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December Maths Puzzle

December Maths Puzzle - 5 December

I am thinking of a 6-digit number. The sum of the digits is 43.

And only two of the following three statements about the number are true:

(1) it's a square number,
(2) it's a cube number, and
(3) the number is under 500000.

For Solution : Click Here


  1. The answer is 499849.And its sum is 43.

    Statement 1 & 3 are true as the number is square of 707 & it is under 5000000.

  2. Fayyaz : how you got the answer

  3. the answer is 499993 and 499993<500000

  4. ridiculous how are you so sure that 1 and 3 are true and not 1 and 2 show that there is no number like that.

  5. 499849.

    You just have to try. Thought You can reduce the amount of possibilities easily.

    I deduced that 3 is true because it wastly reduces the possible solutions as this is only 4/5 difficulty.

    1. Max sum of numbers is 4+9+9+9+9+9, which equals 49, from this You can deduce that the number is fairly close to the 499999.

    2. It may be a square number. So the square is below square of max sqrt(499999)~707.10607. By trying squares below 707.1.... You will find the right answer that is 707^2.

    Yes, I had some luck but got it right the first time I tried.

  6. its only 499849.... i can't understand... why u choose the number 707.....?

  7. 499849 is the answer

  8. option 3 will be correct because we can also write this number as 449989 which is also a 6- digit number.But this number is not a square or cube .

  9. 499849. Nice que

  10. The easiest way is to use a spreadsheet and a little bit of brute force.

    First, assume that it is a square number. (If you can't find an answer then that means the number is not square and you use a similar process with cube numbers).

    Start by working out the possible squares. Work out sqrt(100000) = 316-and-a-bit, and work out sqrt (999999) = 999-and-a-bit.
    So the square roots (r) are 317 ≤ r ≤ 999. In column A of your spreadsheet, put the numbers 317 to 999.
    Then, across the top:
    B1 = A1^2
    C1 = B1^(1/3)
    D1 = LEFT(B1,1)+RIGHT(LEFT(B1,2),1)+RIGHT(LEFT(B1,3),1)+RIGHT(LEFT(B1,4),1)+RIGHT(LEFT(B1,5),1)+RIGHT(B1,1)

    Copy this top row down to the bottom of your list in column A.
    B gives you the 6-digit numbers you are looking for.
    C gives you their cube roots (so you are looking for whole numbers here)
    D gives you the sum of the digits (so you are looking for 43 here)

    Do a conditional format on column D to look for any where the sum of the digits is 43. There are 5 matches, the first of which is 499849, which therefore matches the criteria (a) and (c).

    If you know that there is only one answer, you can stop there. Or you can show there are no other answers by looking through the other four matches, all of which are over 500000, and noting that the cube root goes decimal in each case so none of them are cube numbers, therefore there is no number satisfying both (a) and (b). You can then use the same principle to generate a list of cube numbers, and see that there is no number satisfying both (b) and (c).