# Mental Math Tricks

**Mental math tricks** are a collection of techniques, some based on algebraic manipulation and some on visualization, that aid in large arithmetic computations. They are useful for increasing the speed at which one can do these computations, but they also serve as a useful verification mechanism to help reduce computational errors (as results reached by multiple methods are less likely to be in error).

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## Addition and Subtraction Tricks

For the most part, the key to addition/subtraction tricks is as follows:

The order of additions is irrelevant, so a given sum can be rearranged in any desired way.

The usual application of this is to hunt for numbers in the sum that add up to a multiple of 10, as those are easy to visualize and work with.

Determine the value of

\[3 + 8 + 6 + 2 + 17 + 3 + 5 + 1 + 2 + 4.\]

The addition can certainly be carried out directly without too much trouble, but this process is relatively tedious and susceptible to error. A cleaner method is to rearrange the terms:

\[\begin{align*} \color{blue}{3} + \color{red}{8}+\color{green}{6}+\color{red}{2}+\color{blue}{17}+\color{skyblue}{3}+\color{skyblue}{5}+1+\color{skyblue}{2}+\color{green}{4} &=\color{blue}{3}+\color{blue}{17}+\color{red}{8}+\color{red}{2}+\color{green}{6}+\color{green}{4}+\color{skyblue}{3}+\color{skyblue}{5}+\color{skyblue}{2}+1 \\ &=20 + 10 + 10 + 10 + 1 \\ &=51. \ _\square \end{align*}\]

Given the number 100, you are tasked to do the following:

- First, add 7 to the number
- Second, add 10 to the number
- Third, add 6 to the number
- Fourth, add 10 to the number
- Fifth, add 14 to the number
- Sixth, add 10 to the number
- Lastly, add 13 to the number

**What is the result?**

- Mental Math Tricks wiki
- Check out the set.

Another technique is useful in finding the sum of series following some pattern, usually arithmetic progressions or geometric progressions. The general technique is to manipulate the sum into another, related sum, and use the two together to cancel terms and/or achieve the desired repetition. Here is an example:

Gauss's teacher has asked him to add up all the integers between 1 and 100, inclusive. What is this sum?

The key realization is to rearrange the terms to pair up the numbers:

\[\begin{align} S &= 1 + 2 + 3 + \cdots + 98 + 99 + 100\\ &= (1 + 100) + (2 + 99) + (3 + 98) + \cdots + (50 + 51)\\ &= 101 \times 50 \\ &= 5050. \ _\square \end{align}\]

This technique works well when there is an even number of terms. When there is an odd number of terms, the technique is essentially the same, but with a slightly more clever implementation:

Now Gauss's teacher has asked him to add up all the integers between 1 and 101, inclusive. What is this sum?

The "pairing up the terms" intuition can be formalized:

\[\begin{align} S = &1 + 2 + 3 + \cdots + 99 + 100 + 101\\ = &101 + 100 + 99 + \cdots + 3 + 2 + 1\\\\ 2S = &(1 + 101) + (2 + 100) + \cdots + (50 + 52) \\&+ (51 + 51) + (52 + 50) \cdots + (100 + 2) + (101 + 1) \\ = &102 \times 101\\\\ \Rightarrow S =&51 \cdot 101 \\ =& 5151.\ _\square \end{align}\]

This leads to the general result:

An \(n\)-term arithmetic series with first term \(A\) and last term \(B\) has sum

\[\frac{n(A+B)}{2}.\ _\square\]

A geometric series can be dealt with in a similar way. Here is an example:

Gauss's teacher has now asked him to find the sum of the first 10 non-negative powers of 2. What is this sum?

Similar to before, compute

\[\begin{align} S &= 2^0 + 2^1 + \cdots + 2^9\\ 2S &= 2^1 + 2^2 + \cdots + 2^{10}\\\\ \Rightarrow S &= 2S - S \\ &= 2^{10}-2^{0} \\ &= 1023, \end{align}\]

where the general strategy is to multiply by the common ratio and then subtract. \(_\square\)

This leads to the general result:

An \(n\)-term geometric series with first term \(a\) and common ratio \(r\) has sum

\[a \cdot \frac{r^n-1}{r-1}.\ _\square\]

## Multiplication and Division Tricks

Firstly, it is important to note that multiplying and dividing by powers of 10 is quite easy, as multiplication and division by \(10^n\) is equivalent to moving the decimal point \(n\) steps to the right (in the case of multiplication) or left (in the case of division).

What is \(1024.276 \times 10^{5}?\)

The decimal is moved 5 steps to the right, so the result is 102427600. \(_\square\)

As in addition,

The order of multiplications is irrelevant, so a given sum can be rearranged in any desired way.

The usual application of this is to hunt for numbers in the product that multiply to a power of 10, as those are easy to visualize and work with (as shown above).

Compute the value of

\[2 \cdot 3 \cdot 5 \cdot 3 \cdot 2 \cdot 3 \cdot 5.\]

The addition can certainly be carried out directly without too much trouble, but this process is relatively tedious and susceptible to error. A cleaner method is to rearrange the terms:

\[\begin{align*} \color{red}{2} \cdot \color{green}{3} \cdot \color{red}{5} \cdot \color{green}{3} \cdot \color{blue}{2} \cdot \color{green}{3} \cdot \color{blue}{5} &=\color{red}{2}\cdot \color{red}{5}\cdot \color{blue}{2}\cdot \color{blue}{5} \cdot \color{green}{3} \cdot \color{green}{3} \cdot \color{green}{3} \\ &= 10 \cdot 10 \cdot 27 \\ &= 2700. \ _\square \end{align*}\]

Basic multiplication can also be done visually using **multiplication by lines**:

There are also various time-saving tricks available in special cases, as the following sections show.

## Dividing by Numbers Ending in 9

Dividing by integers ending in 9, i.e. 19, 29, 39, etc., can be easily done with the help of geometric series. The general approach is shown by the below example:

What is the value of \(\frac{6}{19}\) to five decimal places?

Firstly, take the numerator, in this case \(6\). Divide the numerator by \(2\) and write the result as \( \color{red}{n} \text{ remainder r}\), in this case \(\color{red}{n}=3\) and \(r=0\). Now divide \((10r+\color{red}{n})\) by \(2\). Write the result below your first calculation, expressing the result in the same format. Repeat these steps until the required accuracy is achieved (if you need 5 decimal places, do 5+1=6 lines of working):

\[\begin{align}&&\dfrac{6}{2} \quad \, \, &=& \color{red}{3} \text{ remainder 0} \\ \\ \dfrac{10(0)+3}{2} \quad &=& \dfrac{3}{2} \quad \, \, &=& \color{red}{1} \text{ remainder 1} \\ \\ \dfrac{10(1)+1}{2} \quad &=& \dfrac{11}{2} \quad &=& \color{red}{5} \text{ remainder 1} \\ \\ \dfrac{10(1)+5}{2} \quad &=& \dfrac{15}{2} \quad &=& \color{red}{7} \text{ remainder 1} \\ \\ \dfrac{10(1)+7}{2} \quad &=& \dfrac{17}{2} \quad &=& \color{red}{8} \text{ remainder 1} \\ \\ \dfrac{10(1)+8}{2} \quad &=& \dfrac{18}{2} \quad &=& \color{red}{9} \text{ remainder 0} &. \end{align} \]

Here's where the magic happens.

List the values of \(\color{red}{n}\) for each line, and this is the decimal expansion of \(\frac{6}{19}\).Therefore we can say \(\frac{6}{19}=0. \color{red}{315789}= \color{green}{\boxed{0.31579 \, \text{(5 d.p.)}}}\). \(_\square\)

Dividing by larger numbers, such as 29, 39, etc., can be done in a similar method: to divide by \(10n+9\), divide by \(n+1\) (rather than 2) at each step in the above process. It is also worth noting that this trick demonstrates that the resulting decimal expansion is periodic.

Unsurprisingly, this trick doesn't simply work by magic; there is a much deeper reason why it holds:

Consider the following:\[\dfrac{x}{19}=\dfrac{x}{20-1}=\dfrac{\frac{x}{20}} {1-\frac1{20}}.\]

By the geometric sum, we can write this as

\[\begin{align} \dfrac{\frac{x}{20}}{1-\frac1{20}} &=\left(\dfrac{x}{20}\right)\left(1+\dfrac{1}{20}+\dfrac{1}{20^2}+\cdots\right)\\ &=\dfrac{\hspace{3mm} \frac{x}{2^1}\hspace{3mm} }{10^1}+\dfrac{\hspace{3mm} \frac{x}{2^2}\hspace{3mm} }{10^2}+\dfrac{\hspace{3mm} \frac{x}{2^3}\hspace{3mm} }{10^3}+\cdots. \end{align}\]

We see two consecutive terms satisfy

\[\dfrac{\hspace{3mm} \frac{x}{2^n}\hspace{3mm} }{10^n}+\dfrac{\hspace{3mm} \frac{x}{2^{n+1}}\hspace{3mm} }{10^{n+1}}=\dfrac{\hspace{3mm} \frac{x}{2^n}\hspace{3mm} }{10^n}+\dfrac{\hspace{3mm} \frac{\hspace{3mm} \frac{x}{2^n}\hspace{3mm} }2\hspace{3mm} }{10^{n+1}},\]

which means the next digit will be the previous divided by 2 and, in case of remainder, we will take common denominator with the next term which has \( {10}^{n+2} \), so we multiply the remainder by 10. \(_\square\)

## Product of Numbers Close to 100

When two numbers are sufficiently close to 100, their product can be easily calculated "component-wise." In other words, multiplying out the product \((100-a)(100-b)\) results in terms too far apart to affect one another, so they give the digits of the product. More formally,

Let \(a,b\) be integers such that \(a,b<100\). The result of \((100-a)(100-b)\) is formed as follows: the first two digits are \(100-(a+b)\), and the last two digits are the product of \(a\) and \(b\).

This is better demonstrated by examples:

What is \(94 \times 98?\)

We have 2 numbers, 98 and 94. Now, to calculate this rather hard-looking product, we use a trick. First, find the difference between this and 100: \(100-94=6\) and \(100-98=2\). Now notice that \(94-2=98-6=92.\) (This will always be true for any number choice.) Our Last 2 digits are \(6\times 2=12\) and the first two are \(92\), so the answer is \(94 \times 98=9212\). \(_\square\)

What is \(97 \times 94?\)

We have 2 numbers, 97 and 94. Now, to calculate this rather hard looking product, we use a trick. First, find the difference between this and 100: \(100-94=6\) and \(100-97=3\). Now notice that \(94-3=97-6=91\). Our Last 2 digits are \(6\times 3=18\) and the first two are \(91\), so the answer is \(94 \times 97=9118\). \(_\square\)

If you are interested in the proof, check this below:

Assumption: When saying the integer is close to 100, in this example, it means the integers from 91 to 99, so their product will always be 4 digits.Let both \(a\) and \(b\) be some integers from 1 to 9, then we have

\[\begin{align} (100-a)(100-b) &= 10000-100(a+b)+ab\\\\ &=\underbrace{100\big(100-(a+b)\big)}_{\text{first two digits due to multiplication by 100}}+\underbrace{ab}_{\text{last two digits}}.\ _\square \end{align}\]

## Fractions and Decimals Tricks

**Turning recurring decimals into fractions**

If

\[0.12341234\ldots=\dfrac{a}{b}\]

with \(a\) and \(b\) being coprime positive integers, find \(a\) and \(b.\)

We need

- the number before the decimal: \(0\)
- the number before recurrence starts and after the decimal point: \(1234\)
- the length of that number: \(4.\)
Then we have

\[0.12341234\ldots=0+\dfrac{1234}{10^{4}-1}=\dfrac{1234}{9999}.\]

So, \(a=1234\) and \(b=9999\). \(_\square\)

If a number can be written as \(0.aaaaa\ldots\) with a length of \(n,\) then it can be written as\[\sum_{k=1}^\infty \dfrac{a}{10^{kn}}.\]

Using the infinite geometric progression sum, this is

\[\dfrac{a}{10^n-1}.\ _\square\]

## Squaring Numbers

**Numbers that end in 5:**

What is \({75}^{2}?\)

In order to square a number ending in 5, take the digits before the 5 (in this case it's 7, but let's call it \(\color{red}{X}\) so we can generalize the rule). The square number is equal to \(100\color{red}{X}(\color{red}{X}+1) + 25\).

In other words, you take \(\color{red}{X}\), multiply it by \(\color{red}({X}+1)\) and write \(25\) at the end of that number.Using the rule we just read, we can write \(75^2\) as \(100 \times 7 \times 8 + 25 = \color{green}{\boxed{5625}}\). \(_\square\)

As always, there's no magic involved here; some algebraic expansion shows why this result is generally true:

Let the number we are trying to square be an \((n+1)\)-digit number of the form \(\overline { { X }_{ 1 }{ X }_{ 2 }{ X }_{ 3 }\dots { X }_{ n }5 }. \) Then our goal is to prove that\[{ \left( \overline { { X }_{ 1 }{ X }_{ 2 }{ X }_{ 3 }\dots { X }_{ n }5 } \right) }^{ 2 }=100\left( \overline { { X }_{ 1 }{ X }_{ 2 }{ X }_{ 3 }\dots { X }_{ n } } \right) \left( \overline { { X }_{ 1 }{ X }_{ 2 }{ X }_{ 3 }\dots { { X }_{ n } } } +1 \right) +25,\]

where \({ X }_{ 1 },{ X }_{ 2 },{ X }_{ 3 },\dots ,{ X }_{ n }\) represent digits. Let \(\overline { { X }_{ 1 }{ X }_{ 2 }{ X }_{ 3 }\dots { X }_{ n } }=n\). Then we have

\[\begin{align} {\left( \overline { { X }_{ 1 }{ X }_{ 2 }{ X }_{ 3 }\dots { X }_{ n }5 } \right) }^{ 2 } &={(10n+5)}^{2}\\ &=100n(n+1)+25 \\ &= 100\left( \overline { { X }_{ 1 }{ X }_{ 2 }{ X }_{ 3 }\dots { X }_{ n } } \right) \left( \overline { { X }_{ 1 }{ X }_{ 2 }{ X }_{ 3 }\dots { { X }_{ n } } } +1 \right) +25, \end{align}\]

which is the required result and completes the proof. \(_\square\)

\[\]
**Squaring numbers up to 25:**

When \(n\) is this small, the following procedure returns \(n^2\):

- Add the units digit of \(n\) to itself \((\)e.g. if \(n=13\), the result here is \(13+3=16).\)
- Multiply by 10 \((\)e.g. if \(n=13\), the result here is \(10 \cdot 16 = 160)\) or 20, depending on the tens digit of the number.
- Add the square of the units digit \((\)e.g. if \(n=13\), the result here is \(160 + 3^2 = 169).\)

Indeed, 169 is equal to \(13^2\).

Find \(\ 24^2.\)

We follow the procedure above:

- Add the units digit to the number: \(24+4=28.\)
- Multiply it by 20: \(28\times 20=560.\)
- Add the square of the units digit to the number: \(560+4^2=576. \ _\square\)

We have\[\begin{align} n^2 &=10\big(n+(n-10)\big)+(n-10)^2\\ &=20n-100+n^2-20n+100\\ &=n^2\\ \\ n^2&=20\big(n+(n-20)\big)+(n-20)^2\\ &=40n-400+n^2-40n+400\\ &=n^2. \ _\square \end{align}\]

\[\]
**Squaring numbers between 25 and 75:**

For these numbers, a slightly different procedure can be followed:

- Find the difference between the number and 25 and multiply it by 100.
- Find the difference between the number and 50 and square it, using the procedure from the last section (squaring numbers below 25) if necessary.
- Add the two numbers.

Find \({39}^{2}.\)

We follow the above procedure:

- Find the difference between 25 and the number and multiply by 100: \((39-25)\times 100=1400.\)
- Find the difference between the number and 50 and square it: \((50-39)^2=11^2=121.\)
- Add all these: \(1400+121=1521=39^2.\ _\square\)

We have\[n^2=100(n-25)+(n-50)^2=100n-2500+n^2-100n+2500=n^2.\ _\square\]

\[\]
**Squaring numbers between 75 and 125:**

Finally, these numbers can be squared according to the following procedure:

- Find the difference between the number and 100
- Subtract this result from the original number and multiply by 100
- Square the result from step 1, using the section on squaring numbers below 25 if necessary.
- Add the results from steps 2 and 3.

Find \({109}^{2}.\)

We follow the procedure below to obtain

- Find the difference with 100: \(100-109=-9.\)
- Subtract this from the original number and multiply by 100: \(100\times\big(109-(-9)\big)=11800.\)
- Square step 1: \((-9)^2=81.\)
- Add step 2 and 3: \(109^2=11800+81=11881. \ _\square\)

We have\[n^2=100\big(n-(100-n)\big)+(100-n)^2=200n-10000+n^2-200n+10000=n^2.\ _\square\]

## Finding the Next Square

To find \( (n + 1 ) ^{2} \) when \( n^{2} \) is known, the identity

\[ (n + 1)^{2} = n^2+2n+1 = n^{2} + n + (n + 1 ) \]

is incredibly useful.

For example, if \( 85 ^{2} = 7225\), what is \( 86 ^{2}?\) We have

\[\begin{align} 86 ^{2} &= 85^{2} + 85 + 86 \\ \phantom{86^{2}}&= 7225 + 85 + 86 \\ \phantom{86^{2}} &= 7396. \end{align}\]

This is especially useful, for instance, in calculating squares of numbers that *don't* end in 5: since numbers ending in 5 can be easily squared (see the previous section), it may be easier to square a multiple of 5 near the desired number, and then continually find the next square until the desired result is reached.

## Estimating Square Roots

Estimating square roots is useful for a variety of reasons, especially in geometry where approximating lengths (which often involves square roots by virtue of the Pythagorean theorem) is useful. One general procedure gives a fairly good approximation (certainly good enough for "by hand" use):

- Find the closest square to the number being rooted, \(n\).
- Divide \(n\) by the square root of the closest square.
- Take the average of the result and the square root of the closest square.

Find an approximation for \(\sqrt{\color{red}{123}}\).

We follow the above procedure:

- Take the closest square: \(121.\)
- Divide \(\color{red}{123}\) by the square root of the closest square: \[\dfrac{123}{11}=11+\dfrac{2}{11}=11.1818.\]
Use the repeating decimals into fractions trick backwards!- Take the average of the answer above and the square root of the closest square: \[\dfrac{(11+11.1818)}{2}=11.0909=\color{green}{11.091}.\]
We now have a good estimate for \(\sqrt{123}\). \(_\square\)

Note: \(\sqrt{123}=11.090536506\ldots=\color{green}{11.091},\) so we achieved an accuracy of 3 decimal places!

## Calculating Integer Roots

**Square Roots:**

**A.** The trick to working with square roots is to know what range the square root is in. Since squaring numbers ending in 0 and squaring numbers ending in 5 are both easy, the answer can be calculated within a range of 5 relatively easily.

**B.** If the number is known to be a perfect square, then the ending digit can be determined by looking at the last digit of the number. More specifically, if the number ends in a

- 0, then the ending digit is a 0;
- 1, then the ending digit is 1 or 9;
- 4, then the ending digit is 2 or 8;
- 5, then the ending digit is a 5;
- 6, then the ending digit is 4 or 6;
- 9, then the ending digit is 3 or 7.

**C.** After finding the last digit (or possibility between two digits), mentally chop off the last two digits and focus on the remaining digits.

**D.** Now, try to find a range of 5 that the number (in step C) is in. Once you do this, you know the answer using step B. First, find a range of 10, then find out if the answer is in the upper half of the range (i.e. it ends in 5, 6, 7, 8, or 9) or if the answer is in the lower half of the range (i.e. it ends in 0, 1, 2, 3, or 4) by squaring the middle number (the number in the range that ends in 5).

\[\]
**Cube Roots:**

Each units digit of a perfect cube is the same as the units digit of the number being cubed, (except for 2, 3, 7, and 8 which end in 8, 7, 3, and 2, respectively). Using this schema, you can quickly identify the units digit of the cube root given. Then, examine the block of numbers to the left of the hundreds digit. Identify the largest perfect cube that is smaller than this block and take its cube root. Take the result and place the units digit found earlier on the end to get the result.

Quickly compute \(\sqrt[3]{226981}\).

The last digit is 1. Therefore, \(\sqrt[3]{226981}\) ends in 1.

The block of digits to the left of the hundreds place is 226. The largest perfect cube smaller than 226 is 216, which is 6 cubed. Therefore, \(\sqrt[3]{226981}\) begins with 6.

We conclude that \(\sqrt[3]{226981} = 61. \ _\square\)

Quickly compute \(\sqrt[3]{148877}\).

Here, 7 is in the last digit, which means \(\sqrt[3]{148877}\) ends in a 3.

The block of digits on the left form 148. Since 125 is the largest perfect cube smaller than 148, the leading digit is 5.

Therefore, \(\sqrt[3]{148877} = 53. \ _\square\)

## Miscellaneous Tricks

**Add square numbers of a particular form:**

Adding square numbers of the form \((10a + b)^2+ \big[10(b-1) + (10-a)\big]^2 = 101\big(a^2+ b^2\big):\)

**A.** This method is simple once we reduce to this form:

\[(10a + b)^2+ \big[10(b-1) + (10-a)\big]^2 = 101\big(a^2+ b^2\big).\]

**B.** Using numbers instead of variables, we get the following:

- Square the ones digit on the left number.
- Square the tens digit on the left number.
- Add up the results of step 1 and step 2.
- Multiply the result of step 3 by 101 for the answer. See multiplying by 101.

**C.** This method is sometimes hard to recognize. If the inside numbers subtract to 1 and the outside numbers add to 10, then you can use this method.

For example,

\[97^2+61^2=101(9^2+7^2)=13130.\]

**Cite as:**Mental Math Tricks.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/mental-math-tricks/