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Posers and Puzzles

Posers and Puzzles

  1. 12 Apr '05 03:15
    This problem is intended to be solved without the use of a calculator (ie, it's only a good challenge if you don't use a calculator):

    Find a 2-digit number and a 3-digit number such that when you place the 2-digit number to the left of the 3-digit number, the 5-digit number that is created is exactly 9 times the product of the original 2-digit and 3-digit numbers.

    As far as I can tell, the answer is unique.
  2. 12 Apr '05 05:11
    Let m = 10a + b
    Let n = 100z + 10y + z

    Then 9mn = 1000m + n

    or

    m = n / (9n - 1000) >= 10

    So n = 112, m = 14 is the only solution
    and the number is 14112
  3. 12 Apr '05 06:54
    Originally posted by THUDandBLUNDER
    Let m = 10a + b
    Let n = 100z + 10y + z

    Then 9mn = 1000m + n

    or

    m = n / (9n - 1000) >= 10

    So n = 112, m = 14 is the only solution
    and the number is [b]14112

    [/b]
    yes...14112 is correct.

    Out of curiosity: how did you find the answer (m,n) = (14,112) and also determine it is unique? did you substitute in and try to reason out the values of a,b,x,y,z (one of your z's should be an x or something) directly?

  4. 12 Apr '05 09:04 / 2 edits
    Originally posted by davegage
    yes...14112 is correct.

    Out of curiosity: how did you find the answer (m,n) = (14,112) and also determine it is unique? did you substitute in and try to reason out the values of a,b,x,y,z (one of your z's should be an x or something) directly?

    m = n / (9n - 1000) >= 10
    gives
    89n =< 10000
    n =< 112

    and

    9n - 1000 > 0
    gives
    n > 111
  5. 12 Apr '05 16:13
    Originally posted by THUDandBLUNDER
    m = n / (9n - 1000) >= 10
    gives
    89n =< 10000
    n =< 112

    and

    9n - 1000 > 0
    gives
    n > 111
    nice...