so far I've created a spreadsheet witht he following columns, n, n squared, 3/4 n squared. the final column should be the answer I think but it needs to be a cubed number. I don't know any way to check this other than checking visually between a list of cubed numbers and the final coumn of my spreadsheet

Suppose the wanted number is x. Then 3x must be a square, so the square can be decomposed as (3n)(3n). N must be even, otherwise you cannot have a quarter leaving, which means the square can be decomposed as (3*2*m)(3*2*m). X is then 2*m*3*2*m, one quarter of that is 3*m*m, which must be a cube. So the cube contains a factor 3, which means the number to be cubed must also contain a factor 3.

The base number that has to be cubed has to be a multiple of 3. The original number of cats in the herd is then 4/3 * the cube of this number. Try working from there (backwards), to comply with the other limitations set by the puzzle.

12 -- it is a third of a square and 3/4 of it is a cube

The answer is below 1000

1. take n and square it to get m - if m isn't divisible by three, start over with the next n 2. multiply m by 3/4 to get a - if a isn't a cube, start over with the next n 3. if everything's worked so far, m is an acceptable answer

Enter consecutive values for n in column A. Then, in each subsequent column entre the following formulae: B =A1^2 C =B1/3 D =C1*3/4 E =D1^(1/3) Now look for an integer in column E to find the solution in column C. !

3/4 times the original number must be a multiple of 3, so the cube number must be a multiple of 3; and for it to be the least number, a multiple of 3 ONLY. The square number must have an even number of each factor (otherwise it can't be a square number); this means that the cube number must have an even number of factors. You should be able to work it out from here.

But I've only just noticed the fact that 3/4 is a cube that STAYED. From that the answer should be 54x54.

"If a quarter strayed, still a cube will have stayed." I made a list of the first few cube numbers Then in turn I divided each cube by 3 (disregarding those that are not divisible by 3) and added the result (the quarter that strayed) to the cube itself (i.e. 4 quarters). This gave me a possible answer. "They numbered one third of a square" To check if this answer was 'one third of a square' I multiplied it by 3 and checked that the result was a square number by finding its root with a calculator.

I looked for cubes by adding a column that calculated the cube root , which will be a whole number if the starting number was a cube. Use =N^1/3 to get the cube root of N, if you're using Excel (and probably most other spreadsheets). (Spoilered by Rotek, just in case.)