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NCERT Solutions Ganita Prakash (Part 1) Chapter 6 Fast Multiplications Using the Distributive Property — In-text Questions

Book page 144 Updated on2026-09-05

Q1.
Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line.
Answer

Rule. Write the units digit as it is. Then, moving left, write the sum of each pair of neighbouring digits. Finally write the leading digit. Carry as usual.

For … d c b a:
units digit → a
tens digit → a + b
hundreds digit → b + c
thousands digit → c + d
next digit → d
product = d | (c + d) | (b + c) | (a + b) | a (with carries)

Worked line for 3874 × 11, right to left:

StepDigits addedWritten so far
144
27 + 4 = 11 → write 1, carry 114
38 + 7 + 1 = 16 → write 6, carry 1614
43 + 8 + 1 = 12 → write 2, carry 12614
53 + 1 = 442614
Why it happens: N × 11 = N × 10 + N, and the shift by one place makes every column of that addition a pair of adjacent digits of N. Nothing about the length of N is used, so the rule works for a number of any size.
Q2.
Evaluate (i) 94 × 11, (ii) 495 × 11, (iii) 3279 × 11, (iv) 4791256 × 11.
Answer
(i) 9 | 9 + 4 | 4 = 9 | 13 | 4 → carry 1 → 1034
(ii) 4 | 4 + 9 | 9 + 5 | 5 = 4 | 13 | 14 | 5 → 5445
(iii) 3 | 3 + 2 | 2 + 7 | 7 + 9 | 9 = 3 | 5 | 9 | 16 | 9 → 36069
(iv) 4 | 4+7 | 7+9 | 9+1 | 1+2 | 2+5 | 5+6 | 6
= 4 | 11 | 16 | 10 | 3 | 7 | 11 | 6 → 52703816

Long-hand checks: 94 × 11 = 940 + 94 = 1034 ✓; 495 × 11 = 4950 + 495 = 5445 ✓; 3279 × 11 = 32790 + 3279 = 36069 ✓; 4791256 × 11 = 47912560 + 4791256 = 52703816 ✓

Tip: do the carries from the right, exactly as in ordinary addition — in (iii) the 16 gives 6 with a carry of 1 into the 9, making it 10, which gives 0 with a further carry.
Q3.
Can we come up with a similar rule for multiplying a number by 101?
Answer

Yes — for exactly the same reason.

N × 101 = N (100 + 1) = N × 100 + N

Multiplying by 100 shifts every digit two places, so in the addition each column pairs digits that are two places apart instead of adjacent.

Why it happens: the only thing that changes from the 11 rule is the size of the shift. Whenever the multiplier is 10k + 1, the rule becomes “add digits that are k places apart”.
Q4.
Multiply 3874 by 101.
Answer
3874 × 101 = 3874 (100 + 1) = 387400 + 3874 = 391274

Written as a column sum with dcba = 3874:

387400d c b a 0 0
+   3874+     d c b a
391274d, c, (b + d), (a + c), b, a
Q5.
Use this to multiply 3874 × 101 in one line.
Answer

Read off the digits d = 3, c = 8, b = 7, a = 4 and apply the pattern d | c | (b + d) | (a + c) | b | a.

3 | 8 | (7 + 3) | (4 + 8) | 7 | 4
= 3 | 8 | 10 | 12 | 7 | 4
carry: … 12 → 2 carry 1; 10 + 1 = 11 → 1 carry 1; 8 + 1 = 9
= 391274

Check: 387400 + 3874 = 391274 ✓

Q6.
What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for multiplication by 1001, 10001, …
Answer

By 101: copy the last two digits, then in each further place write the sum of the digit there and the digit two places to its right; finish with the leading two digits. In symbols, dcba × 101 = d | c | (b + d) | (a + c) | b | a.

By 1001: the shift is three places, so pair digits three apart:

dcba × 1001 = dcba (1000 + 1) = dcba000 + dcba
= d | c | b | (a + d) | c | b | a

By 10001: the shift is four places. For a four-digit dcba the two blocks do not overlap at all:

dcba × 10001 = dcba0000 + dcba = dcba dcba
Why it happens: in every case the multiplier is 10k + 1, so the product is “the number shifted k places” plus “the number”. Whether the two copies overlap, and by how much, depends only on how k compares with the number of digits.
Tip: the same idea handles 10k – 1. Since 99 = 100 – 1 and 999 = 1000 – 1, we get N × 99 = N00 – N and N × 999 = N000 – N.
Q7.
Use this to find (i) 89 × 101, (ii) 949 × 101, (iii) 265831 × 1001, (iv) 1111 × 1001, (v) 9734 × 99 and (vi) 23478 × 999.
Answer
(i) 89 × 101 = 8900 + 89 = 8989 (two-digit number simply repeats)
(ii) 949 × 101 = 94900 + 949 = 95849
Digit rule: 9 | 4 | (9 + 9) | 4 | 9 = 9 | 4 | 18 | 4 | 9 → 95849 ✓
(iii) 265831 × 1001 = 265831000 + 265831 = 266096831
(iv) 1111 × 1001 = 1111000 + 1111 = 1112111
(v) 9734 × 99 = 9734 (100 – 1) = 973400 – 9734 = 963666
(vi) 23478 × 999 = 23478 (1000 – 1) = 23478000 – 23478 = 23454522
Did you know? Brahmagupta (628 CE) called these shortcuts iṣṭa-guṇana — “multiplication by a chosen (convenient) number”. Sridharacharya (750 CE) and Bhaskaracharya (Lilavati, 1150 CE) discuss them too. Every one of them is distributivity: break the multiplier into parts that are easy to handle.
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