Form the biggest squaring number

Question

You are going to start from any number with distinct digits lower than four digits (such as $a$, $ab$, $abc$) and you will apply the rule below:

Take the square any number you want in the number (you may take the square the number itself too). (such as $1(02)$ -> $14$, or $12(3)$ -> $129$ or $(13)$ -> $169$)
Every number you found has to have distinct digits. (Fail example: $(12)3$-> $1443$ X)
You may apply this as many times as you want. ($(13)$ -> $169$ then $(16)9$->$2569$)

What is the biggest final number you can have?

is there a restriction about how many digits the number which we square can have? like, if I get a number like 1023 as an interim result (even though 1023 itself is not possible), am I allowed to square it all to get to 1046529? — elias, Sep 16 '18 at 17:11
@elias "you may take the square the number itself too". written on the first rule. no restriction. — Oray, Sep 16 '18 at 17:12
$1(69) -> 14761$. Isn't this a violation of the previous rule? — Wais Kamal, Sep 16 '18 at 17:52

elias · Accepted Answer · 2018-09-16 19:06:56Z

Update:

I've found

9857240136 to be maximal (confirmed with programming).

Many possible ways to reach it, here is one starting with...

(4)2 -> 16(2) -> 1(6)4 -> 1(3)64 -> (1964) -> 3857(2)96 -> 3857(496) -> (3)85724016 -> 98572401(6) -> 9857240136

Previous attempt:

My new record is:

9810362574

which can be reached via

(5)74 -> (25)74 -> (625)74 -> 3(9)062574 -> (3)81062574 -> 9810(6)2574 -> 9810362574

First attempt:

Not proven to be maximal at all, but I managed to reach

10 digits with 4817093625

The steps are:

(705) -> 4(9)7025 -> 48170(25) -> 48170(6)25 -> 48170(3)625 -> 481709(6)25 -> 4817093625

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