Mental maths and numerical fluency

2 lessons · 5 equations. Each lesson below gives its formulas and key rules; open the lesson for the full explanation.

Arithmetic that survives a clock

Squares by difference

n^2 = (n + d)(n - d) + d^2

Pick $d$ so that one factor becomes a round number. This turns a square into a product you can do in one step.

Doubling time

t_{\text{double}} \approx \frac{72}{r}

The rule of 72 approximates $\ln 2 / \ln(1+r)$ . Seventy-two is used rather than the more accurate $69.3$ because it divides cleanly by 2, 3, 4, 6, 8, 9 and 12.

Products around a midpoint

(m - d)(m + d) = m^2 - d^2

When two factors sit the same distance either side of a round number, the product is one square you know minus one small square. $63 \times 57$ is $60^2 - 3^2$ ; $98 \times 102$ is $100^2 - 2^2$ . It only works when the two numbers have the same parity, so that the midpoint is a whole number — otherwise shift one factor by one and correct.

Remember

Round to a base and correct; never long-multiply under a clock.

Estimation: decompose, bound, and defend the number

The middle of a range is geometric

\hat x = \sqrt{a\,b}

When a quantity could plausibly be anywhere from $a$ to $b$ across orders of magnitude, the sensible central estimate is the geometric mean, not the arithmetic one. Between $10$ and $1000$ the midpoint is $100$ , not $505$ : you are equally likely to be a factor of ten out either way.

How factor errors combine

\sigma_{\ln N} = \sqrt{\sum_i \sigma_{\ln f_i}^2}

If each factor in a product is uncertain by a multiplicative amount, the errors add in quadrature in logs — they partly cancel. Four factors each good to a factor of two give a total good to a factor of four, not sixteen.

Remember

Decompose into a product of anchorable factors; errors partly cancel.

Arithmetic that survives a clock

Estimation: decompose, bound, and defend the number

Detailed formula cards