In-text — The Divisibility Table — Number Play

1. 15 marksGanita Prakash Cl-8 Part 1, in-text table, page 129

   Fill in the table showing which of 128, 990, 1586, 275, 6686, 639210, 429714, 2856, 3060 and 406839 are divisible by 2, 3, 4, 5, 6, 8, 9, 10 and 11. Find a quick way to do this.

   Hint. Do not divide. Use the tests: last digit for 2, 5, 10; digit sum for 3 and 9; last two digits for 4; last three for 8; combinations for 6; alternating sum for 11.

   The quick way — never divide, just test.

   Divisor	Test
   2	last digit is even
   3	digit sum divisible by 3
   4	last two digits divisible by 4
   5	last digit is 0 or 5
   6	divisible by 2 and 3
   8	last three digits divisible by 8
   9	digit sum divisible by 9
   10	last digit is 0
   11	alternating digit sum divisible by 11

   The big saving is that 6 needs no test of its own — it follows from the 2 and 3 columns, since 6 = 2 × 3 and 2 and 3 share no common factor.

   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 working behind the trickier cells: • 128 — digit sum 11, so not by 3 or 9. Last two digits 28 ÷ 4 = 7 ✓. Whole number 128 ÷ 8 = 16 ✓. Alternating 8 − 2 + 1 = 7, so not by 11. • 990 — digit sum 18, so by both 3 and 9. Last two digits 90 ÷ 4 fails. Alternating 0 − 9 + 9 = 0, so by 11 ✓. • 275 — digit sum 14, not by 3. Alternating 5 − 7 + 2 = 0, so by 11 ✓ (275 = 11 × 25). • 639210 — digit sum 21, by 3 but not 9. Last two 10 ÷ 4 fails. Alternating 0 − 1 + 2 − 9 + 3 − 6 = −11, so by 11 ✓. • 429714 — digit sum 27, so by both 3 and 9. Alternating 4 − 1 + 7 − 9 + 2 − 4 = −1, not by 11. • 2856 — digit sum 21, by 3 not 9. Last three 856 ÷ 8 = 107 ✓. • 3060 — digit sum 9, by 3 and 9. Last three 060 = 60, not divisible by 8. Alternating 0 − 6 + 0 − 3 = −9, not by 11. • 406839 — ends in 9, so not by 2, 5 or 10, and therefore not by 6 either. Digit sum 30, by 3 but not 9.

   ✦ See the completed table above; the fastest route is to test 2, 3, 5 and 11 first, then read off 6 from the 2 and 3 columns, and only check 4, 8, 9 and 10 where they can still succeed.

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.