At first her question was mysterious to us. Then she explained that she knew that 9 + 9 = 18, and she had heard that 9 × 9 = 81, and she had noticed that in 18 and 81, the digits are reversed. So what she wanted to know was, does it work the same way for 7? Given that 7 + 7 = 14, can you conclude that 7 × 7 = 41?

It was a great question! But no, we answered, it doesn't work that way. A little reflection shows that 2 and 9 are the only positive integers for which doubling and squaring yield identical digits, up to ordering. (There will never be a multi-digit case, because squaring a multi-digit number always yields more digits than doubling it.)

However, 434 is an interesting case, because 434 + 434 = 868, and 434 × 434 = 188,356, so 434 has the property that the digits of the double are a subset of the digits of the square, counting multiplicity.

Another example like that is 99, because 99 + 99 = 198 and 99 × 99 = 9,801.

In fact, any number of the form 99...9 works this way. Can you show that the digits of 99...9 + 99...9 are the same as the digits of 99...9 × 99...9, discounting zero digits?

There are 4,747,374 integers

*k*in the range 1 ≤

*k*≤ 10^8 such that the digits of

*k*+

*k*are a subset of the digits of

*k*×

*k*, counting multiplicity. Of these, 83,238 are such that every digit appearing in

*k*×

*k*also appears in

*k*+

*k*.

P.S., Update just to add this graph, which is a scatterplot of (

*x*,

*y*) where

*x*is a number between 1 and 10,000 and

*y*is the number you get by reversing the digits of

*x*. For example, the plot contains the point (46, 64) and the point (2539, 9352).

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