Find the largest 4-digit perfect square number for which all of its digits are also all perfect square numbers.
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Yes, it is just that simple!
First reaction: Is this some kind of trap? Second reaction: Not sure if I should submit this answer. Answered reaction: OMG! I was correct!?
Since 100^2 = 10,000, start squaring numbers beginning with 99 and work downwards; we soon come to 97^2 = 9409, each digit being a square. Ed Gray
I remember that ( 1 0 0 − a ) 2 = 1 0 0 0 0 − 2 0 0 a + a 2 . If the first digit started with 9 then the next digit must be even and hence 4. I then put in a 0 and a 9 and got 9409 which is 1 0 0 0 0 − 2 0 0 × 3 + 3 2 = ( 1 0 0 − 3 ) 2 = 9 7 2
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Haha...I don't have any general solution I just happened to start with 9 9 2 since 1 0 0 2 = 1 0 0 0 0 which is 5-digit. I just realised 9 7 2 = 9 4 0 9 .