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A Counting Puzzle

Reading & CountingA Counting Puzzle - 13 April

If you wrote all of the numbers from 300 to 400 on a piece of paper, how many times would you have written the number 3?

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15 comments:

  1. Each of the numbers from 300 to 399 inclusive has a 3 in the "hundreds" position. That makes 100.

    Each number from 330 to 339 has a 3 in the tens position, That makes 10 more.

    Then there are 10 numbers (303, 313, 323, .... 393) with a 3 in the units position. That's another 10.

    So I make it 120 altogether.

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    Replies
    1. but buddy first you counted 333 in second phase i.e between 330 and 339 ...then again counted 333 in third..i.e 303,313,323,'333'and so on.....so correct ans is 119 as 333 is counted 2 times

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    2. Varun but when we are counting the number at tens place then we are not counting the least significant bit in '333'.So you are counting one less.So total would be 120.

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  2. This comment has been removed by the author.

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  3. 119 as total words=10 x 10 =100;(principle of counting)
    words having 3 only at hundreds place but not in ones and tens place =9 x 9 =81;
    total 3 at hundreds place=100;
    total 3 at tens or hundreds place =100-81;
    so total 3 =100+19=119

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  4. only one time (300-400)

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  5. 120,
    taking the hundred digit you have 100 3's ([3]00-[3]99)
    taking the ten digit you have 10 3's
    (3[3]0, 3[3]1, 3[3]2, 3[3]3, 3[34], 3[3]5, 3[3]6, 3[3]7, 3[3]8, 3[3]9)
    then taking the last digit you have 10 (30[3], 31[3], 32[3], 33[3], 34[3], 35[3], 36[3], 37[3], 38[3], 39[3])

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  6. The subrange is 300..400, which contains 101 numbers.
    The last one does not have any '3' digits; the rest have at least one.
    For the 3xx pattern, there are 100 of them.
    For the x3x pattern, there are 10 of them.
    For the xx3 pattern, there are 10 of them.
    Thus, '3' is written a total of 120 times.
    The distribution is: 1 number has no '3' digit; 81 numbers have exactly one '3' digit; 18 numbers have two '3' digits; and 1 number has three '3' digits.

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