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Number Puzzle: Who could I be?

August 31, 2008 pm31 7:44 pm
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We have six statements about a number, and we know that exactly 1 is false.

  1. I am greater than 50
  2. I am a multiple of 7
  3. I am a perfect square
  4. I am a 3-digit number
  5. I am less than 500
  6. I am a multiple of 17

Are there any numbers that fit? How many? And if they exist, what are they?

Unrelated, interesting note: Nice factoring techniques for solving problems such as

Find n such that n(n+16) is a perfect square

are presented at the Ultimate Quant Marathon Blog for IIM Cat (whatever that means), a brand new blog. I think it’s called Quantologic for short.

from → Math, mathematics, Puzzles
3 Comments leave one →
  1. Sonic Charmer permalink
    August 31, 2008 pm31 8:06 pm 8:06 pm

    I’m assuming ‘number’ = positive. First, which exactly-one is false?

    #1 can’t be false if #4 is true.

    Suppose #2 is false. X is between 100 and 499, a perfect square, and a multiple of 17 but not 7: 17^2 = 289.

    Suppose #3 is false. X is a multiple of 17 and 7 between 101 and 499, not a perfect square. 17*7*i=119,238,357,476.

    Suppose #4 is false. X is between 50 and 99 and a perfect-square multiple of 17 and 7; no such X.

    Suppose #5 is false. X is between 500 and 999 and a perfect-square multiple of 17 and 7, i.e. (17*7*i)^2 for some i. No such X.

    Suppose #6 is false. X is between 100 and 499, a perfect-square multiple of 7 but not 17: X=(7i)^2 = 196, 441.

    Reply
  2. TwoPi permalink
    September 1, 2008 am30 2:33 am 2:33 am

    I had a similar solution, but narrowed down to three cases at the start:

    A number which is a perfect square, and also a multiple of both primes 7 and 17 would have to be a multiple of 7^2×17^2=14161. But clearly no such number can satisfy either condition 4 or condition 5.

    Thus we know that one of 2, 3, or 6 is false, and that 1, 4, and 5 must be true.
    (etc… as in Sonic Charmer’s attack on not #2, not #3, and not #6.)

    Reply
  3. dr rick permalink
    September 2, 2008 pm30 8:47 pm 8:47 pm

    Without reading other solutions:

    7, 17 and “perfect square” are incompatible, since the smallest such is (7*17)^2=14161, which is neither three-digit nor less then 500. So one of these three is the falsehood.

    If “7” is false, the number is a square, multiple of 17, less than 500: it can only be 289.

    If “17” is false, the number is a square multiple of 49 between 100 (three-digit) and 500: 49×4 = 196, 49×9 = 441. 2 possibilities.

    If “square” is false, the number is a multiple of 119 between 100 and 500: 119, 238, 357, 476. 4 possibilities.

    In all that makes seven.

    Reply

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