In-text — The Divisibility Shortcut for 11 — Number Play

1. 11 markGanita Prakash Cl-8 Part 1, in-text, page 128

   If the difference between the two alternating digit sums is 11 or a multiple of 11, what does that say about the remainder obtained when the number is divided by 11?

   Hint. A remainder that is itself a multiple of the divisor is not really a remainder.

   The remainder is zero — the number is exactly divisible by 11.

   Why this follows: the shortcut works because the powers of 10 alternate in what they leave behind when divided by 11. Since 10 = 11 − 1, 100 = 99 + 1, 1000 = 1001 − 1, and so on, each place value contributes its digit with alternating sign. So the whole number leaves the same remainder as the alternating sum of its digits.

   If that alternating sum comes out as 0, 11, 22 or any other multiple of 11, then what is 'left over' is itself a whole number of 11s. Anything that is a complete multiple of the divisor can be absorbed into the quotient rather than counted as a remainder, so nothing at all remains.

   Example: for 90904 the alternating sum is 4 − 0 + 9 − 0 + 9 = 22, a multiple of 11. And indeed 90904 ÷ 11 = 8264 exactly ✓

   ✦ The remainder is zero — the number is divisible by 11, because a leftover that is itself a multiple of 11 forms complete groups and leaves nothing behind.

2. 23 marksGanita Prakash Cl-8 Part 1, in-text, page 128

   Using the alternating-sum shortcut, find whether these are divisible by 11, and give the remainder if not: (i) 158 (ii) 841 (iii) 481 (iv) 5529 (v) 90904 (vi) 857076

   Hint. Starting from the units digit, add and subtract the digits alternately. If the result is negative, add 11.

   The method: starting from the units digit, alternately add and subtract the digits. The result is the remainder on division by 11 (adding 11 first if it comes out negative).

   (i) 158 → 8 − 5 + 1 = 4 Not divisible; remainder 4. (Check: 11 × 14 = 154, and 158 − 154 = 4 ✓)

   (ii) 841 → 1 − 4 + 8 = 5 Not divisible; remainder 5. (Check: 11 × 76 = 836, and 841 − 836 = 5 ✓)

   (iii) 481 → 1 − 8 + 4 = −3 The result is negative, so add 11: −3 + 11 = 8. Not divisible; remainder 8. (Check: 11 × 43 = 473, and 481 − 473 = 8 ✓)

   (iv) 5529 → 9 − 2 + 5 − 5 = 7 Not divisible; remainder 7. (Check: 11 × 502 = 5522, and 5529 − 5522 = 7 ✓)

   (v) 90904 → 4 − 0 + 9 − 0 + 9 = 22, which is a multiple of 11. Divisible by 11. (Check: 90904 ÷ 11 = 8264 ✓)

   (vi) 857076 → 6 − 7 + 0 − 7 + 5 − 8 = −11, a multiple of 11. Divisible by 11. (Check: 857076 ÷ 11 = 77916 ✓)

   The one trap: a negative alternating sum does not mean 'not divisible'. In (vi) the value −11 is a perfectly good multiple of 11, so the number is divisible; in (iii) the value −3 has to be shifted up by 11 to give the true remainder 8.

   ✦ (i) 4 (ii) 5 (iii) 8 (iv) 7 (v) divisible by 11 (vi) divisible by 11.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp105.pdf). The chapter is about divisibility reasoning, digital roots and cryptarithms — questions sit in four 'Figure it Out' blocks (pages 122, 126, 131 and 132-135) plus several in-text drills. Every numeric answer here was independently recomputed before being compared with the book's printed answer key: the full 10-row divisibility table was re-derived rule by rule, every cryptarithm was re-solved from scratch, and the divisibility-by-44 and multiple-of-18 digit pairs were found by exhaustive search over the digits.. Questions are referenced from the NCERT textbook for identification.