I have put eight four-digit numbers together in a 4x4 grid, such that four can be read across and four downwards.
Four of my numbers are odd and four are even. I have one cubed number going across and one going down, and I have one fourth-power number going across and one going down. Two of the other numbers in the grid are squares.
Which two numbers in my grid are not perfect powers?
(In reply to
Solution by Hand by Lisa)
I am Vinod Kagal from Mumbai India and I give a possible solution as follows:-
Of the 4 digit numbers x^4 the only options are 1296, 2401, 4096 and 6561.
Similarly y^3 gives 12 options 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000 1nd 9231 which are the cubes of 10 to 21.
I observed that 4096 is unique to all 3 categories x^4, y^3 and z^2 and it is an even number.
Hence we can select 2401 and 4096 such that '0' is common between row 2 and column 3. This takes care of options of a 4 digit number not to begin with a zero.
I filled the 3rd row with 2197 and the 1st colmn with 1225.
Thus the conditions of two x^4, y^3 and z^2 get filled satisfying the odd/even condition too.
This leaves only 4 unfilled positions as .....Row 1 with col 2 and 4 blank in addition to Row 4 with column 2 and 4 blank. These 4 positions can be filled any reptitive digit such as all 2s or 3s or 9s and ensuring that the given condition are not violated.
Loking forward to your valuable feedback........Vinod