NCERT Solutions Ganita Prakash (Part 1) Chapter 5 A Shortcut for Divisibility by 11 / More on Divisibility Shortcuts — In-text Questions

Book page 129 Updated on2026-09-05

Q1.

Look at the following procedure — 1. Place alternating ‘+’ and ‘–’ signs before every digit starting from the unit’s digit. 2. Evaluate the expression. 3. The result denotes the remainder obtained when the number is divided by 11. Is this method similar to or different from the method we saw just before?

Answer

It is the same method, written more compactly.

Earlier method (320185)	Alternating-sign method
excess = 2 + 1 + 5 = 8 (places 1, 100, 10000)	these digits get '+'
short = 3 + 0 + 8 = 11 (places 10, 1000, 100000)	these digits get '–'
8 – 11 = – 3	– 3 + 2 – 8 + 1 – 0 + 5 = – 3

Why they agree: writing (excess) – (short) means adding some digits and subtracting the others. Which digits get which sign is decided by the place value — and place values alternate between 1 more and 1 less than a multiple of 11. So attaching alternating signs from the units place is the excess-minus-short calculation, just written in one line.

Tip: the compact version is faster and less error-prone, but the first version is the one that explains why. Both give – 3 for 320185: 3 short of, or 8 more than, a multiple of 11.

Q2.

Fill in the following table. Find a quick way to do this? (Numbers: 128, 990, 1586, 275, 6686, 639210, 429714, 2856, 3060, 406839 — divisibility by 2, 3, 4, 5, 6, 8, 9, 10, 11)

Answer

Number	2	3	4	5	6	8	9	10	11
128	Yes	No	Yes	No	No	Yes	No	No	No
990	Yes	Yes	No	Yes	Yes	No	Yes	Yes	Yes
1586	Yes	No	No	No	No	No	No	No	No
275	No	No	No	Yes	No	No	No	No	Yes
6686	Yes	No	No	No	No	No	No	No	No
639210	Yes	Yes	No	Yes	Yes	No	No	Yes	Yes
429714	Yes	Yes	No	No	Yes	No	Yes	No	No
2856	Yes	Yes	Yes	No	Yes	Yes	No	No	No
3060	Yes	Yes	Yes	Yes	Yes	No	Yes	Yes	No
406839	No	Yes	No	No	No	No	No	No	No

The quick way — do only four checks per row and read off the rest:

2 — units digit even. 5 — units digit 0 or 5. 10 — units digit 0 (so 10 = 2 and 5 together).
3 and 9 — one digit sum serves both. If the digit sum is a multiple of 9 it is automatically a multiple of 3.
6 — no new work: it is just 2 and 3.
4 — last two digits; 8 — last three digits.
11 — alternating sum of the digits from the units place.

639210 → digit sum 21 → multiple of 3 but not 9
→ alternating sum 0 – 1 + 2 – 9 + 3 – 6 = – 11 → multiple of 11 ✓
→ last two digits 10 → not a multiple of 4

Note on the book's sample row: the printed table on page 129 marks 128 as not divisible by 4. That is a misprint — 128 = 4 × 32, so the entry should be Yes. (Indeed 128 = 2⁷, so it is divisible by 2, 4, 8, 16, 32, 64 and 128.)

Q3.

How can we find out if a number is divisible by 6?

Answer

Check divisibility by 2 and by 3 — both must hold.

6 = 2 × 3, and 2 and 3 have no common factor
LCM (2, 3) = 6

Why it works: if a number is divisible by 2 and also by 3, then by the rule from page 121 it is divisible by LCM (2, 3) = 6. Conversely, if it is divisible by 6 it is divisible by every factor of 6, so by 2 and 3. The two tests are equivalent.

Q4.

Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts for 2 and 3 on these numbers and divide each number by 6 to verify — 38, 225, 186, 64.

Answer

Yes, it works every time.

Number	÷ 2?	Digit sum → ÷ 3?	Prediction	Actual ÷ 6
38	Yes	11 → No	Not divisible	38 = 6 × 6 + 2 ✓
225	No	9 → Yes	Not divisible	225 = 6 × 37 + 3 ✓
186	Yes	15 → Yes	Divisible	186 = 6 × 31 ✓
64	Yes	10 → No	Not divisible	64 = 6 × 10 + 4 ✓

Notice that 38 and 64 pass the test for 2 but fail for 3, and 225 does the opposite. Both tests must be passed.

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