Reciprocal 9000

Number Theory Level 3

Find the largest 4-digit perfect square number for which all of its digits are also all perfect square numbers.


The answer is 9409.

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Bryan Lee Shi Yang by Bryan Lee Shi Yang , 17, Malaysia

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3 solutions

Noel Lo
May 20, 2015

Haha...I don't have any general solution I just happened to start with 9 9 2 99^2 since 10 0 2 = 10000 100^2 = 10000 which is 5-digit. I just realised 9 7 2 = 9409 97^2 = 9409 .

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Moderator note:

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!?

Gavin LO - 6 years ago
Edwin Gray
Aug 25, 2018

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

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Josh Banister
Jul 22, 2015

I remember that ( 100 − a ) 2 = 10000 − 200 a + a 2 (100-a)^2 = 10000 - 200a + 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 10000 − 200 × 3 + 3 2 = ( 100 − 3 ) 2 = 9 7 2 10000 - 200 \times 3 + 3^2 = (100 - 3)^2 = 97^2

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