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CBSE · Class 8 · Mathematics Part 1

Part 1, Chapter 5: Number Play

5 concepts · 40 practice questions with answers · Free to read

This lesson is also in: मराठी · हिंदी

Sums of consecutive numbers

From daily life

A hall has 14, 15 and 16 seats in three consecutive rows. Instead of adding all three, notice that the middle row has 15 seats: the total is 3 × 15 = 45.

Simple definition

Consecutive numbers follow one another by 1: n, n + 1, n + 2, ... The sum of an ODD number of consecutive numbers = (how many) × (middle number). The sum of the first n counting numbers is n × (n + 1) ÷ 2.

141516

In the figure: Middle number 15, three numbers: 3 × 15 = 45.

Rule 1: Odd count: count × middle — Divide the sum by the count to get the middle number, then build the others around it.

  • Easy: Three numbers, sum 51: middle 17, so 16, 17, 18.
  • Medium: Five numbers, sum 85: middle 17, so 15, 16, 17, 18, 19.
  • Exam-type: Three consecutive even numbers, sum 90: middle 30, so 28, 30, 32.

Rule 2: Sum of the first n numbers — Pair the first with the last: each pair makes n + 1, and there are n ÷ 2 pairs.

  • Easy: 1 + 2 + ... + 20 = 20 × 21 ÷ 2 = 210.
  • Medium: 1 + 2 + ... + 30 = 30 × 31 ÷ 2 = 465.
  • Exam-type: 1 + 2 + ... + 100 = 100 × 101 ÷ 2 = 5050.

Rule 3: Even count and parity — With an even count there is no middle number. Use the first number: n + (n + 1) = 2n + 1, n + ... + (n + 3) = 4n + 6.

  • Easy: Two consecutive numbers with sum 79: 39 and 40.
  • Medium: Six consecutive numbers always add to an odd number (6n + 15).
  • Exam-type: Four consecutive numbers with sum 102: 4n + 6 = 102, n = 24, so 24, 25, 26, 27.
Remember

A number that is a power of 2 (such as 8 or 16) cannot be written as a sum of two or more consecutive positive numbers. Others can, often in several ways: 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.

Common mistake

Using count × middle for an EVEN count of numbers. The rule works only when there is a single middle number.

Shortcut

Odd count? Find the middle and multiply. Memory clue: "Middle times how many".

Spot the difference

Point3 consecutive numbers4 consecutive numbers
Sum formula3n + 3 = 3 × middle4n + 6
Example (start 5)5 + 6 + 7 = 185 + 6 + 7 + 8 = 26
Always divisible by 3?yesno

Practice questions

Easy
  1. 1Find 20 + 21 + 22.
  2. 2Find 1 + 2 + 3 + ... + 10.
  3. 3The sum of two consecutive numbers is 45. Find them.
Medium
  1. 4Five consecutive numbers add up to 120. Find them.
  2. 5Find 1 + 2 + 3 + ... + 40.
  3. 6The sum of three consecutive odd numbers is 45. Find them.
Challenging
  1. 7Four consecutive numbers add up to 58. Find them.
  2. 8Can 8 be written as a sum of two or more consecutive positive numbers?
Show answers (try them yourself first)
  1. 163
  2. 255
  3. 322 and 23
  4. 422, 23, 24, 25, 26
  5. 5820
  6. 613, 15, 17
  7. 713, 14, 15, 16
  8. 8No, 8 is a power of 2

Divisibility properties

From daily life

A teacher packs 36 pencils and 48 erasers into sets of 12 with nothing left over. Without any more work she knows that all 84 items will also pack into sets of 12.

Simple definition

A number a divides b when b = a × k for a whole number k. Rules: if a divides b and c, it divides b + c and b − c; if a divides b, it divides every multiple of b; if a divides b and b divides c, then a divides c.

Rule 1: Sum and difference — If a number divides both parts, it divides their sum and difference. If it divides only one part, it cannot divide the sum.

  • Easy: 91 + 35: both are multiples of 7 (13 × 7 and 5 × 7), so 126 is too.
  • Medium: 120 + 7 and 5: 120 is a multiple of 5 but 7 is not, so the sum is not.
  • Exam-type: If d divides a + b and a, then d divides b, because b = (a + b) − a.

Rule 2: Combining divisors — A number divisible by two numbers is divisible by their LCM. If it is divisible by a product, it is divisible by each factor.

  • Easy: Divisible by 12 means divisible by 3 and by 4.
  • Medium: Divisible by both 4 and 6 means divisible by 12 (their LCM), not necessarily by 24.
  • Exam-type: Divisible by 6 means divisible by 2 and 3; but divisible by 2 and 4 does not mean divisible by 8 (take 4).

Rule 3: Products of consecutive numbers — Among n consecutive numbers there is a multiple of every number up to n.

  • Easy: Two consecutive numbers: the product is always even.
  • Medium: 4 × 5 × 6 = 120, divisible by 6; three consecutive numbers always give a multiple of 6.
  • Exam-type: 3 × 4 × 5 × 6 = 360 = 24 × 15, a multiple of 24 (four consecutive numbers).
Remember

1 divides every number, and every number divides itself and 0. The number of factors of 48 is 10: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

Common mistake

Thinking "divisible by 4 and 6, so divisible by 4 × 6 = 24". Correct: 12 is divisible by both but not by 24.

Shortcut

Split a sum into multiples of the same number: 91 + 35 = 7 × 13 + 7 × 5 = 7 × 18. Memory clue: "Split into multiples, then add".

Spot the difference

PointMultiple + multipleMultiple + non-multiple
Example (divisor 7)35 + 21 = 5635 + 22 = 57
Divisible by 7?yesno
Reason7 × 5 + 7 × 3 = 7 × 87 × 5 + 22: 22 is not 7 × k

Practice questions

Easy
  1. 1Is 63 + 27 divisible by 9?
  2. 2Is 40 + 9 divisible by 4?
  3. 3A number is divisible by 6. Name two smaller numbers that also divide it.
Medium
  1. 4A number is divisible by 4 and by 5. By which number must it be divisible?
  2. 5How many multiples of 7 are there from 1 to 100?
  3. 6How many factors does 36 have?
Challenging
  1. 75 divides a. Does 5 divide a × b for every whole number b?
  2. 8Check with 3 × 4 × 5 × 6 whether the product of four consecutive numbers is divisible by 24.
Show answers (try them yourself first)
  1. 1Yes (90 = 9 × 10)
  2. 2No (49 is not a multiple of 4)
  3. 32 and 3
  4. 420
  5. 514
  6. 69
  7. 7Yes, a × b = 5k × b
  8. 8Yes: 360 ÷ 24 = 15

Divisibility tests (2, 3, 4, 5, 6, 8, 9, 10, 11)

From daily life

A shopkeeper has 4821 sweets and wants to share them equally among 3 shops. Without dividing, he adds the digits 4 + 8 + 2 + 1 = 15, a multiple of 3, so the sweets can be shared equally.

Simple definition

Divisibility tests: 2: last digit even. 5: last digit 0 or 5. 10: last digit 0. 4: last two digits. 8: last three digits. 3 and 9: digit sum. 6: even and digit sum multiple of 3. 11: alternating digit sum.

Rule 1: Tests that look at the end of the number — 100 is a multiple of 4 and 25, and 1000 is a multiple of 8, so only the last 2 or 3 digits matter.

  • Easy: 3128 ÷ 4: last two digits 28 = 4 × 7, so yes.
  • Medium: 5432 ÷ 8: last three digits 432 = 8 × 54, so yes.
  • Exam-type: 4716 ÷ 12: divisible by 3 (digit sum 18) and by 4 (16), so yes.

Rule 2: Tests that add the digits — 10, 100, 1000 are each one more than a multiple of 9, so a number leaves the same remainder as its digit sum.

  • Easy: 4821: 4 + 8 + 2 + 1 = 15, so divisible by 3 but not by 9.
  • Medium: 7254: 7 + 2 + 5 + 4 = 18, so divisible by 9.
  • Exam-type: 47d2 divisible by 9: 13 + d must be 18, so d = 5 (4752).

Rule 3: Test for 11 — Add digits in odd places and in even places separately; if the difference is 0 or a multiple of 11, the number is divisible by 11.

  • Easy: 1331: (1 + 3) − (3 + 1) = 0, so yes.
  • Medium: 2728: (2 + 2) − (7 + 8) = −11, so yes.
  • Exam-type: 10,00,001: digits 1,0,0,0,0,0,1; (1 + 0 + 0 + 1) − (0 + 0 + 0) = 2, so no.
Remember

To test a composite number, test its coprime parts: 6 = 2 × 3, 12 = 3 × 4, 15 = 3 × 5. A number divisible by 3 but not 9 has a digit sum that is a multiple of 3 only.

Common mistake

Testing 6 with only the 3-test: 21 is divisible by 3 but not by 6 because it is odd. Always check evenness too.

Shortcut

Split a number you want to test into coprime parts and test each part. Memory clue: "6 = 2 × 3, 12 = 3 × 4".

Spot the difference

PointTests by end digits (2, 4, 5, 8, 10)Tests by digit sum (3, 9)
What to look atonly the last digitsall the digits added
Example3128 ÷ 4 (28)7254 ÷ 9 (18)
Why it works100 and 1000 are multiples of 4, 25, 810, 100, ... are 1 more than multiples of 9

Practice questions

Easy
  1. 1Is 7358 divisible by 2?
  2. 2Is 6350 divisible by 5?
  3. 3Is 4821 divisible by 3?
Medium
  1. 4Is 5432 divisible by 8?
  2. 5Is 7254 divisible by 9?
  3. 6Find the smallest 4-digit number divisible by 9.
Challenging
  1. 7Which digit * makes 52*8 divisible by 9?
  2. 8Find the largest 3-digit multiple of 11.
Show answers (try them yourself first)
  1. 1Yes (last digit 8)
  2. 2Yes (last digit 0)
  3. 3Yes (digit sum 15)
  4. 4Yes (432 = 8 × 54)
  5. 5Yes (digit sum 18)
  6. 61008
  7. 73
  8. 8990
Live figure

Divisibility tests (2, 3, 4, 5, 6, 8, 9, 10, 11)

In the app you can move this figure with your finger.

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Digit sums, digital roots and remainders

From daily life

Accountants use a quick check called "casting out nines" to catch mistakes in long bills. They shrink every number to a single digit and compare, which is faster than redoing the whole sum.

Simple definition

The digital root of a number is found by adding its digits again and again until one digit is left. It equals the remainder on division by 9 (9 if the remainder is 0). Also: dr(a + b) = dr(dr(a) + dr(b)) and dr(a × b) = dr(dr(a) × dr(b)).

248751

In the figure: Digital roots of 2, 4, 8, 16, 32, 64 ... repeat: 2, 4, 8, 7, 5, 1.

Rule 1: Finding a digital root — Add the digits; if the result has two digits, add again.

  • Easy: 4876: 4 + 8 + 7 + 6 = 25, 2 + 5 = 7.
  • Medium: 123456: 1 + 2 + 3 + 4 + 5 + 6 = 21, 2 + 1 = 3.
  • Exam-type: 15625: 1 + 5 + 6 + 2 + 5 = 19, 1 + 9 = 10, 1 + 0 = 1.

Rule 2: Remainder on dividing by 9 — The remainder is the digital root, with 9 replaced by 0.

  • Easy: 5432: digit sum 14, dr 5, so the remainder is 5.
  • Medium: 7000: dr 7, so the remainder is 7.
  • Exam-type: 1234567: digit sum 28, 2 + 8 = 10, 1 + 0 = 1, so the remainder is 1.

Rule 3: Checking a calculation — Do the same operation on the digital roots. If the result does not match, the answer is certainly wrong.

  • Easy: 47 + 86 = 133: dr 2 + dr 5 = 7 and dr(133) = 7 ✓.
  • Medium: 77 × 12 = 924: 5 × 3 = 15 → 6 and dr(924) = 6 ✓.
  • Exam-type: 238 × 17: 4 × 8 = 32 → 5. The answer 4046 has dr 5 ✓, but 4056 has dr 6, so it is wrong.
Remember

Multiples of 9 always have digital root 9. A perfect square has digital root only 1, 4, 7 or 9, so a number with digital root 2, 3, 5, 6 or 8 is never a perfect square.

Common mistake

Writing the digital root of a multiple of 9 as 0. It is 9. Also, a matching check does not prove the answer is right; it only proves that a mismatch means a mistake.

Shortcut

While adding digits, cross out any 9, and any digits that add to 9 (4 + 5, 1 + 8). Memory clue: "Nines don't count - cast them out".

Spot the difference

PointDigital rootRemainder on dividing by 9
Possible values1 to 90 to 8
For 543255
For 72990

Practice questions

Easy
  1. 1Find the digital root of 3458.
  2. 2Find the digital root of 729.
  3. 3Find the remainder when 41 is divided by 9.
Medium
  1. 4Find the digital root of 2¹⁰ = 1024.
  2. 5Find the digital root of 1,000,001.
  3. 6Find the remainder when 6789 is divided by 9.
Challenging
  1. 7A student writes 56 × 43 = 2418. Use digital roots to check whether it is certainly wrong.
  2. 8Find the smallest 3-digit number with digital root 4.
Show answers (try them yourself first)
  1. 12
  2. 29
  3. 35
  4. 47
  5. 52
  6. 63
  7. 7Yes: dr(56) × dr(43) = 2 × 7 = 14 → 5, but dr(2418) = 6
  8. 8103
Live figure

Digit sums, digital roots and remainders

In the app you can move this figure with your finger.

Open in the app

Digits in disguise and always/sometimes/never

From daily life

In a puzzle on a quiz show, "A + A + A = 21" hides a digit. And a friend says "the sum of two odd numbers is always even". Are such claims true every time, only sometimes, or never?

Simple definition

In letter puzzles each letter is one digit, and AB means 10A + B. Use place value, digit sums and divisibility tests. To judge a statement: ONE counterexample defeats "always"; ONE example defeats "never"; algebra (even = 2k, odd = 2k + 1) can prove "always".

Rule 1: Solving letter puzzles — Turn the puzzle into an equation or a divisibility test and check the answer.

  • Easy: A + A + A = 21 → 3A = 21 → A = 7.
  • Medium: 5A2 divisible by 9: 7 + A must be 9, so A = 2 (522 = 9 × 58).
  • Exam-type: AB with A + B = 9 and AB − BA = 27: 9(A − B) = 27, A − B = 3, so A = 6, B = 3 and AB = 63.

Rule 2: Number and its reverse — AB + BA = 11 × (A + B) and AB − BA = 9 × (A − B). For three digits, ABC − CBA = 99 × (A − C).

  • Easy: AB + BA = 121 gives A + B = 11.
  • Medium: 72 − 27 = 45 = 9 × 5 (A − B = 5).
  • Exam-type: 842 − 248 = 594 = 99 × 6 (A − C = 6).

Rule 3: Always, sometimes or never? — Try small examples (odd and even, 0, 1, 2); look for a counterexample; then try to prove.

  • Easy: Sum of two odd numbers is even: (2a + 1) + (2b + 1) = 2(a + b + 1). Always.
  • Medium: A number divisible by 6 is divisible by 4: 12 yes, 18 no. Sometimes.
  • Exam-type: A number ending in 5 is divisible by 10: multiples of 10 end in 0. Never.
Remember

One counterexample is enough to disprove an "always" claim, but many examples never prove it. 2 + 3 = 5 shows that the sum of two primes is not always even.

Common mistake

Calling a statement "always true" after checking a few examples such as 12 and 24. Check the odd cases and the small numbers too.

Shortcut

Try small numbers, try both odd and even, then try to prove. Memory clue: "Try small, try both, then prove".

Spot the difference

PointChecking examplesAlgebra proof
Can show a claim is false?yes (one counterexample)yes
Can prove "always"?noyes
Speedquickneeds letters like 2k and 2k + 1

Practice questions

Easy
  1. 1Find the digit A if A + A = 14.
  2. 2"The sum of two even numbers is odd." Always, sometimes or never?
  3. 3A two-digit number and its reverse add up to 99. Find the sum of the digits.
Medium
  1. 4Find A so that 3A5 is divisible by 9.
  2. 5"The product of two odd numbers is even." Always, sometimes or never?
  3. 6"The square of an odd number is odd." Always, sometimes or never?
Challenging
  1. 7Find the smallest 3-digit number whose digits add up to 15.
  2. 8"The sum of two primes is odd." Always, sometimes or never?
Show answers (try them yourself first)
  1. 17
  2. 2Never (even + even = even)
  3. 39 (11 × 9 = 99)
  4. 41 (315 = 9 × 35)
  5. 5Never (odd × odd = odd)
  6. 6Always
  7. 7159
  8. 8Sometimes (2 + 3 = 5 odd; 3 + 5 = 8 even)