Circle of Learning • New NCERT textbook "Ganita Manjari" Part II (2026-27) • Exercise Sets 11.1–11.3, Think and Reflect, End-of-Chapter Exercises – step-by-step solutions
Exercise Set 11.1
Add two 4-digit numbers following the steps exactly. Do you get the correct result?
Answer
Example: 4736 + 2589.
Step 1: write 4736 above 2589, aligned on the right.
Step 2: 6 + 9 = 15 (more than 10)
⇒ write 5, carry = 1.
Step 3: 3 + 8 + 1 = 12
⇒ write 2, carry = 1.
Step 3 again: 7 + 5 + 1 = 13
⇒ write 3, carry = 1.
Step 3 again: 4 + 2 + 1 = 7 (less than 10)
⇒ write 7, carry = 0. No digits left.
Step 5: carry is 0, so nothing more. Answer 7325 ✓ – correct.
Careful – a gap in the steps: the steps say "less than 10" and "more than 10", but not what to do when a column adds up to exactly 10. For example, in 4736 + 2574 the units give 6 + 4 = 10, and following the steps literally we do nothing! The case should read "10 or more" (as the textbook's description before the steps says: "equal to or more than 10"). With this fix the algorithm always works.
What happens when adding a 5-digit number to a 3-digit number?
Answer
Example: 45678 + 789. After three columns, the 3-digit number has no digits left, but Step 3 says "add the two digits", and there is only one digit in the column. Strictly following the steps, it is not clear what to do. To handle this, we treat a missing digit as 0 (i.e. write 789 as 00789). Then: 8 + 9 = 17
⇒ 7, carry 1; 7 + 8 + 1 = 16
⇒ 6, carry 1; 6 + 7 + 1 = 14
⇒ 4, carry 1; 5 + 0 + 1 = 6; 4 + 0 = 4. Answer 46467 ✓. So the steps work once we agree to fill empty places with 0 (Step 4 should say "until there are no more digits in either number").
Answer
Why align the columns from right to left?
So that digits of the same place value (units with units, tens with tens, hundreds with hundreds, …) are added together. Numbers can have different numbers of digits, and only the units digits are always in the last position. Aligning on the left would, for example, add the hundreds of one number to the units of another, giving a wrong answer.
Answer
Why can carry never be more than 1?
In any column the largest possible sum is 9 + 9 + 1 (previous carry) = 19, which is less than 20. So the sum is at most one ten plus some units, and the carry is 0 or 1. (By induction: the first carry is at most 1 since 9 + 9 = 18, and if a carry is at most 1, the next column's sum is at most 19.)
Answer
What if we leave out Step 5?
Without Step 5 the final carry would be lost.
It still works when the leftmost column does not produce a carry, e.g. 473 + 325 = 798, or 4736 + 2589 = 7325.
It fails when the leftmost column produces a carry: e.g. 473 + 695 would give 168 instead of 1168; 9 + 1 would give 0 instead of 10.
Think and Reflect (Why the Addition Algorithm Works)
1. Justify by grouping into units, tens, hundreds, …
Write each number by place value, e.g. 473 = 4 hundreds + 7 tens + 3 units and 695 = 6 hundreds + 9 tens + 5 units. Adding is just combining the groups: units 3 + 5 = 8 units; tens 7 + 9 = 16 tens = 1 hundred + 6 tens (the carry moves the new hundred into the hundreds group); hundreds 4 + 6 + 1 = 11 hundreds = 1 thousand + 1 hundred. Total = 1 thousand + 1 hundred + 6 tens + 8 units = 1168. Each column handles one place value, and every time a column reaches ten or more, ten of those units make one unit of the next place – this is exactly the carry. So the digit-by-digit result is the true sum.
How to modify the algorithm for decimals?
Answer
Align the numbers at the decimal point (not at the rightmost digit), and fill missing places after the decimal point with zeros (e.g. 3.5 + 12.75
⇒ 03.50 + 12.75). Then add column by column from the right with carries exactly as before, and put the decimal point in the answer directly below the others. Answer: 16.25.
Think and Reflect (Divisors)
Execute the algorithm for 15, 135 and 775.
divisors(15) = [1, 3, 5, 15] (15 numbers checked)
divisors(135) = [1, 3, 5, 9, 15, 27, 45, 135] (135 numbers checked)
divisors(775) = [1, 5, 25, 31, 155, 775] (775 numbers checked)
Work: The algorithm checks every number from 1 to n, so the work grows in proportion to the size of the number: a number 10 times bigger needs about 10 times as many checks (one more digit = 10 times the work).
Exercise Set 11.2
List 1 and List 2 are in increasing order.
Answer
(i) Elements of List 1 not in List 2:
1. Start with an empty list only-in-1.
For each number x in List 1:
Answer
– if x does not appear in List 2, add x to only-in-1.
3. Report only-in-1.
(Faster version using the order: keep a pointer i at the start of List 1 and j at the start of List 2. While i is within List 1: if List 2 is finished or List1[i] < List2[j], add List1[i] to the answer and move i on; if List1[i] = List2[j], move both i and j on; if List1[i] > List2[j], move j on.)
(ii) Elements of List 2 not in List 1: the same algorithm with the roles of the two lists swapped – for each number y in List 2, if y does not appear in List 1, add y to only-in-2.
Answer
An algorithm for lcm(m, n)
Method 1 (from the definition)
1. Let k = max(m, n).
2. If k is divisible by both m and n, report k as lcm(m, n).
3. Otherwise increase k by 1 (or, faster, by max(m, n) – check only multiples of the larger number) and go back to Step 2.
This must stop, at the latest when k = m × n.
Method 2 (using gcd): compute g = gcd(m, n) by Euclid's/Āryabhaṭa's algorithm and report lcm(m, n) = (m × n)/g. Example: lcm(12, 18) = 216/6 = 36.
Question 3*
(i) Divisors come in pairs (d, n/d), and in each pair the smaller one is at most √n (if both were bigger than √n, their product would exceed n). So we only need to check numbers from 1 to √n. Each time j divides n, record both j and n/j. (If j = √n, record it once.) For n = 18, check only 1 to 4: we get (1, 18), (2, 9), (3, 6).
(ii) The divisors are still all found, but not in increasing order if we just write the pairs one after another (e.g. 1, 18, 2, 9, 3, 6). Our gcd algorithm reported the rightmost common divisor, assuming increasing order, so it would no longer work as described. Fix: either arrange the list in increasing order first (put the small partners in order, followed by the large partners in reverse order), or report the largest element of the common-divisors list instead of the rightmost one.
Exercise Set 11.3
Answer
What changes if divisors are found from n down to 1?
All divisors are still found, but each list is in decreasing order, and so is the list of common divisors. The gcd is then the leftmost (first) element, so Step 5 must say "report the leftmost element of common-divisors".
Answer
Searching for common divisors from min(m, n) down to 1?
Then the first number k that divides both m and n is the greatest common divisor, so we can stop immediately and report it – no list and no "most recent" variable is needed. This can save a lot of work (e.g. for gcd(6, 12) we find 6 straight away). In the worst case (gcd = 1) we still go all the way down to 1.
Āryabhaṭa's Division Algorithm – in-text check
gcd(2k + 1, 2) needs only two reductions: (2k + 1) mod 2 = 1, so gcd(2k + 1, 2) → gcd(2, 1) → gcd(1, 0) = 1. Always two steps, however large k is.
End-of-Chapter Exercises
Compute using the improved (division) version of Euclid's algorithm.
Answer
(i) gcd(375, 825): reverse to gcd(825, 375).
825 = 2 × 375 + 75 → gcd(375, 75)
375 = 5 × 75 + 0 → gcd(75, 0) = 75.
(ii) gcd(51000, 81000): reverse to gcd(81000, 51000).
81000 = 1 × 51000 + 30000
⇒ gcd(51000, 30000)
51000 = 1 × 30000 + 21000
⇒ gcd(30000, 21000)
30000 = 1 × 21000 + 9000
⇒ gcd(21000, 9000)
21000 = 2 × 9000 + 3000
⇒ gcd(9000, 3000)
9000 = 3 × 3000 + 0 → gcd = 3000.
(iii) gcd(1789287, 237656):
1789287 = 7 × 237656 + 125695
237656 = 1 × 125695 + 111961
125695 = 1 × 111961 + 13734
111961 = 8 × 13734 + 2089
13734 = 6 × 2089 + 1200
2089 = 1 × 1200 + 889
1200 = 1 × 889 + 311
889 = 2 × 311 + 267
311 = 1 × 267 + 44
267 = 6 × 44 + 3
44 = 14 × 3 + 2
3 = 1 × 2 + 1
2 = 2 × 1 + 0
⇒ gcd = 1 (the numbers are co-prime).
(iv) gcd(2587392, 157656):
2587392 = 16 × 157656 + 64896
157656 = 2 × 64896 + 27864
64896 = 2 × 27864 + 9168
27864 = 3 × 9168 + 360
9168 = 25 × 360 + 168
360 = 2 × 168 + 24
168 = 7 × 24 + 0 → gcd = 24.
For m ≥ n, show: d divides m and n ⇔ d divides n and (m mod n).
Answer
Write m = qn + r, where r = m mod n and 0 ≤ r < n.
(
⇒ ) If d divides m and n, say m = ad and n = bd, then r = m − qn = ad − qbd = (a − qb)d, so d divides r.
(⇐) If d divides n and r, say n = bd and r = cd, then m = qn + r = qbd + cd = (qb + c)d, so d divides m.
Hence m, n and n, m mod n have exactly the same common divisors, and so gcd(m, n) = gcd(n, m mod n).
Answer
Algorithm prime(n)
1. Compute list-of-divisors = divisors(n).
2. If list-of-divisors has exactly two elements (namely 1 and n), report "n is prime".
3. Otherwise report "n is not prime".
(This correctly reports 1 as not prime, since divisors(1) = [1] has only one element.)
Faster version: if n < 2, report "not prime"; for each j from 2 to √n, if j divides n, report "not prime" and stop; if no such j is found, report "prime".
Question 4*
Algorithm primedivisors(n)
1. Compute list-of-divisors = divisors(n).
2. Start with an empty list prime-divisors.
3. For each number d in list-of-divisors: if prime(d) reports "prime", add d to prime-divisors.
4. Report prime-divisors.
Example: divisors(60) = [1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60] → primedivisors(60) = [2, 3, 5].
Question 5*
Algorithm for prime factorisation of n (n ≥ 2)
1. Start with an empty list factorisation and set p = 2.
While n > 1:
Answer
– set count = 0;
– while p divides n: divide n by p and add 1 to count;
– if count > 0, add the pair (p, count) to factorisation;
– increase p by 1.
3. Report factorisation.
(Only primes ever divide n at the moment they are tested, because all smaller factors have already been divided out.)
(i) 180 = 2² × 3² × 5¹ can be stored as a list of (prime, exponent) pairs: [(2, 2), (3, 2), (5, 1)], in increasing order of the primes.
(ii) Comparing two factorisations (to get the gcd): go through the primes that appear in both lists. For each such prime, take the smaller of the two exponents. Multiply these prime powers to get the gcd (taking the larger exponents over all primes gives the lcm).
Example: 180 = [(2, 2), (3, 2), (5, 1)], 84 = [(2, 2), (3, 1), (7, 1)]. Common primes 2 and 3
⇒ gcd = 2² × 3¹ = 12; lcm = 2² × 3² × 5 × 7 = 1260.
Extra Practice Questions
1. Find gcd(1071, 462) using the division algorithm. 1071 = 2 × 462 + 147; 462 = 3 × 147 + 21; 147 = 7 × 21 + 0 ⇒ gcd = 21
2. How many reductions does Euclid's subtraction algorithm need for gcd(10, 3)? gcd(10,3)→(3,7)→(7,3)→(3,4)→(4,3)→(3,1)→(1,2)→(2,1)→(1,1)→(1,0): many steps, while division needs gcd(3,1)→gcd(1,0): just 2.
3. Why is "algorithm" named after Al-Khwārizmī? His book explaining the Indian methods of calculation was translated into Latin as "Algoritmi …", and his name came to mean such step-by-step methods.