Mental Math – Tips and Tricks for Quick Calculations

This course provides a collection of tips and tricks for performing mental math calculations quickly and efficiently. It is designed for anyone looking to improve their ability to perform arithmetic in their head, focusing on practical methods for (common) mathematical tasks.

Approximating Square Roots

For a rough estimate of the square root of a number, find the nearest perfect squares and use them as reference points.

To estimate $\sqrt{50}$, note that $\sqrt{49} = 7$ and $\sqrt{64} = 8$. $\sqrt{50}$ is closer to 7 since 50 is closer to 49, so in the range of [7, 8] the answer to $\sqrt{50}$ is way closer to 7.

There is also a way to approximate the decimals, even for larger numbers.

• Let’s say you want to find the square root of 333.
• We know that the closest round number is 18, because $18^2 = 324$.
• The difference between $333-324=9$.
• Next, you divide the difference (9) by the round square root (18) and you divide this always by 2, so you have $\frac{9}{18} / 2 = \frac{9}{36}$.
• Finally, you add this number on top of the round root (18), so your answer would be $18 + \frac{9}{36} = 18.25$.
• What’s the real answer? It’s 18.248, which is very close to our approximation.

Difference of Consecutive Squares

To subtract squares of consecutive numbers, use the formula

\begin{equation} N^2 – (N-1)^2 = 2N – 1 \end{equation}

$5000^2 – 4999^2 = 2*5000 – 1 = 9999$

Sum All Numbers from A to B

To sum up all numbers from 1 to N, simply use the formula:

\begin{equation} \frac{N*(N+1)}{2} \end{equation}

To sum up all numbers from 1 to 10, we do

\begin{equation} 10*11/2 = 55 \end{equation}

In case you need to sum up all number from 101 to 200, we do

\begin{equation} 200*201/2 – 100*101/2 = 20100 – 5050 = 15050 \end{equation}

Sum Odd Numbers from A to B

To sum up the odd numbers from 1 to N, where N is the last odd number, use the formula:

\begin{equation} (\frac{N+1}{2})^2 \end{equation}

So, what is the sum of all odd numbers from 100 to 200?

Let’s sum up all odd numbers in the range from 0 to 200. The last odd number is 199. To sum up all odd numbers up to 199 is calculated by 

\begin{equation} (\frac{199+1}{2})^2 = 10000 \end{equation}

Second, let’s sum up all odd numbers in the range from 0 to 100. The last odd number is 99. To sum up all odd numbers up to 99 is calculated by 

\begin{equation} (\frac{99+1}{2})^2 = 2500 \end{equation}

The sum of all odd numbers from 100 to 200 is then given by 10000 – 2500 = 7500.

Multiplying by 9

Use the fact that 9 = 10 – 1. So, for example:

• $9*18 = (10-1)*18 = 180 – 18 = 162$
• $9*37 = (10-1)*37 = 370 – 37 = 333$
• $9*382 = (10-1)*382 = 3820 – 382 = 3438$

The same goes for any number that ends with a nine, for example:

• $29 * 33 = (30-1) * 33 = 990 – 33 = 957$
• $39 * 14 = (40-1) * 14 = 560 – 14 = 546$
• $69 * 26 = (70-1) * 26 = (20*70 + 6*70) – 26 = 1400 + 420 – 26 = 1794$

Multiplying by 11

Use the fact that 11 = 10 + 1. So, for example:

• $11*18=(10+1)*18=180+18=198$
• $11*37=(10+1)*37=370+37=407$
• $11*246=(10+1)*246=2460+246=2706$

The same goes for any number that ends with a one, for example:

• $21*14=(20+1)*14=280+14=294$
• $31*52=(30+1)*52=1560+52=1612$
• $51×87=(50+1)×87=\frac{100*87}{2} + 87 = \frac{8700}{2} + 87 =4350 + 87 =4437$

Multiplying by 5

To multiply a number by 5, first multiply it by 10 and then divide by 2.

To calculate $39*5$, do $39*10$ and then $390/2=195$

Dividing by 5

To divide a number by 5, multiply it by 2 and then divide by 10.

To calculate $160/5$, multiply the numerator and denominator by 2, to get $320/10$.

Squaring Numbers Close to 100

For numbers of the form 100 + x, with x being either positive or negative, we have

\begin{equation} (100+x)^2 = 10000 + 200x + x^2 \end{equation}

For example, $97^2$ could be calculated as

\begin{equation} 97^2 = 10000 + 200*(-3) + (-3)^2 \end{equation}

\begin{equation} 97^2 = 10000 – 600 + 9 = 9409 \end{equation}

Percent Tricks

Some percentages should be remembered as follows:

• 1% of a number is just the number divided by 100
• 10% of a number is just the number divided by 10
• 25% of a number is just the number divided by 4
• 50% of a number is just the number divided by 2

From here, every other number can be easily constructed:

• 15% of 200 = 10% of 200 plus half of 10% of 200 = 20 + 20/2 = 30
• 75% of 400 = 3 times 25% of 400 = 3*100 = 300
• 30% of 600 = 3 times 10% of 600 = 3*60 = 180

Quick Fraction Division to Integer Multiplication

When a number is divided by a fraction, it’s good to remember that:

• if a number is divided by $\frac{1}{2}$, it’s the same as multiplying it by 2
• if a number is divided by $\frac{1}{4}$, it’s the same as multiplying it by 4
• if a number is divided by $\frac{1}{X}$, it’s the same as multiplying it by X
• if a number is divided by $\frac{Y}{X}$, it’s the same as multiplying it by X and dividing it by Y

For example:

• $4 / \frac{1}{2} = 4*2 = 8$
• $8 / \frac{2}{7} = 8*7/2 = 28$

Approximating Factorials

Factorials grow too fast to multiply out in your head, so estimate their natural logarithm with Stirling’s approximation:

\begin{equation} \ln(n!) \approx n \ln n − n + \frac{1}{2} \ln(2 \pi n) \end{equation}

In a test there is no time to derive it, so know it by heart, but the idea behind it takes three steps:

1. Turn the product into a sum
   The logarithm of a product is the sum of the logarithms, so $\ln(n!) = \ln 1 + \ln 2 + \dots + \ln n$.
2. Replace the sum by an area
   Between $x = k − 1$ and $x = k$, the area under the curve $\ln x$ is close to $\ln k$, the height at the right edge. So the sum is close to the area under $\ln x$ from 1 to $n$. That area is $(n \ln n − n) − (1 \times \ln 1 − 1) = n \ln n − n + 1$, because the derivative of $x \ln x − x$ is $\ln x + 1 − 1 = \ln x$. This gives the first two terms.
3. Correct the estimate
   A closer estimate of each strip is the average of its two edges, $\frac{1}{2}(\ln(k − 1) + \ln k)$. Adding these averages counts every term of the sum once, except $\ln 1 = 0$ and $\ln n$, which are counted half. So the sum is about $\frac{1}{2}\ln n$ larger than the area. A more careful calculation turns the $+1$ from the area into $\frac{1}{2}\ln(2\pi) \approx 0.92$, and $\frac{1}{2}\ln n + \frac{1}{2}\ln(2\pi) = \frac{1}{2}\ln(2\pi n)$ is the last term.

To use it without a calculator, learn three logarithms by heart:

• $\ln 2 \approx 0.693$
• $\ln 10 \approx 2.303$
• $\ln(2\pi) \approx 1.84$

Three rules for logarithms turn these into the values you need:

• $\ln(a \times b) = \ln a + \ln b$ and $\ln \frac{a}{b} = \ln a − \ln b$. For example, $\ln 100 = \ln 10 + \ln 10 = 4.605$, and $\frac{1}{2}\ln(2\pi n) = \frac{1}{2}\ln(2\pi) + \frac{1}{2}\ln n \approx 0.92 + \frac{1}{2}\ln n$.
• $\ln(a^k) = k \ln a$, because $a^k$ is $k$ copies of $a$ multiplied together.
• $\log_2 x = \frac{\ln x}{\ln 2}$. To see why, write $x = 2^y$ with $y = \log_2 x$. The previous rule gives $\ln x = y \ln 2$.

For example, to estimate $\log_2(100!)$:

1. Work out $n \ln n$
   With $\ln 100 = 4.605$, this is $100 \times 4.605 = 460.5$.
2. Subtract $n$
   This leaves $460.5 − 100 = 360.5$.
3. Add the last term
   Using the shortcut above, $\frac{1}{2}\ln(200\pi) \approx 0.92 + \frac{1}{2} \times 4.605 = 0.92 + 2.30 = 3.22$. So $\ln(100!) \approx 360.5 + 3.22 \approx 363.7$.
4. Convert to base 2
   Divide by $\ln 2 \approx 0.693$. Since $0.693 \times 500 = 346.5$, that leaves $363.7 − 346.5 = 17.2$, and $0.693 \times 25 \approx 17.3$ is just over that. So $\log_2(100!)$ is just under 525: about 524.8.

The exact answer is 524.76. Without the last term you would get $360.5 / 0.693 \approx 520$, so keep it in when the question asks for a decimal.

The last term matters most when factorials are divided by each other, because the large terms cancel. Take the probability of exactly 50 heads in 100 fair coin flips. There are $\binom{100}{50} = \frac{100!}{50! \times 50!}$ ways to choose which 50 of the 100 flips are heads, and each of those sequences has probability $\frac{1}{2^{100}}$, so the probability is

\begin{equation} \frac{100!}{50! \times 50! \times 2^{100}} \end{equation}

Take the logarithm, which turns the division into subtractions: $\ln(100!) − 2 \ln(50!) − 100 \ln 2$. Then apply Stirling’s approximation to both factorials:

1. Cancel the large terms
   The first two terms give $(100 \ln 100 − 100) − 2 \times (50 \ln 50 − 50) = 100 \ln 100 − 100 \ln 50$, because the $−100$ and the $+2 \times 50$ cancel. Adding $−100 \ln 2$ gives $100 \times (\ln 100 − \ln 50 − \ln 2) = 100 \ln \frac{100}{50 \times 2} = 100 \ln 1 = 0$.
2. Simplify the last terms
   What is left is $\frac{1}{2}\ln(200\pi) − 2 \times \frac{1}{2}\ln(100\pi) = \frac{1}{2}\ln(200\pi) − \ln(100\pi)$. Since $\ln(100\pi) = \frac{1}{2}\ln((100\pi)^2)$, this equals $\frac{1}{2}\ln\frac{200\pi}{10000\pi^2} = \frac{1}{2}\ln\frac{1}{50\pi} = −\frac{1}{2}\ln(50\pi)$.
3. Undo the logarithm
   The probability is $e^{−\frac{1}{2}\ln(50\pi)} = (50\pi)^{−\frac{1}{2}} = \frac{1}{\sqrt{50\pi}}$. Since $50\pi \approx 157$ and $12.5^2 = 156.25$, that is about $\frac{1}{12.5} = 0.08$.

The exact answer is 0.0796. Without the last terms the logarithm would be 0, which is a probability of $e^0 = 1$.