Why Do the ‘Vedic Mathematics’ Tricks Work?

Here is your cousin’s subtraction. To take 386 from 1,000, take each digit from 9, except the last, which you take from 10:

Digit of 386	3	8	6
Taken from	9	9	10
Gives	6	1	4

So 1,000 − 386 = 614. In the book called Vedic Mathematics, the rule is a sutra, nikhilam navatashcaramam dashatah, “all from 9 and the last from 10”.

Why it works

Write 1,000 as 999 + 1. Then

$$1{,}000 - 386 = (999 - 386) + 1.$$

Taking a number from 999 never needs a borrow, since no digit is bigger than 9: each column is just 9 minus the digit, 613. Adding the 1 back changes only the last digit, 614, as if it had been taken from 10.

The same holds for any number of digits. If N has n digits (zeros in front allowed), then 10ⁿ − N = (99…9 − N) + 1, with n nines. The number 99…9 − N, every digit taken from 9, is called the nines’ complement of N; one more than it, 10ⁿ − N, is its tens’ complement. When N ends in zeros, say 340, the zeros come out first: 1,000 − 340 = 10 × (100 − 34), which is why the trick skips them. Our program checked the rule, zeros and all, for every number below 100,000.

The book’s main use for the same sutra is multiplying numbers near 10, 100 or 1,000. If you have read How Can You Multiply 98 × 97 in Your Head?, you know it: for 98 × 97, both 2 and 3 short of 100, the answer is 95 hundreds and 2 × 3 = 6, so 9,506. The book sets it out like this:

Number	Short of 100
98	−2
97	−3
95	06

The left answer is either number less the other’s shortfall (98 − 3 or 97 − 2); the right is the shortfalls multiplied, filling as many places as the base has zeros.

The identity behind it

Call the base x and the shortfalls a and b. Expanding the brackets, as in Grade 8,

$$(x - a)(x - b) = x^2 - ax - bx + ab = x\,(x - a - b) + ab.$$

The left part, x − a − b, is the “left answer”, counted in units of the base; ab is the right. This is an identity: true for every value of x, a and b, so the trick can be trusted for any base, and for numbers over the base too, where a shortfall is negative. Three cases from one line:

• Base 1,000. 996 × 988: shortfalls 4 and 12, so 984 thousands and 48 in three places, 048: 984,048.
• Over the base. 104 × 107: shortfalls −4 and −7, so 104 + 7 = 111 hundreds and (−4)(−7) = 28: 11,128.
• One over, one under. 103 × 98: shortfalls −3 and 2, so 101 hundreds and (−3)(2) = −6: 10,100 − 6 = 10,094.

The base needn’t be a power of ten: with 50, 49 × 48 is 47 fifties and 1 × 2 = 2, so 2,350 + 2 = 2,352.

Where it stops helping

The identity never fails, but the trick only helps when the corner ab is easy. For 63 × 58 near 100 the corner is 37 × 42 = 1,554, harder than the question you started with. Roughly, the trick wins when both numbers are within about 10 of the same base.

Now the one-line multiplication, urdhva-tiryagbhyam, “vertically and crosswise”. For 43 × 21:

1. Vertically on the right: 3 × 1 = 3. Write 3.
2. Crosswise: 4 × 1 + 3 × 2 = 10. Write 0, carry 1.
3. Vertically on the left: 4 × 2 = 8, plus the carry, 9. Write 9.

Answer: 903. With bigger digits the carries pile up. For 57 × 46: 7 × 6 = 42 (write 2, carry 4), then 5 × 6 + 7 × 4 = 58, plus 4 is 62 (write 2, carry 6), then 5 × 4 = 20, plus 6 is 26: 2,622.

Why it works: a number is a polynomial

The number 43 means 4 × 10 + 3. Write the ten as x, and 43 is the polynomial 4x + 3; 21 is 2x + 1. Multiply the polynomials, as in Optional Math’s Polynomial:

$$(4x + 3)(2x + 1) = 8x^2 + (4 \times 1 + 3 \times 2)\,x + 3 = 8x^2 + 10x + 3.$$

The three columns are the three coefficients: the x² term comes from the two left digits (vertically), the x term from the two crossed pairs (crosswise), the number term from the right digits. Put x = 10 back: 800 + 100 + 3 = 903. In letters, for any two-digit numbers,

$$(ax + b)(cx + d) = ac\,x^2 + (ad + bc)\,x + bd.$$

For three digits, (ax² + bx + c)(dx² + ex + f) has five coefficients, and the column for xᵏ collects every pair of digits whose places add up to k. That is the whole sutra: multiplying polynomials, with x = 10, and the carrying done as you go. The widget’s red lines are those pairs.

Same work, different bookkeeping

Count the work. Two n-digit numbers have n × n pairs of digits, and each pair is multiplied once whichever method you use: long multiplication goes row by row, the one-line method column by column. What differs is where the adding happens: on paper at the end, or in your head as you go. For 9,999 × 9,999 the middle column is 81 + 81 + 81 + 81 = 324, plus the carry, held in your head while you keep your place among the crossed lines.

That is exactly what the mathematician S. G. Dani found when he looked at the whole book in 1993: the methods “turn out to be quicker only for certain special classes of examples. For a general example the amount of effort involved (for instance, the count of the individual operations needed to be performed with digits, in arriving at the final answer) is about the same as required by the standard methods”. For two-digit numbers with small digits, one line is quicker. For four digits, it is the same work with less paper and more to remember: “faster” in a demonstration usually means “fewer things written down”.

Fewer than n × n?

In 1960 a 23-year-old student in Moscow, Anatoly Karatsuba, saw that the crosswise sum needs only one new product, not two:

$$ad + bc = (a + b)(c + d) - ac - bd.$$

For 57 × 46: ac = 20, bd = 42, (5 + 7)(4 + 6) = 120, so the middle is 120 − 20 − 42 = 58. Three products instead of four; used again and again on halves of long numbers, it was the first method ever found that beats n × n.

The polynomial view explains a famous pattern. Square 11, 111, 1,111…

	Square
11²	121
111²	12,321
1,111²	1,234,321
11,111²	123,454,321
111,111,111²	12,345,678,987,654,321

This is the twist in every “digit” trick. The polynomial identity never breaks; what breaks is reading the answer off as digits, the moment a coefficient reaches 10. Every carry in the one-line method is that moment.

Here are the book’s most famous tricks side by side. The first two columns are true for every number; the last two say when the trick is worth using.

Trick	The identity (x = 10, 100, …)	Quick when	No help when
All from 9, the last from 10	10ⁿ − N = (99…9 − N) + 1	always: it is just subtracting from 1,000…	subtracting from a number that isn’t a power of ten
Near a base (nikhilam)	(x − a)(x − b) = x(x − a − b) + ab	both numbers near the same base	the shortfalls are big
Vertically and crosswise	(ax + b)(cx + d) = acx² + (ad + bc)x + bd	two digits, small digits	many digits: same products, bigger sums in your head
Ending in 5 (ekadhikena purvena, “by one more than the one before”)	(10t + 5)² = 100t(t + 1) + 25	squares of numbers ending in 5	any other last digit

So: the tricks work, and some are clever; several have older Indian relatives; the book is from the 20th century. Dani, who wrote the sharpest criticism of its title, still said that “some of the observations could be used in teaching in schools. They are entertaining and could to some extent enable children to enjoy mathematics”, best used to teach the ideas behind them, as here, “rather than like a series of tricks of magic”.

On 18 June 1980, at Imperial College London, Shakuntala Devi was given two 13-digit numbers picked by the college’s computer department, 7,686,369,774,870 and 2,465,099,745,779, and in 28 seconds gave the product, 18,947,668,177,995,426,462,773,730: a world record (our program agrees with every digit). She had trained since childhood, giving shows with her father.

Some of her famous feats are easier than they look. In 1988 the psychologist Arthur Jensen, in California, handed her two cards: the cube root of 61,629,875 and the seventh root of 170,859,375. She answered 395 and 15 before he could write the questions down. You can learn to do the first kind.

Cube roots in two looks

Two facts make it work.

• The last digit. The last digit of a product depends only on the last digits of the numbers multiplied. The cubes of 0 to 9 end in 0, 1, 8, 7, 4, 5, 6, 3, 2, 9: ten different digits. So the last digit of a cube names the last digit of its root: 2 and 8 swap, 3 and 7 swap, the rest stay. A cube ending in 1 has a root ending in 1.
• The first digit. If the root is 10t + u, its cube is between (10t)³ = 1,000t³ and (10t + 10)³ = 1,000(t + 1)³. Cover the last three digits of the cube and what is left lies between t³ and (t + 1)³. For 357,911: 357 lies between 7³ = 343 and 8³ = 512, so t = 7, and the last digit 1 gives 71.

Jensen’s 61,629,875 has a three-digit root, so the middle digit needs a further step, but the start is the same: it ends in 5, so the root does, and 61,629,875 is just under 64,000,000 = 400³, so the root is just under 400: 395. As the writer Peter Magnusson points out, the problem was easy for the method, if not for Jensen’s calculator.

The 23rd root

The most famous story is from 1977, at Southern Methodist University in Dallas: Devi gave the 23rd root of a 201-digit number, 546,372,891, in 50 seconds, and the reports said a UNIVAC computer took 62. Treat it with care. Magnusson notes that the UNIVAC 1101 named was a machine of the early 1950s, so that detail is likely wrong, and that the computer had the harder job: raising a number to the 23rd power, not rooting a number known to be a perfect power.

The method is the cube root’s, stretched. The last four digits of the power fix the last four of the root: our program checked that among numbers ending in 1, 3, 7 or 9, no two different last-four-digit endings give the same ending when raised to the 23rd power, and 2891²³ ends in 6771, as the 201-digit number did. The first few digits come from logarithms, as in How Did a Book of Numbers Make Multiplying Easy?: the number is about 9.167 × 10²⁰⁰, its logarithm about 200.962, and a 23rd of that is about 8.7375, so the root is about 10⁸·⁷³⁷⁵, roughly 546 million. Only the middle digits are truly hard: a craft, not a mystery.

Memory, not magic

In 2001 a team led by Mauro Pesenti in Belgium scanned the brains of a German mental calculator, Rüdiger Gamm, and six people with ordinary maths skills while they calculated. The expert did not use the same brain areas harder: he used different ones, switching between effortful short-term holding of numbers and long-term memory, where he stored and fetched the steps of a long sum. Gamm, a news report of the study notes, showed no aptitude for maths until he was 20 and trained four hours a day. The psychologist Brian Butterworth’s reading of it: “The study shows that anyone can develop exceptional capacities if they work at it.”

• How Can You Multiply 98 × 97 in Your Head? (Grades 6–8) The near-base rule and squaring tricks as pictures of rectangles cut up and moved, with Bhaskara’s own examples.
• How Did People Multiply Before Calculators? (Grades 6–8) Doubling and halving, the lattice, Napier’s bones, and Karatsuba’s three products.
• Is the Answer Roughly Right? (Grades 6–8) Checking any trick’s answer in seconds by casting out nines.
• How Did a Book of Numbers Make Multiplying Easy? Logarithms, which give the first digits of roots like the 23rd root here.
• How Did a Student Beat a Great Mathematician at Multiplying? (Computing, Going Further 9–12) Karatsuba’s method, and how computers multiply numbers with millions of digits. (Coming soon.)
• Grade 10 Optional Math: Polynomial Multiplying and dividing polynomials, the algebra every trick here rests on.