# Finding roots of numbers quickly

## Computing cube roots of two digit numbers

In summer 2018, I was teaching a math class for high school students on problem solving and algebra at Harvard. A student of mine mentioned a trick he knew for calculating a cube root of a number quickly in his head. The entire class was interested, and he gave us the following instructions:

• Choose any two digit number you like and cube it on a calculator.
• Tell me the number you cubed.

I pulled out my phone, cubed a number, and told the student “my number cubed is 148,877”. In a split second, he responded back “the number you cubed was 53”. I was shocked that he solved it so quickly! For a few more minutes, other students called out cubes, and the student would respond back with what number was cubed. After amazing all of us, the student was kind enough to teach us his trick. In this blog post, I’ll show the same steps so that you too can pull off this trick to impress your friends/family.

To begin with, the only bits of information you need to be able to pull off this trick are knowing the 1-digit cubes in your head. If you don’t know them, they are:

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

One important thing to notice in this list is that the units digit of every cube is unique.

Since the units digit are unique, if we know what the units digit of $x^3$, we can look at our table above to determine the units digit of $x$. For example, if $x^3 =148,877$, then since $x^3$ has a units digit of $7$, we know the units digit of $x$ is $3$ by looking at the table above.

Now that we know the units digit of $x$, all that’s left to determine is the tens digit (since we started with a $2$ digit number). For this, we need to determine what two cubes the digits in the thousands place of $x^3$ lie between. For example, with $148,877$, the digits in the thousands place are $148$. We see that $125<148<216,$ therefore,

$50^3=125,000<148,8777<216,000=60^3.$

Hence, we know the tens digit of $x$ is $5$, so our original cube must be $53$.

Let’s try another quick example, and then I’ll leave some exercises for the reader.

Suppose we want to find the two digit number, which, when cubed gives us $373,248$. We begin by looking at the units digit of our cube, which is $8$. We therefore know the units digit of the two digit number must be $2$. Furthermore, we see that $343<373<512$, therefore, the tens digit of the cube must be $7$. Hence, our two digit number is

$72^3=373,248$.

### Practice Problems

Here are some exercises for the reader to try out themselves:

• Find the two digit number, such that its cube is equal to $19,683$.
• Find the two digit number, such that its cube is equal to $29,791$.
• Find the two digit number, such that its cube is equal to $74,088$.
• Find the two digit number, such that its cube is equal to $185,193$.
• Find the two digit number, such that its cube is equal to $328,509$.
• Find the two digit number, such that its cube is equal to $614,125$.
• Find the two digit number, such that its cube is equal to $970,299$.

## Computing 5th roots of numbers ending in 1, 3, 7, or 9 quickly

On Dr. Ali Gurel’s YouTube channel, COOL MATH, he shows a neat trick for finding fifth roots of odd numbers up to 200, not ending in a 5. I recommend watching his video first to see some examples of the trick, and below I’ll explain the math behind why the trick works and also extend it to any three digit number.

There are four cases for the units digit of the number: either a 1, 3, 7, or 9. Notice that

$1^5\equiv 1\pmod{10}, 3^5\equiv 3\pmod{10}, 7^5\equiv 7\pmod{10}, 9^5\equiv 9\pmod{10}$.

Therefore, the units digit of $x$ and $x^5$ are the same.

To use this trick on 5th roots of numbers between $1$ and $200$, you only need to memorize that $3^5=243$. For larger 5th roots, you’ll need to use a table to establish bounds.

### Units digit 1 case

In the case that the units digit is 1, then we can represent our number in the form $10x+1$. For now, we’ll tackle the same case as in Dr. Ali’s video and assume that $0\le x\le 19$.

Using the binomial theorem,

$(10x+1)^5=10^5x^5+\binom{5}{1}10^4x^4+\binom{5}{2}10^3x^3+\binom{5}{3}10^2x^2+\binom{5}{4}10x+\binom{5}{5}.$

Since $\binom{5}{3}=10$, we see every term is a multiple of $1000$, except for the last two. Therefore, when we consider the last three digits of $(10x+1)^5$, we get $(10x+1)^5\equiv 50x+1\pmod{1000}$. Since $0\le x\le 19$, we see that this remainder mod $1000$ is unique.

To find the original number, we take the last three digits and perform the calculation:

$\frac{\text{Last Three Digits}-1}{5}+1=\frac{50x+1-1}{5}+1=10x+1.$

For example, $21^5=4,084,101,$ which has the last three digits $101$. We therefore see that the fifth root of this number was $\frac{101-1}{5}+1=21.$

### Units digit 3 case

Now, if the number we started with ends in a $3$, then we can write the number in the form $10x+3$, where we assume for now that $0\le x\le 19$. Using the binomial theorem,

$(10x+3)^5=10^5x^5+\binom{5}{1}10^4x^4\cdot 3+\binom{5}{2}10^3x^3\cdot 3^2+\binom{5}{3}10^2x^2\cdot 3^3+\binom{5}{4}10^1x^1\cdot 3^4+3^5.$

Again, we only consider the last three digits of this number, and therefore

$(10x+3)^5\equiv 50x\cdot 3^4+3^5\pmod{1000}.$

Notice that $50\cdot 3^4=4050$, therefore we can further reduce to get

$(10x+3)^5\equiv 50x+3^5\pmod{1000}$.

Now if $0\le x\le 15$, then $50x+3^5<1000$, and we can use the same trick again:

$\frac{\text{Last Three Digits}-3^5}{5}+3=\frac{50x+3^5-3^5}{5}+3=10x+3.$

However, if $16\le x\le 19$, then the last three digits of our number will be $(50x+3^5-1000)$, so we have to add $1000$ to get our original number:

$\frac{\text{Last Three Digits}+1000-3^5}{5}+3.$

As an example of this, if we want to find the fifth root of $2,073,071,593$, since $593>3^5=243,$ we know it’s the first case and we get the fifth root as $\frac{593-3^5}{5}+3=73$.

Alternatively, if we want to find the fifth root of $205,236,901,143$, we see that $143<3^5$, therefore, we have the second case and get the fifth root as $\frac{143+1000-3^5}{5}+3=183$.

### Units digit 7 case

If our number ends in a $7$, then it can be represented as $10x-3$ where $0\le x\le 19$. By similar logic to the calculations before, we only consider the last three digits:

$(10x-3)^5\equiv \binom{5}{4}\cdot 10x\cdot (-3)^4+\binom{5}{5}\cdot (-3)^5\pmod{1000}.$

As before, since $\binom{5}{4}\cdot (-3)^4=4050$, this further reduces as

$(10x-3)^5\equiv 50x-3^5\pmod{1000}.$

Now if $19\ge x\ge 5$, the last three digits of $(10x-3)^5$ will be $50x-3^5$, so we perform the calculation

$\frac{\text{Last Three Digits}+3^5}{5}-3=\frac{(50x-3^5)+3^5}{5}-3=10x-3.$

As an example, if we wish to find the fifth root of $8,587,340,257$, we take the last three digits, $257$, and using the formula above get $\frac{257+3^5}{5}-3=97$, which is indeed the fifth root.

However, if $0\le x\le 4$, then $50x-3^5<0$, and therefore the last three digits will be $(50x-3^5+1000)$. In this case, we subtract $1000$ from our initial calculation to get:

$\frac{\text{Last Three Digits}-1000+3^5}{5}-3.$

As an example, if we want to find the fifth root of $69,343,957$, since $957+3^5>1000$, we use the calculation from the second case to get a fifth root of $\frac{957-1000+3^5}{5}-3=37.$

### Units digit 9 case

Finally, if a number ends in a $9$, then it can be represented in the form $10x-1$, where $0\le x\le 19$. By the same logic as before, we see that

$(10x-1)^5\equiv \binom{5}{4}\cdot 10x\cdot (-1)^4+\binom{5}{5}\cdot (-1)^5\pmod{1000}\equiv 50x-1\pmod{1000}.$

Therefore, to recover the original number, we can perform the following calculation:

$\frac{\text{Last Three Digits}+1}{5}-1=\frac{50x-1+1}{5}-1=10x-1$.

As an example, if we want to find the fifth root of $282,475,249$, we take the last three digits, $249$, and perform the calculation above to get a fifth root of $\frac{249+1}{5}-1=49$.

### Practice Problems

Here are some practice problems to test out your understanding of this method. I highly recommend reviewing the sections above as you work through these tricky examples! As a hint, pay attention tot he last three digits!

• Find the fifth root of $161,051$.
• Find the fifth root of $6,436,343$.
• Find the fifth root of $90,224,199$.
• Find the fifth root of $14,348,907$.
• Find the fifth root of $16,850,581,551$.
• Find the fifth root of $41,615,795,893$.

### Generalizing to larger numbers

This trick will work for all odd fifth powers ending in a $1,3,7,$ or $9$ up until $200^5=3.2 \cdot 10^{11}$. If we start with a number between $200$ and $400$, then the range for the fifth power is between $3.2\cdot 10^{11}$ and $1.024\cdot 10^{13}$. In this case, we can use the same trick as above, to get a number between $1$ and $200$, and then add $200$ to the final result.

For example, if someone tells us to find the fifth root of $3,973,195,810,651$, we see this number is approximately $3.9\cdot 10^{12}$, therefore the fifth root must be between $200$ and $400$. Using the same trick as before, we see that the number mod $200$ must be $\frac{651-1}{5}+1=131$, and adding $200$ to bring us into the correct range gives us $331$ (you can verify this is the fifth root of the original number using WolframAlpha).

In general, the fifth powers fall into the ranges given in the table below.

As a quick example, suppose you’re told to find the fifth root of $377,571,474,600,343$. Using the table above, you know this number is approximately $3.776\cdot 10^{14}$, which is in the bottom-most row, therefore the number which was cubed was between $800$ to $1000$. We look at the last three digits, $343$, and since the number ends in $3$, perform the calculation $\frac{343-3^5}{5}+3=23$ to determine the mod $200$ residue of our number. We then add $800$ to get our number into the desired range, and find the fifth root is $823$.

### Practice Problems with Larger Numbers

• Find the fifth root of $481,170,140,857$.
• Find the fifth root of $3,303,341,057,599$.
• Find the fifth root of $14,195,130,030,907$.
• Find the fifth root of $78,410,163,603,001$.
• Find the fifth root of $872,095,812,856,093$.

While these problems all look daunting at first, with some practice using the technique above, you should be able to compute the fifth roots in a matter of a few seconds! Give the method some practice and patience, and hopefully you too should be an expert on computing cube and fifth roots in your head quickly!

# Quick Divisibility Test for Primes up to 67

While there’s known divisibility tests for certain primes (see my notes), the following test very quickly determines if a number is divisible by any prime up to 67 by hand! The original method is due to John Conway, and I learned it from Arthur Benjamin.

Here’s a sample problem along with explanation:

63779 is the product of three primes, each less than $71$. Determine their sum.

We set up the equation $63779=pqr$, where $p, q,$ and $r$ are primes.

We begin by finding the factor of 2000 closest to 63, namely 64,000. We can therefore rewrite this as

$221=2000\cdot 32-63,779.$

We can quickly determine that since $2, 5\mid 2000$ and $2, 5\nmid 221$, $2, 5\nmid 63,779$. We now add $32$, the quotient in the divisibility above to both sides of the equation to produce:

$253=2001\cdot 32-63,779.$

Notice the factor of $32$ and the left hand side have both increased, while our original number, 63779 has remained the same. Now, as any good contest math student should have the years close to the current year memorized, we know $2001=3\cdot 23\cdot 29.$ Therefore $3\mid 63779\iff 3\mid 253$, however, $3\nmid 253$. Similarly $23\mid 63779\iff 23\mid 253$. This time, $253=23\cdot 11$, therefore $23\mid 63779$. We’ve found one prime divisor: $p=23$.

We could now divide $63779$ by $23$, however, this gives the equation $2773=qr$, and we don’t have any better ideas for how to factorize $2773$. Therefore, instead, we’ll use the prime factorizations of some other numbers close to $2000$:

$1998=2\cdot 3^3\cdot 37, 2002=2\cdot 7\cdot 11\cdot 13, 2006=2\cdot 17\cdot 59, 2009=41\cdot 49$

and $2010=2\cdot 3\cdot 5\cdot 67,2013=3\cdot 11\cdot 61, 2015=5\cdot 13\cdot 31, 2021=43\cdot 47.$

For instance, to determine if our number is divisible by $37$, we would take the original equation and subtract $2\cdot 32$ to produce $157=1998\cdot 32-63,779.$ Then, we know $37\mid 63779\iff 37\mid 157$, however, clearly $37\nmid 157$.

Similarly, to test if our number is divisible by $7, 11,$ or $13$, we simply add $2\cdot 32$ to our original equation to produce $285=2002\cdot 32-63779$. Therefore, again $11\mid 285\iff 11\mid 63779$, however, clearly $11\nmid 285$, hence the same is true for $63779$.

Next, to determine if the original number is divisible by $17$ or $59$, we add $6\cdot 32$ to our original equation (or more simply, $4\cdot 32$ to the equation in the last paragraph) to get

$413=2006\cdot 32-63779.$

Since $2006=2\cdot 17\cdot 59$, we see $17\mid 413\iff 17\mid 63779$, however, clearly $17\nmid 413$. Similarly, $59\mid 413\iff 59\mid 63779$. Now since $420=60\cdot 7$, we see $413=59\cdot 7$, hence $59\mid 413$, and $59\mid 63779$. We’ve therefore found a second prime factor: $q=59$.

At this point that we have two prime factors, we could continue on with our test, but we have a pretty good idea of what the third factor is by division: $63779=23\cdot 59\cdot r$. One way to quickly compute this in our heads would be to notice

$59\cdot 23=(60-1)(20+3)=1200+180-20-3=1357.$

Since we know $r$ is an integer, we attempt to bound it. We know $r=\frac{63779}{1357}$. We notice $40\cdot 1357=52480$, while $50\cdot 1357=67850$, therefore, $40.  We then subtract $40$ from both sides of our equation:

$r-40=\frac{63779-52480}{1357}.$

While we could compute the numerator mentally, since we know $1\le r-40\le 9$, we know it must be a single numerical digit! We therefore just consider the unit digit of the numerator which is $9$. Since we’re dividing by a number which ends in $7$, we know $r-40$ must be $7$, since $7\cdot 7\equiv 49\equiv 9\pmod{10}$. Therefore $r=47$.

Hence our desired sum is $p+q+r=23+59+47=\boxed{129}$.

Alternatively, we could verify $47\mid 63779$ using the prime factorization of next year, 2021. I’ll leave this as an exercise for anyone interested.

While this method isn’t super practical, I think it’s interesting that you can essentially compute prime factors up to 67 with a quick trick in your head. The most difficult part is memorizing the prime factorizations of the numbers close to 2000, but hopefully this is familiar for any seasoned contest mathematician (who might be interested in using this).

References: Arthur Benjamin, Factoring Numbers with Conway’s 150 Method
The College Mathematics Journal, Vol. 49, No. 2, pp. 122-125, March 2018.

# Similar to Fermat’s Last Theorem

Let $n$ be a positive integer. Prove that there exist distinct positive integers $x, y, z$ such that $x^{n-1} + y^n = z^{n+1}.$ (Source: 1997 IMO Shortlist N6)

The equation $x^{n}+y^n=z^n$ is known to have no nontrivial solutions for $n\ge 3$ due to Fermat’s Last Theorem (if you’ve not read Fermat’s Enigma I highly recommend checking it out). This very similar looking equation, however, has an infinite number of solutions. In this blog post, we’ll examine how this difficult looking problem can be solved with one simple observation: $1+8=9$.

For a concrete example of this, consider the case $n=3$ giving the equation $x^{2}+y^{3}=z^{4}.$ Assume we multiply the equation by a number of the form $2^{\alpha}\cdot 3^{\beta}$.  We then will have: $2^{\alpha}\cdot 3^{\beta}+2^{\alpha+3}\cdot 3^{\beta}=2^{\alpha}\cdot 3^{\beta+2}.$ Since we want the first part to be a perfect square, we should have $2\mid \alpha$ and $2\mid \beta$. Since the second part should be a perfect cube, we should have $3\mid (\alpha+3)$ and $3\mid \beta$. Finally, since the last part should be a perfect fourth power, $4\mid \alpha$ and $4\mid (\beta+2)$. Solving these system of linear congruences gives $\alpha=0$ and $\beta=6$! We get back the equation $3^6+3^6\cdot 2^3=3^8$, from which we read off the solution $(x,y,z)=(3^3, 3^2\cdot 2, 3^2)=(27, 18, 9)$

We now attempt to generalize this technique. We begin by multiplying $1+8=9$ by $2^{\alpha}\cdot 3^{\beta}$ again to obtain $2^{\alpha}\cdot 3^{\beta}+2^{\alpha+3}\cdot 3^{\beta}=2^{\alpha}\cdot 3^{\beta+2}.$ We then want the exponents on the leftmost term to be divisible by $n$, on the middle term to be divisible by $n+1$, and on the rightmost term to be divisible by $n+2$. This is equivalent to:

$\begin{cases} \alpha&\equiv 0\pmod{n} \\ \alpha&\equiv -3\pmod{n+1} \\ \alpha&\equiv 0\pmod{n+2} \end{cases}$   and    $\begin{cases} \beta&\equiv 0\pmod{n} \\ \beta&\equiv 0\pmod{n+1} \\ \beta&\equiv -2\pmod{n+2}. \end{cases}$

Notice that in the first case, the first and third conditions are equivalent to $\alpha\equiv 0\pmod{n(n+2)}$. Since $\gcd(n, n+2)=\gcd(n+1, n+2)=1$, we can use the Chinese Remainder Theorem to know that $\alpha$ has a unique solution mod $\text{lcm}(n, n+1, n+2)$

On the other hand, for $\beta$, in the case that $n$ is even, $\gcd(n, n+2)=2$, therefore, we have to use an extended version of the Chinese Remainder Theorem. Thankfully, since $\gcd(n, n+2)=2$ in this case and $0\equiv -2\pmod{2}$, we are still guaranteed a unique solution for $\beta$ mod $\text{lcm}(n, n+1, n+2)$.

Substituting these solutions in for $\alpha$ and $\beta$ gives a solution $(x,y,z)$ to the original equation. Once we have the congruences for $\alpha$ and $\beta$, we can then find an infinite parameterized version of solutions for $(x,y,z)$.

For instance, in the case of $n=3$ above, we had $\alpha\equiv 0\pmod{12}, \beta\equiv 6\pmod{12}$. If we in general write $\alpha=12\alpha'$ and $\beta=12\beta'+6$, we then find

$2^{12\alpha'}\cdot 3^{12\beta'+6}+2^{12\alpha'+3}\cdot 3^{12\beta'+6}=2^{12\alpha'}\cdot 3^{12\beta'+8}.$

This then translates into $(x,y,z)=(2^{6\alpha'}\cdot 3^{6\beta'+3}, 2^{4\alpha'+1}\cdot 3^{2+4\beta'}, 2^{3\alpha'}\cdot 3^{3\beta'+2})$. Notice for $\alpha'=\beta'=0$, this gives the solution triple $(x,y,z)=(27, 18, 9)$ found above.

It’s pretty interesting to me how when you change one small thing in the formulation of the problem (subtract $1$ from the exponent on $x$ and add $1$ to the exponent of $z$) gives an entirely new problem, where we go from no solutions to an infinite number!

# The Search for Mersenne Primes

To be written later…