Why the methods work

You can already add, subtract, multiply and divide numbers with lakhs and crores in them. You carry, you borrow, you shift rows, you bring digits down. This layer asks a harder question: why are those moves allowed?

That matters for three reasons. First, a method you understand is a method you can repair: if you forget a step, you can rebuild it. Second, once you see why it works, you can invent faster methods of your own, and spot when a "trick" is really a trap. Third, the reasons are beautiful. Every column method in this layer rests on just two ideas: place value (each place is worth ten of the place to its right) and a handful of properties of the operations, especially the distributive property.

At the end we meet the people who worked these methods out, many of them in India, more than a thousand years ago.

Take 2,47,368 + 1,85,974. Write each number in expanded form, place by place:

• 2,47,368 = 2 lakh + 4 ten-thousands + 7 thousands + 3 hundreds + 6 tens + 8 ones
• 1,85,974 = 1 lakh + 8 ten-thousands + 5 thousands + 9 hundreds + 7 tens + 4 ones

Because addition can be done in any order and in any grouping (the commutative and associative properties), we are allowed to add like places together:

• ones: 8 + 4 = 12 ones
• tens: 6 + 7 = 13 tens
• hundreds: 3 + 9 = 12 hundreds
• thousands: 7 + 5 = 12 thousands
• ten-thousands: 4 + 8 = 12 ten-thousands
• lakhs: 2 + 1 = 3 lakhs

That is already the correct answer, but it is written untidily: no place is allowed to hold more than 9. Carrying is simply tidying up. 12 ones = 1 ten + 2 ones, so 1 ten moves up to the tens place. Then the tens hold 13 + 1 = 14 tens = 1 hundred + 4 tens, and so on.

In 5,003 − 1,867, the ones column asks for 3 − 7, which is impossible with whole numbers. Borrowing (also called decomposition or regrouping) rewrites the top number without changing its value:

5,003 = 5 thousands + 0 hundreds + 0 tens + 3 ones = 4 thousands + 9 hundreds + 9 tens + 13 ones

Check: 4,000 + 900 + 90 + 13 = 5003. Same number, different packing. Now every column works: 13 − 7 = 6, 9 − 6 = 3, 9 − 8 = 1, 4 − 1 = 3, giving 3,136.

The long chain of borrowing across zeros, which so many people find confusing, is just unpacking one thousand into 9 hundreds, 9 tens and 10 ones in one go.

There is a completely different way to subtract, called equal additions. Some European schools still teach it instead of borrowing (it is also called the Austrian or additions method), and older relatives may well have learnt to subtract this way. It uses the fact from the Investigate layer: adding the same amount to both numbers does not change the difference.

For 5,003 − 1,867:

• Ones: 3 − 7 won't go. Add 10 ones to the top (13 − 7 = 6) and add 1 ten to the bottom (the 6 tens become 7 tens). Both numbers grew by 10.
• Tens: 0 − 7 won't go. Add 10 tens to the top (10 − 7 = 3) and 1 hundred to the bottom (8 hundreds become 9).
• Hundreds: 0 − 9 won't go. Add 10 hundreds to the top (10 − 9 = 1) and 1 thousand to the bottom (1 thousand becomes 2).
• Thousands: 5 − 2 = 3.

Answer: 3,136, with no crossing-out of zeros anywhere.

The distributive property says that multiplying a sum is the same as multiplying each part and adding: a × (b + c) = a × b + a × c. A tiffin example: 6 children each get 3 idlis and 2 vadas. Total items = 6 × (3 + 2) = 6 × 3 + 6 × 2 = 18 + 12 = 30.

Long multiplication is this property used again and again. Split each number into its places: 347 = 300 + 40 + 7 and 26 = 20 + 6. Then every place of one meets every place of the other, like the cells of a grid. The picture is a rectangle 347 long and 26 wide, cut into six smaller rectangles whose areas add up to the whole.

Now compare with the usual layout:

  347
 × 26
-----
2,082
6,940
-----
9,022

The first row, 2,082, is the bottom row of the grid: 347 × 6. The second row, 6,940, is the top row: 347 × 20. Many people write it as 347 × 2 = 694 and "shift one place left" or "put a zero first". That shift is not a rule to memorise: it happens because the 2 in 26 is really 2 tens, and multiplying by 20 is multiplying by 2 and then by 10. Multiplying by 10 moves every digit one place to the left, because each place is worth ten times the one to its right.

The most basic meaning of 8,736 ÷ 24 is: how many times can you take 24 away from 8,736? You could subtract 24 again and again, 364 times. That works, but it is slow. Long division is the same idea with big chunks: take away 24 hundreds at a time, then 24 tens at a time, then single 24s.

Now look at the usual long-division layout for 8,736 ÷ 24. It hides the zeros, but it is doing exactly the same chunks.

• "24 into 87" really means 24 into 87 hundreds: 3 hundreds, since 24 × 3 = 72. Take away 72 hundreds (7,200). 87 − 72 = 15 hundreds are left.
• "Bring down the 3" means: 15 hundreds and 3 tens make 153 tens. 24 into 153 tens is 6 tens (24 × 6 = 144). 153 − 144 = 9 tens left.
• Bring down the 6: 9 tens and 6 ones make 96 ones. 24 × 4 = 96, nothing left.

So "bring down" is not a mysterious move. It is combining what is left over with the next place, because leftover hundreds are worth ten times as many tens.

Every division of whole numbers, whether it comes out exactly or not, fits one sentence. For any whole number a (the dividend) and any whole number d bigger than 0 (the divisor), there is exactly one pair of whole numbers q (the quotient) and r (the remainder) with

a = d × q + r, where r is 0 or more and less than d.

This fact is called the division algorithm (although it is really a theorem, a proven statement, rather than a method). It says two things: such a q and r exist, and they are unique: there is only one correct pair.

Why a pair exists. List the multiples of d: 0, d, 2d, 3d, … They go up in steps of d and eventually pass a. Let d × q be the last multiple that is not bigger than a. Then r = a − d × q is 0 or more. And r must be less than d: if r were d or more, then d × (q + 1) would still not be bigger than a, so d × q was not the last one after all.

Why the pair is unique. Suppose two different pairs worked: a = d × q + r and a = d × Q + R, with both remainders between 0 and d − 1. Subtracting, d × (q − Q) = R − r. The right side is a gap between two remainders, so it is smaller than d (and bigger than −d). The left side is a multiple of d. The only multiple of d strictly between −d and d is 0, so q = Q and r = R. There was only one pair all along.

Why you cannot divide by zero. Dividing is the inverse of multiplying: 12 ÷ 3 = 4 because 3 × 4 = 12. So 12 ÷ 0 would have to be a number q with 0 × q = 12. But 0 times anything is 0, never 12. No q works. The division has no answer.

What about 0 ÷ 0? Now we need 0 × q = 0, and every number q works: 0 × 5 = 0, 0 × 1,000 = 0. An answer that could be anything is no answer at all. So mathematicians say division by zero is undefined, in both cases. Notice that the division algorithm already quietly excluded it by demanding that d is bigger than 0.

Long before calculators, accountants and astronomers in India, the Arab world and Europe checked their sums with a trick called casting out nines. Add the digits of a number, and keep adding the digits of the result until one digit is left (treat 9 as 0). Call this the digit sum (or digital root).

For 347 × 26 = 9,022:

• digit sum of 347: 3 + 4 + 7 = 14, then 1 + 4 = 5
• digit sum of 26: 2 + 6 = 8
• multiply the digit sums: 5 × 8 = 40, digit sum 4
• digit sum of the answer 9,022: 9 + 0 + 2 + 2 = 13, then 4

The two results match (4 = 4), so the multiplication passes the check. The same works for addition (add the digit sums) and subtraction (subtract them).

Why does it work? Look at what happens to powers of ten when you divide by 9:

10 = 9 + 1, 100 = 99 + 1, 1,000 = 999 + 1, and so on. Every place value is a multiple of 9, plus 1.

So 347 = 3 × 100 + 4 × 10 + 7 = 3 × (99 + 1) + 4 × (9 + 1) + 7 = (a multiple of 9) + 3 + 4 + 7.

That means a number and its digit sum leave the same remainder when divided by 9. The digit sum is really "the remainder on division by 9". And remainders behave well: the remainder of a product is the remainder of the product of the remainders (the same tidying you saw with 53 + 38 and 6 in the last chapter). So if the digit sums disagree, the calculation is definitely wrong.

"12 notebooks cost ₹540. What do 7 notebooks cost?" The unitary method goes through one: 1 notebook costs ₹540 ÷ 12 = ₹45, so 7 notebooks cost 7 × ₹45 = ₹315. It is called unitary because it finds the value of one unit first.

Indian mathematicians had a name for this kind of problem more than 1,400 years ago: trairāśika, the "rule of three", because three quantities are known (12 notebooks, ₹540, 7 notebooks) and the fourth is found. Aryabhata described it around 499 CE, and Brahmagupta and Bhāskara II both used it. Traders carried the rule to the Arab world and then to Europe, where for centuries it was taught as the most important rule in commercial arithmetic.

Some problems are proportional in the opposite direction. "6 workers build a wall in 10 days. How long would 15 workers take?" More workers means fewer days. Count the total work in worker-days: 6 × 10 = 60 worker-days. Shared among 15 workers: 60 ÷ 15 = 4 days. This is called inverse proportion. (It still assumes every worker is equally fast and nobody gets in anyone's way, which real builders might dispute.)

Hard word problems are rarely hard because of the arithmetic. They are hard because there are several steps and it is not obvious which comes first. Two strategies from strong problem-solvers help.

A bar model draws each quantity as a strip. Equal parts get equal strips. The picture shows you which operation to use without any keyword hunting.

Working backwards starts from the end of the story and undoes each step with its inverse operation: undo "spent ₹40" by adding ₹40, undo "halved" by doubling.

India's 2011 Census counted 1,21,08,54,977 people (about 121 crore) in its published final totals. Of these, 62,32,70,258 were male and 58,75,84,719 female. Check the addition yourself: 62,32,70,258 + 58,75,84,719 = 1,21,08,54,977. ✓

The difference, 62,32,70,258 − 58,75,84,719 = 3,56,85,539, is about 3.6 crore, more than the whole population of many countries. Numbers this large are where careful column work and good estimation both matter: a single slipped digit in the crores place is an error of crores of people.

In the 2024 Lok Sabha election, India had about 96.8 crore registered voters (electors), the largest electorate in the world, spread across 543 constituencies. On average that is about 96,80,00,000 ÷ 543 ≈ 17,82,689 electors per constituency (about 17.8 lakh).

That average hides huge differences: some Himalayan and island constituencies have well under 2 lakh electors, while some city constituencies have well over 30 lakh. An average is just the total divided equally, which never happens in real life. Still, it is the right number for a first estimate: for example, if each polling station serves roughly 1,000 to 1,500 electors, you would need something like 96.8 crore ÷ 1,000 ≈ 9.7 lakh polling stations at most, and indeed the Election Commission set up about 10 lakh.

Every method in this topic depends on place value with a zero: the idea that the same digit is worth different amounts in different places, and that an empty place needs a symbol. That idea took shape in India, probably by the 5th or 6th century CE, and it changed calculation forever. With Roman numerals (MMCDLXXV × CCCVI) column arithmetic is almost impossible; with Indian numerals it is something a ten-year-old can learn.

Here is a problem written in the spirit of the Līlāvatī — not one taken from the book itself: "A merchant buys 3 pearls for 2 coins and sells 5 pearls for 4 coins. How many pearls must he trade to make a profit of 14 coins?"

Cost of one pearl: 2 ÷ 3 coin. Selling price of one pearl: 4 ÷ 5 coin. To keep everything whole, think in groups of 15 pearls (15 is a common multiple of 3 and 5): 15 pearls cost 10 coins and sell for 12 coins, a profit of 2 coins. For 14 coins of profit he needs 7 such groups: 7 × 15 = 105 pearls. The trick of choosing a common multiple is exactly what the LCM topic studies.