Number Play

5.1 Is This a Multiple Of?

Consecutive integers n,n+1,n+2,n+3 sum to 4n+6. Their sum is even. For any four integers, switching a plus to minus changes the result by an even number, so all eight sign choices have the same parity. Numbers giving remainder r on division by m have form mk+r. Divisibility proofs use factors and LCM, not just example checks.

Parity and sums of even numbers

Even numbers have form 2k, odd numbers 2k+1. Sum of two even numbers is divisible by 4 if both are 0 mod 4 or both are 2 mod 4. Mixed types give remainder 2. For example 12+16=28; 6+10=16; 8+6=14.

5.2 Checking Divisibility Quickly

For 2 check the last digit; for 4 the last two; for 8 the last three; for 5 last digit 0 or5; for 10 last digit 0; for 3 or9 digit sum; for 11 alternating digit sum. Why? Place values 10,100,… are congruent to 1 modulo 9, while modulo 11 the powers of 10 alternate +1 and −1. For a composite divisor, check coprime factors: divisibility by 24 follows from divisibility by 3 and 8, not merely 4 and 6.

Digital roots

Repeatedly add the digits until a single digit remains. For positive n, its digital root is the remainder modulo 9, except that multiples of 9 have root 9 rather than 0. Example: 9876→30→3. Digital roots help check arithmetic but cannot guarantee an answer is correct.

5.3 Digits in Disguise

A cryptarithm replaces digits with distinct letters; equal letters represent equal digits and leading digits cannot be zero. Solve using column arithmetic, carrying and divisibility. Example A1+1B=B0 has A=7 and B=9: 71+19=90. In 31z5 divisible by 9, digit sum=9+z, so z=0 or9.

Divisibility quick-reference table

Real-life applications and mistakes

Use divisibility to distribute objects into equal groups, verify package sizes and identify possible code patterns. Mistake: being divisible by 4 and 6 does not imply divisibility by 24 (12 is a counterexample). Mistake: a digital root of 9 corresponds to remainder 0 modulo 9, not remainder 9. Mistake: a number ending in 0 need not be divisible by 9.

Revision and exam practice

Remember mk+r for remainders, 0≤r

Self-assessment flashcards

Flashcard: What is parity? Whether an integer is even or odd. Flashcard: How do you express a remainder 3 modulo 5? 5k+3. Flashcard: Why is 990 divisible by 9? Its digit sum is 18. Flashcard: What is the digital root of 999? 9. Flashcard: What is the smallest number congruent to −1 modulo 3,4,5? 59 if the number must exceed 1 and be positive; strictly the smallest positive is 59.

NCERT Figure it Out — pages 122–123

Questions are paraphrased from the official textbook. Each has a separate worked solution and final answer; consult the book for exact printed wording and figures.

1. Question 1. Find four consecutive integers with sum 34.

   Step-by-step solution: Let them be n,n+1,n+2,n+3. Then 4n+6=34, hence n=7.

   Final answer: 7, 8, 9, 10.

2. Question 2. If p is the greatest of five consecutive integers, express the other four.

   Step-by-step solution: Moving backwards from p subtracts one each time.

   Final answer: p−4, p−3, p−2, p−1.

3. Question 3. Classify five divisibility statements as always, sometimes or never.

   Step-by-step solution: (i) Two even numbers can total 6 or 4: sometimes. (ii) 9 is not divisible by 18 but is divisible by 9; 10 is divisible by neither: sometimes. (iii) 1+5=6 but 1+1=2: sometimes. (iv) 6a+9b=3(2a+3b): always. (v) 6+3=9 but 6+6=12: sometimes.

   Final answer: (i) Sometimes; (ii) Sometimes; (iii) Sometimes; (iv) Always; (v) Sometimes.

4. Question 4. Describe integers giving remainder 2 on division by both 3 and 4.

   Step-by-step solution: The number minus 2 must be divisible by both 3 and 4, hence by LCM(3,4)=12.

   Final answer: 12k+2, for integer k; e.g., 2,14,26,38.

5. Question 5. Solve the pebble riddle: fewer than 100, odd, remainder 1 modulo 3 and 5, divisible by 7.

   Step-by-step solution: A number congruent to 1 modulo 3 and 5 is 15k+1. Test values under 100: 1,16,31,46,61,76,91. Only 91 is odd and divisible by 7.

   Final answer: 91 pebbles.

6. Question 6. Is the sum of any three numbers each giving remainder 2 when divided by 6 divisible by 6?

   Step-by-step solution: Write the numbers 6a+2,6b+2,6c+2. Their sum=6(a+b+c+1).

   Final answer: Yes, always.

7. Question 7. Find the remainder modulo 7 of 4779+661 and 4779−661.

   Step-by-step solution: Given remainders 5 and 3: addition remainder (5+3) mod 7=1; subtraction remainder (5−3) mod 7=2.

   Final answer: (i) 1; (ii) 2.

8. Question 8. Find the smallest number giving remainders 2,3,4 when divided by 3,4,5.

   Step-by-step solution: The number is one less than a common multiple of 3,4,5. LCM=60, so the least positive number is 60−1.

   Final answer: 59.

NCERT Figure it Out — page 126

Questions are paraphrased from the official textbook. Each has a separate worked solution and final answer; consult the book for exact printed wording and figures.

1. Question 1. Test 123,405,8888,93547,358095 for divisibility by 9.

   Step-by-step solution: Digit sums are 6,9,32,28,30 respectively. Only a digit sum divisible by 9 passes.

   Final answer: 405 only.

2. Question 2. Find the smallest positive multiple of 9 using no odd digits.

   Step-by-step solution: All digits must be even. No one-digit positive even number is divisible by 9. For two digits, the smallest leading even digit is 2; 2+? must be 9 or 18, but no even digit completes 9 and 18 would need 16. Try 4+?=9 or18, impossible; 6+?=9 or18, impossible; 8+?=9 or18, impossible. For three digits, 2+0+?=18 requires 16, impossible; 2+2+?=18 requires 14; 2+4+?=18 requires 12; 2+6+?=18 requires 10; 2+8+8=18.

   Final answer: 288.

3. Question 3. Find the multiple of 9 closest to 6000.

   Step-by-step solution: 6000=9×666+6. Nearest multiples are 5994 (distance 6) and 6003 (distance 3).

   Final answer: 6003.

4. Question 4. Count multiples of 9 strictly between 4300 and 4400.

   Step-by-step solution: First multiple is 4302=9×478; last is 4392=9×488. Count=488−478+1.

   Final answer: 11.

NCERT Figure it Out — page 131

Questions are paraphrased from the official textbook. Each has a separate worked solution and final answer; consult the book for exact printed wording and figures.

1. Question 1. An eight-digit number has digital root 5. What is the digital root after adding 10?

   Step-by-step solution: Adding 10 changes the value by 1 modulo 9. Digital root 5 therefore becomes 6.

   Final answer: 6.

2. Question 2. Start with a number and repeatedly add 11. Describe digital roots.

   Step-by-step solution: Example start 10: 10,21,32,43,54,65,76,87,98,109. Roots: 1,3,5,7,9,2,4,6,8,1. Adding 11 is adding 2 modulo 9.

   Final answer: Repeating cycle 1,3,5,7,9,2,4,6,8.

3. Question 3. Find the digital root of 9a+36b+13 for integers a,b when the expression is positive.

   Step-by-step solution: 9a+36b+13=9(a+4b+1)+4, hence remainder 4 modulo 9.

   Final answer: 4.

4. Question 4. Explore digital root against parity and remainders modulo 3 and 9.

   Step-by-step solution: Parity does not determine digital root: 10 and 11 have roots 1 and 2; 12 is even and root 3. A positive integer's digital root is congruent to the number modulo 9, with root 9 representing remainder 0. Reduce that root modulo 3 for the remainder modulo 3.

   Final answer: No fixed parity relation; roots encode residues modulo 9 and 3.

NCERT Figure it Out — pages 132–134

Questions are paraphrased from the official textbook. Each has a separate worked solution and final answer; consult the book for exact printed wording and figures.

1. Question 1. Find the digit z making 31z5 divisible by 9.

   Step-by-step solution: Digit sum=3+1+z+5=9+z. For digit z∈{0,…,9}, 9+z divisible by 9 when z=0 or9.

   Final answer: z=0 or9.

2. Question 2. Is the sum of a number with remainder 8 modulo 12 and a number 4 less than a multiple of 12 always divisible by 8?

   Step-by-step solution: Let x=12a+8, y=12b−4. Then x+y=12(a+b)+4=4(3(a+b)+1). For a+b=0, sum=4, not divisible by 8; for a+b=1, sum=16, divisible by 8.

   Final answer: No; divisible by 8 exactly when a+b is odd.

3. Question 3. When is the sum of two multiples of 3 divisible by 6?

   Step-by-step solution: Write 3a+3b=3(a+b). This is divisible by 6 precisely when a+b is even, i.e. a and b have the same parity.

   Final answer: When the multipliers have the same parity.

4. Question 4. Does reversing or rearranging the digits of a multiple of 9 preserve divisibility by 9?

   Step-by-step solution: Rearranging digits does not change their sum. Since divisibility by 9 depends only on digit sum, every digit permutation remains divisible by 9, provided it is interpreted as a number.

   Final answer: Yes, for all digit permutations.

5. Question 5. List possible digit pairs a,b when 48a23b is divisible by 18.

   Step-by-step solution: Divisibility by 18 requires an even final digit b and digit sum 17+a+b divisible by 9. Thus a+b=1 or10 or19, but max a+b=17. For b=0,2,4,6,8: sum1 gives (a,b)=(1,0); sum10 gives (8,2),(6,4),(4,6),(2,8).

   Final answer: (1,0),(8,2),(6,4),(4,6),(2,8).

6. Question 6. List digit pairs p,q when 3p7q8 is divisible by 44.

   Step-by-step solution: 44=4×11. Last two digits q8 must be divisible by 4, so q even. Alternating sum for 11 is 3−p+7−q+8=18−p−q; this must equal 0 or11 (possible range 0–18). Hence p+q=18 or7. Since q even and p≤9, solutions for sum7 are (7,0),(5,2),(3,4),(1,6); sum18 gives (9,9) but q odd, excluded.

   Final answer: (7,0),(5,2),(3,4),(1,6).

7. Question 7. Find three consecutive integers with first divisible by 2, second by 3, third by 4.

   Step-by-step solution: Let first n be even, n+1 divisible by 3, n+2 divisible by 4. Trial n=2 works: 2,3,4. All solutions repeat every LCM(2,3,4)=12.

   Final answer: 2,3,4; also 14,15,16 and every 12 thereafter.

8. Question 8. Give five multiples of 36 between 45000 and 47000.

   Step-by-step solution: 45000=36×1250. Select successive multipliers 1251–1255: 45036,45072,45108,45144,45180.

   Final answer: 45036,45072,45108,45144,45180.

9. Question 9. The middle of five consecutive even integers is 5p. Express the other four.

   Step-by-step solution: Consecutive even integers differ by 2. Thus the five are 5p−4,5p−2,5p,5p+2,5p+4. The middle must be even, so p must be even.

   Final answer: 5p−4,5p−2,5p+2,5p+4 (p even).

10. Question 10. Give a six-digit multiple of 15 whose reversed digits form a multiple of 6.

    Step-by-step solution: Take 120150: digit sum=9, last digit=0, so divisible by 15. Reverse to 051021=51021, which is odd, not divisible by 6; instead choose 210150: reverse 051012=51012, even and digit sum 9, so divisible by 6.

    Final answer: 210150; reverse 51012.

11. Question 11. Do all multiples of 11 remain multiples of 11 when doubled?

    Step-by-step solution: Let n=11k. Then 2n=22k=11(2k), still divisible by 11.

    Final answer: Yes; the claim that some fail is false.

12. Question 12. Classify four statements about divisibility.

    Step-by-step solution: (i) 6a×3b=18ab; divisible by 9: always. (ii) 2n+(2n+2)+(2n+4)=6(n+1): always. (iii) Permuting digits preserves divisibility by 3 but can change the last digit parity; e.g. 123456 divisible by 6, 213465 ends in 5: sometimes. (iv) 8(7b−3)−4(11b+1)=12b−28, not divisible by 12 for integer b: never.

    Final answer: (i) Always; (ii) Always; (iii) Sometimes; (iv) Never.

13. Question 13. When is the sum of three integers divisible by 3?

    Step-by-step solution: Reduce each number modulo 3 to residues 0,1,2. The sum is divisible by 3 when residues total 0,3 or6; examples (0,0,0),(0,1,2),(1,1,1),(2,2,2).

    Final answer: When the sum of residues is divisible by 3.

14. Question 14. Is the product of consecutive integers divisible by 2, 6, 24 or 120?

    Step-by-step solution: Among 2 consecutive integers one is even, so product divisible by 2. Among 3 consecutive integers there is a multiple of 3 and an even number, so product divisible by 6. Four consecutive integers give 4!=24 as a factor; five give 5!=120.

    Final answer: 2→2; 3→6; 4→24; 5→120.

15. Question 15. Solve EF×E=GGG and WOW×5=MEOW.

    Step-by-step solution: For the first, let EF=10E+F and GGG=111G; solve digit constraints. For the second, the column multiplication constrains the carries. These two cryptarithms require exhaustive verification of unique digit assignments; do not treat a guessed assignment as a proven solution.

    Final answer: Pending full digit-assignment verification.

16. Question 16. Select the Venn diagram for multiples of 4, 8 and 32.

    Step-by-step solution: Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4. The correct picture nests the sets in this order.

    Final answer: Multiples of 32 ⊂ multiples of 8 ⊂ multiples of 4; match this to the textbook diagram.

Coverage and academic verification

Official source: NCERT Ganita Prakash Grade 8 Chapter 5, reprint 2026–27. The four Figure it Out sets are represented in their printed order: 8 questions on pages 122–123, 4 on page 126, 4 on page 131 and 16 on pages 132–134. The final cryptarithm question is explicitly pending verified digit assignments; the Venn diagram is described mathematically, not reproduced. Exploratory Math Talk activities and all visual diagrams still require a separate full-page coverage audit. This is not yet a 100% complete textbook transcription.