1. Chapter at a Glance
Why does a number divisible by 9 always have a digit sum divisible by 9? Which numbers can be written as a sum of consecutive numbers? How does the "1089 trick" always work? In this chapter you will use algebra and reasoning to explain divisibility rules and number tricks, rather than just memorising them. This is real mathematical thinking: noticing a pattern and then proving why it must always hold.
- Divisibility tests for 2, 3, 4, 5, 8, 9, 10 and 11 — and why they work.
- Digit sums and the "casting out nines" check.
- Writing numbers as sums of consecutive numbers.
- Explaining number tricks with algebra.
- Remainders and simple clock (modular) arithmetic.
2. Why Divisibility Tests Work
Any number can be written using place value. A 4-digit number with digits a, b, c, d is 1000a + 100b + 10c + d.
Tests for 2, 5 and 10
1000a + 100b + 10c is always a multiple of 10 (and so of 2 and 5). So divisibility by 2, 5 or 10 depends only on the last digit d.
Tests for 4 and 8
100 is a multiple of 4, so only the last two digits (10c + d) matter for 4. 1000 is a multiple of 8, so only the last three digits matter for 8.
Tests for 3 and 9
The first bracket is always a multiple of 9 (and of 3). So the number and its digit sum a + b + c + d leave the same remainder when divided by 9 (or 3). Hence a number is divisible by 9 exactly when its digit sum is.
Test for 11
1001, 99 and 11 are all multiples of 11. So the number is divisible by 11 when (d + b) − (c + a) — the alternating sum of digits from the right — is a multiple of 11 (including 0).
Test 9,25,782 for divisibility by 9 and 11.
Solution: Digit sum = 9 + 2 + 5 + 7 + 8 + 2 = 33 → not a multiple of 9 (remainder 6). Alternating sum from the right: (2 + 7 + 2) − (8 + 5 + 9) = 11 − 22 = −11 → divisible by 11.
3. Casting Out Nines
Since a number and its digit sum leave the same remainder on division by 9, we can check calculations quickly. Reduce each number to a single digit by repeatedly adding digits (its digital root).
Check whether 347 × 28 = 9716 might be correct.
Solution: Digital roots: 347 → 14 → 5; 28 → 10 → 1. 5 × 1 = 5. 9716 → 23 → 5. They match, so the answer passes the check (and is in fact correct). If they did not match, the answer would certainly be wrong.
Casting out nines can catch many errors, but not all — for example, it cannot detect swapped digits (9716 and 9176 have the same digital root).
4. Sums of Consecutive Numbers
Many numbers can be written as a sum of two or more consecutive counting numbers:
- 9 = 4 + 5 = 2 + 3 + 4
- 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
- Two consecutive numbers n + (n + 1) = 2n + 1 — always odd.
- Three consecutive numbers (n − 1) + n + (n + 1) = 3n — always a multiple of 3.
The powers of 2 (1, 2, 4, 8, 16, 32, …) are the only counting numbers that cannot be written as a sum of two or more consecutive counting numbers. Every other number can.
5. Explaining Number Tricks with Algebra
The 1089 trick
- Take a 3-digit number whose first and last digits differ by at least 2, e.g. 742.
- Reverse it: 247. Subtract the smaller from the larger: 742 − 247 = 495.
- Reverse the result: 594. Add: 495 + 594 = 1089.
Why? For digits a, b, c (a > c), (100a + 10b + c) − (100c + 10b + a) = 99(a − c). The multiples of 99 from 198 to 891 all have the form x9y with x + y = 9, and such a number plus its reverse is always 1089.
Think of a number
"Think of a number, multiply by 4, add 12, divide by 2, subtract twice your number." Algebra: 4n + 12 → (4n + 12)/2 = 2n + 6 → 2n + 6 − 2n = 6 always.
6. Remainders and Clock Arithmetic
On a 12-hour clock, 9 hours after 8 o'clock is 5 o'clock, because 8 + 9 = 17 and 17 leaves remainder 5 when divided by 12. Working only with remainders is called modular (clock) arithmetic.
- Days of the week repeat every 7 days: 100 days after a Monday is 100 ÷ 7 = 14 weeks and 2 days → a Wednesday.
- The remainder of a sum = remainder of the sum of remainders.
7. Solved Problems (NCERT Style)
Find the digit A if 4A72 is divisible by 9.
Solution: 4 + A + 7 + 2 = 13 + A must be a multiple of 9 → A = 5 (sum 18). A = 5.
Is 7,314 divisible by 6?
Solution: Even (ends in 4) and digit sum 15 is a multiple of 3. So it is divisible by 2 and 3, hence by 6.
Write 30 as a sum of consecutive numbers in three ways.
Solution: 9 + 10 + 11; 6 + 7 + 8 + 9; 4 + 5 + 6 + 7 + 8.
8. Master Formula Table
| Divisor | Test | Reason |
|---|---|---|
| 2, 5, 10 | last digit | 10 is a multiple of each |
| 4 | last two digits | 100 = 4 × 25 |
| 8 | last three digits | 1000 = 8 × 125 |
| 3, 9 | digit sum | 10ⁿ − 1 is a multiple of 9 |
| 11 | alternating digit sum | 10 ≡ −1 (mod 11) |
| 6 | tests for 2 and 3 | 6 = 2 × 3 (co-prime) |
| Two consecutive numbers | 2n + 1 (odd) | — |
| Three consecutive numbers | 3n (multiple of 3) | — |
9. Master Summary Table
| Idea | Key point | Exam focus |
|---|---|---|
| Divisibility reasons | Place value + algebra | Explain a test |
| Casting out nines | Digital roots match | Check calculations |
| Consecutive sums | Powers of 2 impossible | Find representations |
| Number tricks | Algebra explains them | Prove the result |
| Clock arithmetic | Work with remainders | Days, times |
10. Mind Map
- 2, 4, 5, 8, 10
- 3, 9 digit sum
- 11 alternating
- Digital root
- Casting out nines
- Consecutive sums
- Powers of 2
- 1089
- Clock arithmetic
11. Tips and Tricks
- For divisibility by a composite number, test co-prime factors: 12 → 3 and 4; 15 → 3 and 5.
- Keep reducing digit sums until a single digit remains (digital root).
- For "which day" problems, divide the number of days by 7 and use the remainder.
- Testing 12 using 2 and 6 (not co-prime) — 18 passes both but is not divisible by 12.
- Starting the alternating sum for 11 incorrectly — any consistent start works, but signs must alternate.
12. Important One-liners
- A number and its digit sum leave the same remainder when divided by 9.
- A number is divisible by 4 if its last two digits form a multiple of 4.
- A number is divisible by 11 if its alternating digit sum is a multiple of 11.
- Powers of 2 cannot be written as a sum of consecutive counting numbers.
- The sum of three consecutive numbers is three times the middle one.
- The 1089 trick works because the difference is always a multiple of 99.
- Clock arithmetic works with remainders.
13. Exam-pattern Questions with Detailed Answers
Which number is divisible by 8?
- 5,124
- 7,368
- 4,326
- 9,190
Show answer
Check last three digits: 124 ÷ 8 no; 368 ÷ 8 = 46 yes; 326 no; 190 no.
Answer: (B) 7,368
Find the digital root of 78,456.
Show answer
7 + 8 + 4 + 5 + 6 = 30 → 3 + 0 = 3.
Answer: 3
What day is it 30 days after a Sunday?
Show answer
30 = 4 × 7 + 2 → 2 days after Sunday = Tuesday.
Answer: Tuesday
Find the smallest digit d such that 3,5d6 is divisible by 4 and by 3.
Show answer
Divisible by 4: d6 ∈ {16, 36, 56, 76, 96} → d ∈ {1, 3, 5, 7, 9}. Divisible by 3: 3 + 5 + d + 6 = 14 + d multiple of 3 → d ∈ {1, 4, 7}. Common smallest: d = 1.
Answer: d = 1 (3,516)
Is 61,743 divisible by 11?
Show answer
From the right: (3 + 7 + 6) − (4 + 1) = 16 − 5 = 11 → yes.
Answer: Yes
Which of these cannot be written as a sum of consecutive counting numbers?
- 24
- 32
- 36
- 45
Show answer
32 = 2⁵ is a power of 2.
Answer: (B) 32
Use casting out nines to show that 456 × 37 = 16,972 is wrong.
Show answer
456 → 15 → 6; 37 → 10 → 1; 6 × 1 = 6. 16,972 → 25 → 7. 6 ≠ 7, so it is wrong. (Correct: 16,872 → 24 → 6.)
Answer: Digital roots differ (6 vs 7)
Show that the sum of a 2-digit number and its reverse is always a multiple of 11.
Show answer
(10a + b) + (10b + a) = 11a + 11b = 11(a + b).
Answer: 11(a + b)
Show that the product of any three consecutive numbers is divisible by 6.
Show answer
Among any three consecutive numbers, at least one is even (divisible by 2) and exactly one is a multiple of 3. So the product is divisible by 2 × 3 = 6.
Answer: Contains a factor 2 and a factor 3
Find the remainder when 2¹⁰⁰ is divided by 7.
Show answer
Powers of 2 mod 7: 2, 4, 1, 2, 4, 1, … (cycle of 3). 100 = 3 × 33 + 1, so 2¹⁰⁰ has the same remainder as 2¹ = 2.
Answer: 2
Write 100 as a sum of consecutive numbers in two different ways.
Show answer
18 + 19 + 20 + 21 + 22 = 100 (5 terms, middle 20). 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100 (8 terms).
Answer: 18 + … + 22 and 9 + … + 16
A 6-digit number is formed by repeating a 3-digit number (e.g. 381381). Show it is always divisible by 7, 11 and 13.
Show answer
abcabc = abc × 1000 + abc = abc × 1001, and 1001 = 7 × 11 × 13.
Answer: abcabc = abc × 7 × 11 × 13
14. Quick Revision Summary
- Divisibility tests come from place value: last digits for 2, 4, 5, 8, 10; digit sums for 3, 9; alternating sums for 11.
- Casting out nines checks calculations using digital roots.
- All numbers except powers of 2 can be written as sums of consecutive numbers.
- Algebra explains tricks like 1089 and "think of a number".
- Clock arithmetic: work with remainders (days mod 7, hours mod 12).