# Quick way to find the square root of 123.45654321?

If $$n^2= 123.45654321$$, then what will be the exact value of $$n$$?

Is there a neat/efficient way to solve this without using the long division method? Since this is a GRE question, I believe there must be a shorter solution, without guessing or already knowing a pattern.

• This isn't really worth an answer in its own right, but above and beyond the answers given I'll note that my first hook in would be that $n^2$ is very close to $11^2=121$ (and having the squares memorized at least up through 15 or so comes in surprisingly handy in various situations) and so I was looking for a number just slightly more than 11. Since the square is about 2% more (121+2) then I'm looking for a square root that's about 1% more than 11, i.e. 11.1... and at that point my suspicions are well engaged. – Steven Stadnicki Mar 3 at 23:09
• "...without already knowing a pattern." That's rather vague. – Servaes Mar 4 at 15:24
• The quick way is to punch it into a calculator – Mason Wheeler Mar 4 at 15:25

The answer could require an intuitive understanding of long multiplication (as opposed to long division).

$$11^2$$
$$= 11(10 + 1)$$
$$= 110 + 11$$ $$= 121$$

Similarly

$$111^2$$ $$= 111(100 + 10 + 1)$$
$$= 11100 + 1110 + 111$$ $$= 12321$$

From the above two examples, you can notice a pattern: the middle number is equal to the length of the number, i.e., the number of ones, and then the natural count of numbers are mirrored from the middle number through 1. This is due to the stacking of numbers like so in long multiplication

  111
111
111
—————
12321


By looking at the individual digits of the number in question, it seems to be a natural sequence from 1 through 6 then back to 1. The middle number is 6, hence

the number is a square of six consecutive ones, i.e., $$111111$$

As the other answers have stated, the decimal point is placed before an even number of digits (8), the square root should hence

divide the above number by 10 raised to half the number of displaced decimal places.

$$111111 \times 10^{-4} = 11.1111$$

If you know what happens when you multiply numbers with small digits, then you might know $$111111 \times 111111 = 12345654321$$, so $$11.1111 \times 11.1111 = 123.45654321$$.

Here's how to think about this, if you don't already know the neat solution and don't have enough computational power to divide into prime factors.

• $$n^2= 123.45654321$$ is an (even) power of $$10$$ times an integer, so it's enough to find the square root of $$12345654321$$. That's going to be a six-digit number. Since the number we have is so beautiful and symmetric, it seems likely there might be a beautiful and symmetric answer too.
• How do we calculate the square of a number from its digits? Using the binomial theorem: $$(10b+a)^2=100b^2+20ab+a^2$$ $$(100c+10b+a)^2=10000c^2+2000bc+100(2ac+b^2)+20ab+a^2$$ $$(1000d+100c+10b+a)^2=1000000d^2+200000cd+10000(2bd+c^2)+2000(ad+bc)+100(2ac+b^2)+20ab+a^2$$

$$and\quad so\quad on$$

• So basically we want to find a number $$100000f+10000e+1000d+100c+10b+a$$, take its square, and match up the coefficients: $$f^2=1=a^2$$, $$2ef=2=2ab$$, and so on. Already now it looks like we're going to want to take lots of the digits equal to 1, and a quick check at the pattern emerging above will confirm that if they're all 1 then the digits of the square increase step by step to the middle and then decrease again.

$$111111^2=12345654321$$ giving your answer as $$n=11.1111$$.
My first small idea here is to notice that our number this is actually $$12345654321 \cdot 10^{-8}$$ so actually to find its square root we can divide it into parts $$\sqrt{12345654321} \cdot \sqrt{10^{-8}}$$ which makes us work with the integer number $$12345654321$$ instead of float. Then we can proceed with any factorization technique which can be used to factor on paper (starting with just trying out division on all small prime numbers). A list of several methodologies for fast factorization you can find at https://en.wikipedia.org/wiki/Integer_factorization.
In this case actually any methodology will converge fast enough as $$12345654321$$ $$= 3^2 \cdot 7^2 \cdot 11^2 \cdot 13^2 \cdot 37^2$$ so actually $$12345654321$$ $$= (3 \cdot 7 \cdot 11 \cdot 13 \cdot 37)^2$$ so answer to the global question will be $$(3 \cdot 7 \cdot 11 \cdot 13 \cdot 37) * 10^{-4}$$ $$= 11.1111$$.