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

Book page 127 Updated on2026-09-05

Q1.
Using these observations, can you tell whether the number 462 is divisible by 11?
Answer

Yes, 462 is divisible by 11.

Use the alternating behaviour of the place values: 1 and 100 are 1 more than a multiple of 11, while 10 is 1 less.

PartNearest multiple of 11Excess / short
400 (4 hundreds)396 = 11 × 364 more
60 (6 tens)66 = 11 × 66 short
2 (2 units)0 = 11 × 02 more
Total excess – total short = (4 + 2) – 6 = 0

Nothing is left over, so 462 is a multiple of 11. Check: 462 = 11 × 42 ✓

Q2.
What could be a general method or shortcut to check divisibility by 11?
Answer

Here is the shortcut, in the two forms the chapter uses.

Form 1 — excess and short.

  1. Add the digits sitting in the places 1, 100, 10000, … (these are 1 more than a multiple of 11). Call this the excess.
  2. Add the digits sitting in the places 10, 1000, 100000, … (these are 1 less than a multiple of 11). Call this the short.
  3. Compute excess – short. If it is 0 or a multiple of 11, the number is divisible by 11; otherwise it tells you how far off you are.

Form 2 — alternating signs. Put '+' before the units digit, '–' before the tens digit, '+' before the hundreds digit, and so on. Add up. Same answer, less writing.

462 → + 2 – 6 + 4 = 0 → divisible ✓
320185 → + 5 – 8 + 1 – 0 + 2 – 3 = – 3 → 3 short of a multiple of 11
Why it works: 1 = 11 × 0 + 1, 10 = 11 × 1 – 1, 100 = 11 × 9 + 1, 1000 = 11 × 91 – 1, and this alternation continues for every higher place. So each digit contributes either + itself or – itself to the leftover, and the alternating sum is the leftover.
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