Lazymaths Blog

Last week we asked our fans this question – Can u make the nos 123456789 = 100 by only using + – / x and keeping the nos in the same order e.g.: 123+45-67+8-9=100

Here’s the solution to the puzzle. In fact there are 100 solutions!! Take your pick…

         1     123+45-67+8-9
         2     123+4-5+67-89
         3     123+4*5-6*7+8-9
         4     123-45-67+89
         5     123-4-5-6-7+8-9
         6     12+34+5*6+7+8+9
         7     12+34-5+6*7+8+9
         8     12+34-5-6+7*8+9
         9     12+34-5-6-7+8*9
        10     12+3+4+5-6-7+89
        11     12+3+4-56/7+89
        12     12+3-4+5+67+8+9
        13     12+3*45+6*7-89
        14     12+3*4+5+6+7*8+9
        15     12+3*4+5+6-7+8*9
        16     12+3*4-5-6+78+9
        17     12-3+4*5+6+7*8+9
        18     12-3+4*5+6-7+8*9
        19     12-3-4+5-6+7+89
        20     12-3-4+5*6+7*8+9
        21     12-3-4+5*6-7+8*9
        22     12*3-4+5-6+78-9
        23     12*3-4-5-6+7+8*9
        24     12*3-4*5+67+8+9
        25     12/3+4*5-6-7+89
        26     12/3+4*5*6-7-8-9
        27     12/3+4*5*6*7/8-9
        28     12/3/4+5*6+78-9
        29     1+234-56-7-8*9
        30     1+234*5*6/78+9
        31     1+234*5/6-7-89
        32     1+23-4+56+7+8+9
        33     1+23-4+56/7+8*9
        34     1+23-4+5+6+78-9
        35     1+23-4-5+6+7+8*9
        36     1+23*4+56/7+8-9
        37     1+23*4+5-6+7-8+9
        38     1+23*4-5+6+7+8-9
        39     1+2+34-5+67-8+9
        40     1+2+34*5+6-7-8*9
        41     1+2+3+4+5+6+7+8*9
        42     1+2+3-45+67+8*9
        43     1+2+3-4+5+6+78+9
        44     1+2+3-4*5+6*7+8*9
        45     1+2+3*4-5-6+7+89
        46     1+2+3*4*56/7-8+9
        47     1+2+3*4*5/6+78+9
        48     1+2-3*4+5*6+7+8*9
        49     1+2-3*4-5+6*7+8*9
        50     1+2*34-56+78+9
        51     1+2*3+4+5+67+8+9
        52     1+2*3+4*5-6+7+8*9
        53     1+2*3-4+56/7+89
        54     1+2*3-4-5+6+7+89
        55     1+2*3*4*5/6+7+8*9
        56     1-23+4*5+6+7+89
        57     1-23-4+5*6+7+89
        58     1-23-4-5+6*7+89
        59     1-2+3+45+6+7*8-9
        60     1-2+3*4+5+67+8+9
        61     1-2+3*4*5+6*7+8-9
        62     1-2+3*4*5-6+7*8-9
        63     1-2-34+56+7+8*9
        64     1-2-3+45+6*7+8+9
        65     1-2-3+45-6+7*8+9
        66     1-2-3+45-6-7+8*9
        67     1-2-3+4*56/7+8*9
        68     1-2-3+4*5+67+8+9
        69     1-2*3+4*5+6+7+8*9
        70     1-2*3-4+5*6+7+8*9
        71     1-2*3-4-5+6*7+8*9
        72     1*234+5-67-8*9
        73     1*23+4+56/7*8+9
        74     1*23+4+5+67-8+9
        75     1*23-4+5-6-7+89
        76     1*23-4-56/7+89
        77     1*23*4-56/7/8+9
        78     1*2+34+56+7-8+9
        79     1*2+34+5+6*7+8+9
        80     1*2+34+5-6+7*8+9
        81     1*2+34+5-6-7+8*9
        82     1*2+34-56/7+8*9
        83     1*2+3+45+67-8-9
        84     1*2+3+4*5+6+78-9
        85     1*2+3-4+5*6+78-9
        86     1*2+3*4+5-6+78+9
        87     1*2-3+4+56/7+89
        88     1*2-3+4-5+6+7+89
        89     1*2-3+4*5-6+78+9
        90     1*2*34+56-7-8-9
        91     1*2*3+4+5+6+7+8*9
        92     1*2*3-45+67+8*9
        93     1*2*3-4+5+6+78+9
        94     1*2*3-4*5+6*7+8*9
        95     1*2*3*4+5+6+7*8+9
        96     1*2*3*4+5+6-7+8*9
        97     1*2*3*4-5-6+78+9
        98     1*2/3+4*5/6+7+89
        99     1/2*34-5+6-7+89
       100     1/2*3/4*56+7+8*9

Yesterday, we asked “How many whole numbers less than 100 have 7 in them?”

The correct answer is 19.

Here’s the explanation – There are 10 numbers of the form 7, 17, 27… etc. and 10 numbers of the form 70, 71, 72, …So the total would be 20 numbers, but because you double counted 77 (which belongs to both the lists), you reduce the total by 1 (to make 77 count only once). Hence the final answer is 19.

Yesterday we asked the question – How many three digit numbers are divisible by 17?

Many of you gave the correct answer of 53. For those who are interested in the solution, here it is -

Well, the first number greater than 100 and divisible by 17 is  17×6 = 102. (since 100/17=5.88235. . .)  We don’t need to list these three digit numbers. The last number less than 1000 and divisible by 17 is 58 x 17  = 986 (since 1000/17=58.8235. . .) So all the numbers from 6 (times 17) through 58 (times 17) are all the three digit numbers divisible by 17.  6 through 58 (both inclusive) = 53. So there are 53 such numbers!!