[Math] Why does this cube root trick work

elementary-number-theoryproof-verification

So, I found that you can work out cube roots between $1$ and $100$ using this method:

For example the number: $185193$

If the last digit $x$ is $2$, $8$, $3$, or $7$, the last digit in the answer should be $10 – x$; if otherwise, then $x$. In this case:

$$10-3 = 7$$

We now got $\_7$. Remove the last three digits so that we get only $185$. Then, check in this list:

$$
1^3 = 1 \\
2^3 = 8 \\
3^3 = 27 \\
4^3 = 64 \\
5^3 = 125 \\
6^3 = 216 \\
7^3 = 343 \\
8^3 = 512 \\
9^3 = 729
$$

Check which two cubes that number is between, the lowest one should be the first digit in the answer:

$$5^3 < 185 < 6^3$$

Therefore, we get:

$$57$$

I found this method in this video: https://www.youtube.com/watch?v=Al7PllobwR4

But how does this method really work, and why?

Best Answer

If $10n \leq x < 10(n+1)$, then $1000n^3 \leq x^3 < 1000(n+1)^3$, and therefore $$n^3 \leq \frac{x^3}{1000} < (n+1)^3.$$ Moreover, since $n^3$ and $(n+1)^3$ are integers you can ignore the fractional part of $\frac{x^3}{1000},$ which is essentially what you do when you drop the last three digits of $x$.

For a two-digit number there are only a few choices for $n$; so by examining your table of values of $n^3$ you can figure out which value of $n$ makes the equation above true. So the second half of your "trick" will correctly compute the tens digit of the cube root of any positive integer less than one million.

The first half of the trick depends on the assumption that the number you start with is a perfect cube. It relies on the fact that in your table of cubes of a single digit, no two results on the right-hand side have the same last digit. So if $b$ is a digit and you know the last digit of $b^3$, you can tell what $b$ is. (The trick gives a mysterious rule for computing $b$, but all it really has to do is to give the correct answer for nine different input values. Ten, if you include zero.)

The trick also relies on the fact that $(10a + b)^3 = 10(100a^3 + 30 a^2 b + 3ab^2) + b^3.$ That is, the last digit of $(10a+b)^3$ is just the last digit of $b^3$.

The trick fails for numbers that are not perfect cubes to begin with. The first digit will be correct but not the rest. For example, the trick says the cube root of $185194$ is $54$, but you know it has to be greater than $57 = \sqrt[3]{185193}.$

Related Solutions

Elementary Number Theory – Divisibility Criteria of 24

The problem here might be something like $12$. You see, we have that $12$ is divisible by both $6$ and $4$, but it's not divisible by $24$. The reason they suggest $3$ and $8$ is because they are relatively prime, meaning that you can't have the sort of overlap in the case of $6$ and $4$.

This all has to do with the Fundamental Theorem of Arithmetic, which says that each number can be written uniquely as a product of primes, and primes have the special characteristic (or as Marvis points out, they are defined to be exactly those numbers with the characteristic) that if $p|ab$, then $p|a$ or $p|b$. So if $3$ and $8$ divide a number, then $24$ divides that number. But $6$ and $4$ dividing a number doesn't even guarantee that $8$ divides that number.

[Math] Finding the last digit of $3^{729}$

Every digit 0 thru 9 has a pattern of 4 digits when raised to an exponential power. Simply divide the power by 4 and the remainder shows you where you are at in the pattern. Here are the patterns for all 10 digits. Note that remainder 1 corresponds to the first digit in the pattern and remainder 2 corresponds to the second digit in the pattern. Remainder 3 corresponds to the third digit and remainder 0 (the power is divisible by 4) corresponds to the fourth digit in the pattern. Try a few yourself using a calculator and you will get the hang of it.

0,0,0,0   
1,1,1,1   
2,4,8,6   
3,9,7,1   
4,6,4,6   
5,5,5,5   
6,6,6,6   
7,9,3,1   
8,4,2,6   
9,1,9,1

Best Answer

Related Solutions

Elementary Number Theory – Divisibility Criteria of 24

[Math] Finding the last digit of $3^{729}$

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