For which digits d, is it possible to add d to every digit of a square and get another square?
For example, adding 3 to each digit of 16 gives 49.However, adding zero to each digit in this manner is NOT permissible.
For which digits d are there infinitely many such squares?
*** Digit sums greater than 9 are not allowed. For example, you could not add 8 to the digits of 81 to get 169.
(In reply to
computer solution by Charlie)
I know what you are talking about, but 333334^2 will actually turn into 666667^2, while 3333334^2 will turn into 6666667^2. The number of 3's in the first number must be the same as the number of 6's in the second number.
4^2=16->7^2=49
34^2=1156->67^2=4489
334^2=111556->667^2=444889
3334^2=11115556->6667^2=44448889
33334^2=1111155556->66667^2=4444488889
333334^2=111111555556->666667^2=444444888889
3333334^2=11111115555556->6666667^2=44444448888889
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Posted by Math Man
on 2024-04-16 15:02:27 |
re: computer solution
