number between the two digits of the number m, then a three-digit number is obtained which is 11 times the number m.
2) If the number n is used as a number before the decimal number of the number m, then a three-digit number is obtained that is 21 times as large as the number n.

  • Math -

    If I understand this correctly, then if the digits of m are a and b, we have

    m = 10a+b
    and so
    100a+10n+b = 11(10a+b)
    100n+10a+b = 21n

    But this comes out with negative solutions. Maybe you can fix my logic and take it where it's supposed to go.

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