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Droppable Square numbers Question
A number $\large X$ is defined to be a Droppable Square number if the ordinary base-10 representation of $\large X^2$ can be formed by inserting the digit $\large 2$ anywhere in the base-10 representation of $\large X$, even at the beginning or end of $\large X.$
Rearranging the digits of $\large X$ at any time is not allowed.
$\large 11$ is a Droppable Square number because $\large 11^2 \ = \ 121$ can be formed by inserting the digit $\large 2$ into the middle of $\large 11$ to form $\large 121.$
$\large 5$ is also a Droppable Square number because $\large 5^2 \ = \ 25$ can be formed by inserting the digit $\large 2$ at the beginning of $\large 5$ to form $\large 25.$
Find another Droppable Square number.
3 answers
Key observation! Click here to reveal
Note that $9^2 < 92 < 10^2.$
From this observation we can get a Droppable Square Number $(DSN)$ of the form $9+X$ where $0 < X < 1.$
Details! Click here to reveal
We want $(9+X)^2 \ = \ 92+X$
Solving we get:
$(9+X)^2 \ = \ 92+X$
$81+18X+X^2 \ = \ 92+X$
$X^2+17X-11 \ = \ 0$
Using the Quadratic formula, $X \ = \ \Large \frac{-17 \pm \sqrt{17^2+44}}{2}$ which has only one positive solution.
So $DSN \ = \ 9 + \Large \frac{-17 + \sqrt{333}}{2}$ is a Droppable Square Number if $DSN < 10.$
Evaluating numerically we get:
$DSN \ \ = \ \ \ 9.624143795 \cdots \ $ and
$DSN^2 \ = \ 92.624143795 \cdots$
So $9 + \Large \frac{-17 + \sqrt{333}}{2}$ $\ = \ 9.624143795 \cdots$ is indeed a Droppable Square Number.
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The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| will.octagon.gibson | (no comment) | May 13, 2026 at 05:00 |
Key insights
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The number of digits in the square must be exactly one more than the number of digits in X. A one-digit integer can have a two-digit square; those are easy to check by inspection, and the example given in the question is the only case here. A two-digit integer can have a three-digit square, but that maxes out at 31 (and nothing in that range qualifies, by inspection). 32^2 is 4 digits, so from there up, it is not possible to assemble N^2 by inserting a 2 into N.
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... If we're limited to integers. But X is a number, not specifically an integer.
Two more solutions
I actually worked this out from first principles before noticing the similarity to the examples in the question.
1.1 squared is 1.21, and 0.5 squared is 0.25, both of which satisfy the requirement. These are, it turns out, just the original examples divided by 10. Dividing instead by 100 doesn't work though unless you're willing to be inconsistent about zeros -- 0.11 squared is .0121, which technically works, but for consistency it's really 0.0121, which doesn't.
Because of what I said about the number of digits in the previous section, I think 11, 5, 1.1, and 0.5 are the only solutions involving rational numbers.
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Inspired by will.octagon.gibson's answer and Monica Cellio's answer (which are $2$ very different approaches), I've found another. Spoilers for both of those answers are hidden so that you can try finding a solution yourself.
Spoiler from Monica Cellio's answer
The rational number $1.1$ is a Droppable Square number.
Spoiler from will.octagon.gibson's answer
There are irrational Droppable Square numbers.
My new Droppable Square number (includes spoilers for both of the other answers)
I used will.octagon.gibson's approach to find a Droppable Square number between $1$ and $1.1$. I chose to drop the $2$ into the second decimal place (since dropping the $2$ into the first decimal place would just give the same $1.1$ solution again). So the solution needs to be of the form:
$$1.0...^2=1.02...$$Representing the missing part of the left side as $X$, the missing part of the right side will be $\frac{X}{10}$ (since it is shifted $1$ decimal place to the right by the insertion of the $2$):
$$(1+X)^2=1.02+\frac{X}{10}$$$$X^2+2X+1=1.02+\frac{X}{10}$$Multiplying both sides by $50$ for the convenience of making the coefficients integer:
$$50X^2+100X+50=51+5X$$$$50X^2+95X-1=0$$Using the quadratic formula $X= \large \frac{-b\pm\sqrt{b^2-4ac}}{2a}$:
$$X=\frac{-95\pm\sqrt{95^2-4(50)(-1)}}{2(50)}$$$$X=\frac{-95\pm\sqrt{9225}}{100}$$I'm looking for a number greater than zero, so the $\pm$ can be changed to a $+$:
$$X=\frac{-95+\sqrt{9225}}{100}$$$$X=\sqrt{\frac{9225}{10000}}-\frac{95}{100}$$$$X=\sqrt{\frac{369}{400}}-\frac{19}{20}$$So $1+X=\sqrt{\frac{369}{400}}+\frac{1}{20}$ is a Droppable Square number.
In decimal form this is $1.0104686356149273...$
Squaring this gives $1.0\color{red}{2}\color{black}{104686356149273}...$
Estimating the total number of Droppable Square numbers (a large number of spoilers)
Instead of $1.02...$ being chosen as the square of the number, $1.002...$ or $1.0002...$ could have been chosen. Would these also give new Droppable Square numbers? Is there a limit to how many extra zeroes could be inserted this way?
Since the square is greater than $1$, the number must also be greater than $1$. This means that the number is less than its square, leading to the conclusion that the same number of zeroes can be added to the right of the decimal point of the number as were added to the right of the decimal point of its square. This gives starting points such as the following: $$1.00...^2=1.002...$$ $$1.000...^2=1.0002...$$ $$1.0000...^2=1.00002...$$
Since the infinite string of digits represented here by "$...$" will always be shifted $1$ decimal place to the right upon squaring (due to the insertion of the $2$), the same approach to solving can be used again, generalised to:
$$(1+X)^2=1.0...02+\frac{X}{10}$$Here the "$...$" represents a string of all zeroes.
A similar process to before gives:
$$X^2+\frac{19X}{10}-0.0...02=0$$Since the $a$ and the $c$ in the quadratic formula are positive and negative respectively, the $b^2-4ac$ will always be positive so there will always be a real square root. Since in this case $b^2-4ac>b^2$ it follows that $X$ will always have $1$ positive value.
Will this positive value always be small enough to not overwrite any of the zeroes?
$$X^2+\frac{19X}{10}-0.0...02=0$$$$X^2+\frac{19X}{10}=0.0...02$$$$X(X+\frac{19}{10})=0.0...02$$$$X=\frac{0.0...02}{X+\frac{19}{10}}$$Since $X$ is positive, $X+\frac{19}{10}>1$, so it follows that:
$$X<0.0...02$$This means $X$ will never overwrite any of the zeroes, so there will be a Droppable Square number for every number of zeroes that can be inserted.
This is a family of Droppable Square numbers with infinitely many members, but it is not all of the Droppable Square numbers. The $2$ linked answers both show numbers outside this family. I suspect there are other families, perhaps other infinite families, so there is still plenty of room for more answers.
Bonus number thanks to will.octagon.gibson
A comment points out that multiplying my new Droppable Square number by 10 gives another:
$$10\times1.0104686356149273...=10.104686356149273...$$Squaring this gives $10\color{red}{2}\color{black}{.104686356149273}...$
I wondered if this might extend into a new infinite family, but multiplying the other members of the infinite family by $10$ gives numbers where the $2$ is added on the right hand side of the decimal point, so the left hand side of the decimal point changes from $10$ to $100$, rather than from $10$ to $102$, so the extra zero prevents any of them from being Droppable Square numbers.
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