What is 5.2 Checking Divisibility Quickly in NCERT Class 8 Mathematics Chapter 5?

5.2 Checking Divisibility Quickly is the topic of Part 2 of Chapter 5 (Ch 5 — Number Play) in NCERT Class 8 Mathematics. This lesson builds intuition through examples, visuals, and graded practice.

TOPIC 15 OF 22

5.2 Checking Divisibility Quickly

Q: What is Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool in NCERT Class 8 Mathematics?

Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool is a key concept covered in NCERT Class 8 Mathematics, Chapter 5: Number Play. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

Q: How do I solve problems on Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool step by step?

To solve problems on Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty u2014 start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

Q: Is Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool important for the Class 8 board exam?

Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

Q: What mistakes should students avoid in Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool?

Common mistakes in Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Q: Where can I find more NCERT practice questions on Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool?

End-of-chapter NCERT exercises for Part 2 u2014 Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

🎓 Class 8 Mathematics CBSE Theory Ch 5 — Number Play ⏱ ~18 min

🌐 Language: [gtranslate]

🧠 AI-Powered MCQ Assessment ▲

This MCQ module is based on: 5.2 Checking Divisibility Quickly

No. of Questions:

Difficulty Level:

📐 Maths Assessment ▲

This mathematics assessment will be based on: 5.2 Checking Divisibility Quickly
Targeting Class 8 level in Number Theory, with Basic difficulty.

Question Type:

Number of Questions:

Upload Additional Content (Optional):

Upload images, PDFs, or Word documents to include their content in assessment generation.

5.2 Checking Divisibility Quickly

You have learnt simple divisibility tests^? for 2, 4, 5, 8 and 10. Let us revisit them:

Quick Rules

÷ by 2 → last digit is 0, 2, 4, 6, 8.
÷ by 5 → last digit is 0 or 5.
÷ by 10 → last digit is 0.
÷ by 4 → last two digits form a multiple of 4.
÷ by 8 → last three digits form a multiple of 8.

Why these rules work (place-value view)

Any 5-digit number \(abcde\) can be expanded as:
\(a \times 10000 + b \times 1000 + c \times 100 + d \times 10 + e\).

Since 10000, 1000, 100, 10 are all multiples of 2 (and of 5), only the units digit \(e\) matters for ÷2 and ÷5. For ÷4, only \(de\) (last two digits) matters because 100 is divisible by 4. For ÷8, only \(cde\) matters because 1000 is divisible by 8.

A Shortcut for Divisibility by 9

Can you, without actually calculating, check if these numbers are divisible by 9? 999, 909, 900, 90, 990, 9999 — yes, all of them.

Can we say that any number made up only of the digits '0' and '9', in any order, is divisible by 9? Yes.

Key Identity

Since 9, 99, 999, 9999 are all multiples of 9: \(10 = 9+1\), \(100 = 99+1\), \(1000 = 999+1\). So in any digit's place-value expansion, the "9-part" contributes a multiple of 9 and the "+1" leaves behind only the digit itself.

For example, for the remainder of 7309 when divided by 9, we can just add all the digits: \(7 + 3 + 0 + 9 = 19\). This can be seen as a sum of multiples of 9 plus the sum of digits:

Place-value decomposition shows: remainder of number ÷ 9 = remainder of digit sum ÷ 9.

Rule for ÷ 9: A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

🔵 Which are correct? (i) If sum of digits is divisible by 9, number is divisible by 9. ✓ correct.
(ii) If number is divisible by 9, sum of digits is divisible by 9. ✓ correct.
(iii) If sum of digits is not divisible by 9, number is not divisible by 9. ✓ correct.
(iv) If sum of digits is divisible by 9, but number is not divisible by 9. ✗ impossible.

Figure it Out (Divisibility by 9)

Q1. Find, without dividing, whether the following numbers are divisible by 9: (i) 123 (ii) 405 (iii) 93547 (iv) 358095

(i) 1+2+3 = 6, not ÷9, so no. (ii) 4+0+5 = 9 ✓ yes. (iii) 9+3+5+4+7 = 28, not ÷9, no. (iv) 3+5+8+0+9+5 = 30, not ÷9, no.

Q2. Find the smallest multiple of 9 with no odd digits.

We need digits from {0, 2, 4, 6, 8} summing to a multiple of 9. Smallest digit sum = 18 (using 2+4+6+8+? or 6+2+ pairs). Best: digits 2, 8 → 28 (no). Use sum 18: try 288 → 2+8+8=18 ✓. Check smaller: 198 has odd 1, no. 288.

Q3. Find the multiple of 9 which is closest to the number 6000.

6000 ÷ 9 = 666.67. So 9 × 666 = 5994, 9 × 667 = 6003. Closest: 6003 (difference 3 vs 6).

Q4. How many multiples of 9 are there between the numbers 4300 and 4400?

First multiple ≥ 4300: 9 × 478 = 4302. Last ≤ 4400: 9 × 488 = 4392. Count = 488 − 478 + 1 = 11 multiples.

A Shortcut for Divisibility by 3

Since 9 is a multiple of 3, and 10, 100, 1000 all leave remainder 1 when divided by 3, the same argument applies:

Rule for 3

A number is divisible by 3 if and only if the sum of its digits is divisible by 3. Examples: 15 (1+5=6), 33 (3+3=6), 60 (6+0=6), 87 (8+7=15).

A Shortcut for Divisibility by 11

Check divisibility of 11 by using place-value with alternating signs. The key identity:

Place	Value	How related to 11
Units (1)	1	1 = 0×11 + 1
Tens (10)	10	10 = 1×11 − 1
Hundreds (100)	100	100 = 9×11 + 1
Thousands (1000)	1000	1000 = 91×11 − 1
Ten-thousands (10000)	10000	10000 = 909×11 + 1

Place values alternate: one more than a multiple of 11, one less. This leads to the alternating-sum rule.

Procedure

Place alternating '+' and '−' signs before every digit starting from the units digit.
Evaluate the expression. If the result is 0 or a multiple of 11, the number is divisible by 11.

Example 328105: \(-3 + 2 - 8 + 1 - 0 + 5 = -3\). Since −3 is not 0 or a multiple of 11, 328105 is not divisible by 11.

Alternating-sign test for divisibility by 11.

🔵 Using these observations, can you tell whether the number 462 is divisible by 11? Check: \(-4 + 6 - 2 = 0\). Zero is a multiple of 11, so yes, 462 is divisible by 11 (462 = 42 × 11).

Activity: Divisibility Detective

L3 Apply

Materials: 20 number cards (4–6 digit numbers), pen, paper

Predict: Without long division, can you sort 20 numbers into groups by which of {3, 9, 11} divides them in under 5 minutes?

Write 20 numbers of 4–6 digits on cards.
For each card, compute digit sum (for ÷3, ÷9) and alternating sum (for ÷11).
Sort cards into groups. Verify a few by long division.
Time yourself. Swap shortcuts for ÷2, ÷4, ÷5, ÷8 too.

Shortcuts make this 5–10× faster than long division. For numbers with 6+ digits, shortcut tests are the only practical method without a calculator.

More on Divisibility Shortcuts

Divisibility by 6: Since 6 = 2 × 3, a number is divisible by 6 iff it is divisible by BOTH 2 and 3.

Divisibility by 12: Since 12 = 4 × 3 (with 4 and 3 coprime), a number is divisible by 12 iff it is divisible by both 4 and 3. Caution: Checking 6 and 2 does not work since their LCM is only 6. For example, 18 is divisible by 2 and 6 but not 12.

Divisibility by 24: 24 = 8 × 3 (coprime). So test ÷8 and ÷3. Checking ÷4 and ÷6 does NOT work (example: 12 divisible by 4 and 6, not 24).

General Principle

If \(n = a \times b\) where \(a, b\) are coprime (no common factor other than 1), then a number is divisible by \(n\) iff it is divisible by both \(a\) and \(b\). If \(a\) and \(b\) share a factor, this rule fails.

Digital Roots

Take a number. Add its digits repeatedly until you get a single-digit number. This single-digit number is called the digital root^? of the number.

Example 489710: \(4+8+9+7+1+0 = 29\); \(2+9 = 11\); \(1+1 = 2\). Digital root = 2.

Key Property

The digital root equals the remainder when divided by 9 (unless the number is a multiple of 9, in which case the digital root is 9). This is why the ÷9 shortcut works!

🔵 What property do you think this digital root will have? Between the numbers 600 and 700, which numbers have the digital root (i) 5 (ii) 7 (iii) 3? Answer: digital root \(r\) means \(N \equiv r \pmod 9\). In 600–700 range (100 numbers): each residue appears 11 times.

🔵 Write the digital roots of any 12 consecutive numbers. What do you observe? They cycle through 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3 (and so on) — a cycle of 9.

🔵 The digital roots of multiples of 9 are always 9. ✓ The digital roots of consecutive multiples of (i) 3, (ii) 4, (iii) 6? (i) cycle 3, 6, 9 (ii) cycle 4, 8, 3, 7, 2, 6, 1, 5, 9 (iii) cycle 6, 3, 9.

Find the Digital Root (Exercise)

Q1. The digital root of an 8-digit number is 5. What will be the digital root of this number?

Digital root is already 5 — the single-digit result from repeated digit summing. Answer: 5.

Q2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers?

Adding 11 repeatedly increases by 11 each step; 11 ≡ 2 (mod 9). So digital roots increase by 2 each time (mod 9): e.g., if starting at 5 → 5, 7, 9, 2, 4, 6, 8, 1, 3, 5, ... — cycle of 9 values.

Q3. What will be the digital root of the number 9 × 36 × 137?

Since 9 is a factor, the product is a multiple of 9 → digital root = 9.

Q4. Make conjectures by examining if there are any patterns on relations between:
(i) the parity of a number and its digital root
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.

(i) No fixed relation — an even number can have any digital root 1–9. E.g., 10 → 1 (odd), 12 → 3 (odd), 14 → 5 (odd), 20 → 2 (even). (ii) Digital root mod 9 = remainder on ÷9. Digital root mod 3 = remainder on ÷3.

Competency-Based Questions

Scenario: A warehouse manager must verify the total serial count of shipments. A 6-digit serial code \(N = 4\,7\,a\,3\,9\,b\) is printed on a batch. Records show \(N\) must be divisible by 99.

Q1. What must be true for \(N\) to be divisible by 99?

L3 Apply

(a) divisible by 11 only
(b) divisible by 9 only
(c) divisible by both 9 and 11
(d) digit sum = 99

Answer: (c). 99 = 9 × 11, and 9 & 11 are coprime. So need divisibility by both.

Q2. Determine all valid pairs (a, b) with digits 0–9 such that \(N\) is divisible by 99.

L4 Analyse

÷9: Digit sum = 4+7+a+3+9+b = 23+a+b ≡ 0 (mod 9) → a+b ≡ 4 (mod 9), so a+b = 4 or 13.
÷11 (alternating from units): +b − 9 + 3 − a + 7 − 4 = b − a − 3. Need ≡ 0 (mod 11) → b − a = 3 or b − a = −8.
Case a+b=4, b−a=3: a=½ invalid. Case a+b=13, b−a=3: a=5, b=8 ✓. Case a+b=13, b−a=−8: a=10.5 invalid. Case a+b=4, b−a=−8: negative a. Only (a=5, b=8).

Q3. Evaluate: A colleague claims the shortcut "divisibility by 99 = divisibility by 9 and 11" fails when the number shares common factors. Evaluate this claim.

L5 Evaluate

Claim is false. The shortcut works because \(\gcd(9, 11) = 1\). The rule fails only when the two factors themselves share a common factor (e.g., testing ÷12 with 4 and 6 — both multiples of 2). Since 9 and 11 are coprime, the shortcut is always valid.

Q4. Design a 7-digit serial format \(d_1 d_2 d_3 d_4 d_5 d_6 d_7\) such that any valid code is divisible by 1001. Give the factorisation and describe the necessary conditions.

L6 Create

1001 = 7 × 11 × 13 (all coprime). Need divisibility by all three. Simple design: use any 3-digit pattern \(d_1 d_2 d_3\) and repeat as \(d_1 d_2 d_3 \, d_1 d_2 d_3\) (6 digits: e.g. 123123). Such numbers = \(\overline{d_1 d_2 d_3} \times 1001\), automatically ÷1001. Add a check digit at front. Many designs possible.

Assertion–Reason Questions

Assertion (A): A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
Reason (R): The place values 1, 10, 100, 1000, ... all leave remainder 1 when divided by 3.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Every place value \(10^k \equiv 1 \pmod 3\). So the number's remainder mod 3 equals the digit-sum mod 3.

Assertion (A): A number is divisible by 12 if it is divisible by both 4 and 6.
Reason (R): \(12 = 4 \times 6 / \gcd(4, 6)\), and dual divisibility gives divisibility by their LCM.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — LCM(4,6) = 12. A number divisible by 4 and 6 is divisible by 12. R correctly explains A.

Assertion (A): Divisibility by 24 can be tested by checking ÷4 and ÷6.
Reason (R): 24 = 4 × 6, and this equals LCM(4, 6).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d) — A is false: LCM(4, 6) = 12, not 24. Example: 12 is divisible by 4 and 6 but not 24. R is also false (24 ≠ LCM(4,6)). Correct test: ÷8 and ÷3 (since \(\gcd(8, 3) = 1\) and \(8 \times 3 = 24\)).

Frequently Asked Questions — Number Play

What is Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool in NCERT Class 8 Mathematics?

Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool is a key concept covered in NCERT Class 8 Mathematics, Chapter 5: Number Play. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool step by step?

To solve problems on Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 5: Number Play?

The essential formulas of Chapter 5 (Number Play) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool important for the Class 8 board exam?

Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool?

Common mistakes in Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool?

End-of-chapter NCERT exercises for Part 2 — Divisibility Shortcuts, Digit Sums & Digital Roots | Class 8 Maths | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

AI Tutor

Mathematics Class 8 — Ganita Prakash Part I

Ready

Hi! 👋 I'm Gaura, your AI Tutor for 5.2 Checking Divisibility Quickly. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.