MathsCBSE Class 9 · The World of Algorithms

Column Addition, Lists and Algorithm Efficiency: Class 9

Updated 17 October 2026 · Fact-checked

An algorithm is a precise step-by-step procedure. Column addition is correct because place value lets each column be added separately, and a carry of at most 1 moves to the next column (the largest column sum is $9 + 9 + 1 = 19$). Its work grows with the number of digits, not with the size of the numbers. Keeping intermediate results in a sorted list lets an algorithm find common elements in one pass instead of comparing every pair.

Add one column at a time from the right: write the units digit, carry the 1, and see the carry can never be more than 1 because 9+9+1=19.

Illustration drawn for QuestionBuddy.

Understand Column Addition, Lists and the Efficiency of Algorithms

*Note: this chapter of the textbook series is not listed in CBSE's Class 9 course structure for 2026-27, so check with your school whether it is examined. It develops computational thinking.*

What an algorithm is. An algorithm is a precise sequence of steps, each one a basic action we can already do, that solves every problem of a certain kind. It uses repetition (do the same step for each column) and conditions (if the sum is 10 or more, carry 1). Two questions apply to every algorithm: *is it correct for all inputs?* and *how much work does it do as the input grows?*

Column addition. To add $4768 + 5385$ , align the digits by place value and add from the right, writing the units digit of each column sum and carrying the tens digit:

units: $8 + 5 = 13$ , write $3$ , carry $1$ ;
tens: $6 + 8 + 1 = 15$ , write $5$ , carry $1$ ;
hundreds: $7 + 3 + 1 = 11$ , write $1$ , carry $1$ ;
thousands: $4 + 5 + 1 = 10$ , write $0$ , carry $1$ ;
the final carry becomes a new leading digit: $10153$ .

A shorter number is padded with zeros on the left: $7 + 9856$ is treated as $0007 + 9856 = 9863$ .

Why it is correct. Place value separates a number into units, tens, hundreds and so on. Adding column by column adds each place separately; when a column total reaches ten or more, ten units of one place make one unit of the next place, which is exactly the carry. The carry can never exceed $1$ because the largest possible column sum is $9 + 9 + 1 = 19$ . Leaving out the final carry gives a wrong answer: $5000 + 5000$ would give $0000$ . The sum of two numbers has at most one more digit than the longer number: $99999 + 999 = 100998$ .

Why it is fast. Suppose, as a made-up illustration, that adding two numbers by counting dots takes one second per dot and adding one column takes 15 seconds. Counting for numbers near $9999 + 9999 = 19998$ takes $19998$ seconds, which is $5$ hours $33$ minutes $18$ seconds, while four columns take about $60$ seconds. Doubling the number of digits to $8$ doubles the column work to $120$ seconds, but the counting work multiplies by about ten thousand. So the work of column addition grows with the number of digits, not with the value.

Lists as data structures. Intermediate results can be stored in an organised list: a data structure. A list of divisors written in increasing order helps. To find the divisors of $36$ , test numbers only up to $6$ (since $6 \times 6 = 36$ ) and pair each divisor with its partner: $(1, 36), (2, 18), (3, 12), (4, 9), (6, 6)$ gives all $9$ divisors in $6$ checks. For $60$ , test up to $7$ : pairs $(1, 60), (2, 30), (3, 20), (4, 15), (5, 12), (6, 10)$ give $12$ divisors in $7$ checks.

Walking two sorted lists. Take the sorted divisor lists of $36$ and $60$ :
$36$ : $1, 2, 3, 4, 6, 9, 12, 18, 36$ ; $60$ : $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$ .
Place a marker at the start of each list; compare the two numbers; if equal, record it and advance both; otherwise advance the marker at the smaller number. In one pass we get the common divisors $1, 2, 3, 4, 6, 12$ , so $g cd (36, 60) = 12$ and $lcm = \frac{36 \times 60}{12} = 180$ . Comparing every pair would need $9 \times 12 = 108$ comparisons, while the marker method needs at most $9 + 12 = 21$ . For two sorted lists $[2, 3, 5, 7, 11, 13]$ and $[1, 3, 4, 5, 9, 11, 12]$ the one-pass method finds the common elements $3, 5, 11$ in $9$ comparisons, instead of the $6 \times 7 = 42$ pairs.

Writing an algorithm. Use numbered steps, name the intermediate values, state the stopping condition, and run one example in a table. To find an lcm: (1) find $g cd (m, n)$ by the division method; (2) compute $\frac{mn}{g c d}$ . For $18$ and $24$ , $g cd = 6$ and $lcm = 72$ .

Column Addition, Lists and the Efficiency of Algorithms: formulas to remember

Carry in addition: $column sum \leq 9 + 9 + 1 = 19 \Rightarrow carry \leq 1$; The carry is 0 or 1.
Length of a sum: $digits (a + b) \leq max (digits (a), digits (b)) + 1$; One extra digit at most.
Work of column addition: $steps \approx number of columns$; Grows with digits, not value.
Divisor pairs: $d \leq n \Rightarrow pair (d, n / d)$; Test only up to $n$ .
Two sorted lists: $comparisons \leq ∣ A ∣ + ∣ B ∣$; One pass with two markers.
Using the gcd: $lcm (m, n) = \frac{mn}{g c d ( m , n )}$; From the previous page.

How to solve Column Addition, Lists and the Efficiency of Algorithms questions

Trace the algorithm step by step in a table and state why it works.

1Write the numbers one under the other with digits aligned by place value; pad shorter numbers with zeros.
2For each column, compute (digit + digit + carry), write the units digit and carry the tens digit.
3After the last column, write any final carry as a new leading digit.
4For lists, sort the intermediate results and walk through them with markers.
5Count the steps or comparisons and compare with the naive method.
6Give a one-line justification of correctness and a note on how the work grows.

Column Addition, Lists and the Efficiency of Algorithms in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Column addition

Suppose a cashier adds two bills of 4768 rupees and 5385 rupees using the column method.

Show the carries and the total.

Try it first, then show the solutionSolution

Units: $8 + 5 = 13$ , write 3, carry 1.
Tens: $6 + 8 + 1 = 15$ , write 5, carry 1.
Hundreds: $7 + 3 + 1 = 11$ , write 1, carry 1.
Thousands: $4 + 5 + 1 = 10$ , write 0, carry 1; the final carry is the leading 1.

Answer: The total is 10153 rupees.

Memory hook: Write the units digit, carry the rest.

Tricks to remember Column Addition, Lists and the Efficiency of Algorithms

Short memory aids. Say each one aloud once, then use it on a question the same day.

Carry is 0 or 1
The largest column sum is 19, so you never carry more than 1.
Use it when: Justifying the algorithm.
Digits, not value
Column addition takes one step per digit, however large the number.
Use it when: Efficiency answers.
Two markers, one pass
On sorted lists, move the marker at the smaller number.
Use it when: Finding common elements.

Column Addition, Lists and the Efficiency of Algorithms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Column addition with carries · 3 marks · Short answerBoard level

Add $4768 + 5385$ using the column algorithm. Show the digit written and the carry in each column.

Try it first, then show the solutionSolution

Units: $8 + 5 = 13$ , write $3$ , carry $1$ . (½)
Tens: $6 + 8 + 1 = 15$ , write $5$ , carry $1$ . (½)
Hundreds: $7 + 3 + 1 = 11$ , write $1$ , carry $1$ . (½)
Thousands: $4 + 5 + 1 = 10$ , write $0$ , carry $1$ . (½)
The final carry becomes the leading digit: $10153$ . Check: $4768 + 5385 = 10153$ . (1)

Answer: $4768 + 5385 = 10153$ .

Example 2: Why the carry is at most 1 · 3 marks · Short answerBoard level

Explain why the carry in column addition is never more than 1, and add $7 + 9856$ by padding.

Try it first, then show the solutionSolution

Each digit is at most $9$ and the carry-in is at most $1$ (by induction from the start with carry $0$ ). (1)
So the largest column sum is $9 + 9 + 1 = 19$ , which gives units digit $9$ and carry $1$ . (1)
Padding: $7 = 0007$ ; units $7 + 6 = 13$ , write 3 carry 1; tens $0 + 5 + 1 = 6$ ; hundreds $8$ ; thousands $9$ : $9863$ . (1)

Answer: The largest column sum is 19, so the carry is at most 1; $7 + 9856 = 9863$ .

Example 3: Counting versus columns · 5 marks · Long answerHarder board level

Suppose, as an illustration, that counting takes one second per dot and adding one column takes 15 seconds. Estimate the time to add numbers near $9999 + 9999$ by counting and by the column method, and then for numbers with 8 digits each.

Try it first, then show the solutionSolution

Counting for $9999 + 9999 = 19998$ dots takes $19998$ seconds. (1)
$19998 = 5 \times 3600 + 33 \times 60 + 18$ , that is $5$ h $33$ min $18$ s. (1)
Column method for 4-digit numbers: $4 \times 15 = 60$ seconds. (1)
For 8-digit numbers: columns $8 \times 15 = 120$ seconds, while counting would be about $199999998$ seconds, over $6$ years. (1)
Doubling the digits doubles the column work, but the counting work grows with the value: the algorithm's work depends on the number of digits. (1)

Answer: About 5 h 33 min against 60 s; 120 s against more than 6 years for 8 digits.

Example 4: Common elements of sorted lists · 3 marks · Short answerHarder board level

Two sorted lists are $A = [2, 3, 5, 7, 11, 13]$ and $B = [1, 3, 4, 5, 9, 11, 12]$ . Find the common elements with two markers in one pass, count the comparisons, and compare with testing all pairs.

Try it first, then show the solutionSolution

Start at $2$ and $1$ : $1 < 2$ so advance in $B$ ; $2 < 3$ advance in $A$ ; $3 = 3$ record $3$ and advance both. (1)
Continue the pass: $5$ vs $4$ advance $B$ ; $5 = 5$ record; $7 < 9$ advance $A$ ; $11 > 9$ advance $B$ ; $11 = 11$ record; $13 > 12$ advance $B$ and $B$ ends. The common elements are $3, 5, 11$ after $9$ comparisons (the first two were $2$ vs $1$ and $2$ vs $3$ ). (1)
All pairs would need $6 \times 7 = 42$ comparisons. (1)

Answer: The common elements are 3, 5, 11, found in 9 comparisons instead of 42.

Example 5: Divisor lists and the gcd · 4 marks · Case-basedHOTS / competency

Case: to find the gcd and lcm of 36 and 60, a student builds sorted lists of divisors using divisor pairs. (i) Show the pairs and the number of checks for 36 and for 60. (ii) List the common divisors by walking the two lists. (iii) Give the gcd and the lcm. (iv) Compare the number of checks with testing every number up to 36 and up to 60.

Try it first, then show the solutionSolution

(i) For $36$ test $1$ to $6$ (6 checks): pairs $(1, 36), (2, 18), (3, 12), (4, 9), (6, 6)$ give $1, 2, 3, 4, 6, 9, 12, 18, 36$ . For $60$ test $1$ to $7$ (7 checks): pairs $(1, 60), (2, 30), (3, 20), (4, 15), (5, 12), (6, 10)$ give $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$ . (1½)
(ii) Common divisors: $1, 2, 3, 4, 6, 12$ . (1)
(iii) $g cd = 12$ and $lcm = \frac{36 \times 60}{12} = 180$ . (½)
(iv) The pairs method uses $6 + 7 = 13$ checks against $36 + 60 = 96$ checks. (1)

Answer: (i) as listed; (ii) 1, 2, 3, 4, 6, 12; (iii) gcd 12, lcm 180; (iv) 13 checks against 96.

Where students lose marks in Column Addition, Lists and the Efficiency of Algorithms

The slips examiners see most often, and what to write instead.

Mistake: Dropping the final carry
Stopping at the last column.
Wrong working
$5000 + 5000 = 000$ .
Right working
The final carry gives $10000$ .
Do this: Look at the carry after the leftmost column.
Usually costs: The answer.
Mistake: Misaligned digits
Aligning from the left instead of the right.
Wrong working
$7 + 9856$ with 7 under the 9.
Right working
Align units: $0007 + 9856$ .
Do this: Pad with zeros on the left.
Usually costs: The answer.
Mistake: Confusing work with value
Saying bigger numbers always need more steps.
Wrong working
Adding $1 0^{9}$ numbers takes a billion steps.
Right working
It takes about as many steps as there are digits.
Do this: Count digits, not size.
Usually costs: 1 mark.

Exam tips

Write a carry row above the columns.
State the stopping condition and what happens to the final carry.
Compare algorithms by counting steps, not by timing guesses.
Label any made-up rates (such as seconds per step) as illustrations.

Practice questions on Column Addition, Lists and the Efficiency of Algorithms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In adding two whole numbers by the column method, the largest possible carry from one column to the next is
(a) 1
(b) 2
(c) 9
(d) 10
Hint
The largest column sum is $9 + 9 + 1$ .
Answer
(a) 1. The sum is at most 19, so the carry is at most 1.
Q2 · 3 marks · Short answerBoard level
Add $38579 + 4687$ by the column algorithm and list the digit written and the carry in each column.
Hint
Pad $4687$ to $04687$ .
Answer
Units: $9 + 7 = 16$ , write 6, carry 1. Tens: $7 + 8 + 1 = 16$ , write 6, carry 1. Hundreds: $5 + 6 + 1 = 12$ , write 2, carry 1. Thousands: $8 + 4 + 1 = 13$ , write 3, carry 1. Ten-thousands: $3 + 0 + 1 = 4$ , write 4, carry 0. Total $43266$ .
Q3 · 3 marks · Short answerHarder board level
How many digits can the sum of a 5-digit number and a 3-digit number have? Give an example for each possibility.
Hint
Think of the largest possible sum and the effect of the final carry.
Answer
At most $5 + 1 = 6$ digits, and at least $5$ . Five digits: $10000 + 123 = 10123$ . Six digits: $99999 + 999 = 100998$ , because the carry from the leading column becomes a new digit.
Q4 · 5 marks · Long answerHarder board level
Write an algorithm to find the lcm of two numbers using the division method for the gcd. Apply it to 18 and 24, and to 45 and 75.
Hint
Number your steps and state the stopping rule.
Answer
Algorithm: (1) set $(m, n)$ with $m \geq n$ ; (2) while $n \neq = 0$ , replace $(m, n)$ by $(n, m mod n)$ ; (3) the gcd is $m$ ; (4) the lcm is $\frac{product of the original numbers}{gcd}$ . For $18, 24$ : $24 mod 18 = 6$ , $18 mod 6 = 0$ , gcd $6$ , lcm $\frac{432}{6} = 72$ . For $45, 75$ : $75 mod 45 = 30$ , $45 mod 30 = 15$ , $30 mod 15 = 0$ , gcd $15$ , lcm $\frac{3375}{15} = 225$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: three students use different algorithms to add $4768 + 5385$ . Asha follows the column method correctly. Binu forgets to write the final carry. Charu adds each column separately and writes the column sums side by side as 9, 10, 14, 13 without carrying. (i) What is the correct answer? (ii) What does Binu get and by how much is he wrong? (iii) Why is Charu's answer not a number? (iv) Which feature of the number system is the carry using?
Hint
Binu's thousands digit is 0 with the leading 1 missing.
Answer
(i) $10153$ . (ii) Binu writes $0153$ , that is $153$ , which is wrong by $10000$ (the missing leading digit). (iii) The column sums $9, 10, 14, 13$ include $10$ , $14$ and $13$ , which are not single digits, so writing them side by side does not represent a number in place value. (iv) Place value: ten units of one place make one unit of the next place.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Largest possible carry?Show answer
1.
2.Largest column sum with a carry-in?Show answer
19.
3.What does column addition's work depend on?Show answer
The number of digits.
4.Why test divisors only up to the square root?Show answer
Every divisor above the square root pairs with one below it.
5.What is a data structure?Show answer
An organised way of storing intermediate values, such as a sorted list.

Column Addition, Lists and the Efficiency of Algorithms: frequently asked questions

Can an algorithm be correct but slow?

Yes. Counting dots is correct, but its work grows with the value of the numbers, so it is far slower than column addition.

Why do we pad with zeros?

To line up the same place values in each column so that units are added to units, tens to tens, and so on.

Why is an algorithm described as precise?

Anyone, or a computer, following the steps exactly must reach the right answer without guessing.