✨ Term 1 Maths · Chapter 5

Number Play

Can you tell if a HUGE number is divisible by 9 or 11 — without dividing at all? Discover the shortcuts, crack cryptarithm puzzles, and try the live divisibility checker below!

Is this a multiple of…?

Hi, it's Patto! Take any 4 consecutive numbers, like 3, 4, 5, 6. Put + or − signs between them any way you like. No matter which signs you choose, the answer is always even! Let's find out why.

📖 Why this always works

Flipping just ONE sign (say, changing +b to −b) changes the total by 2b — and 2 times ANYTHING is always even. So flipping signs can never change whether the result is even or odd — it always stays the same parity!

Rule
odd ± odd = even
even ± even = even
odd ± even = odd

💡 Key divisibility rules (proved with algebra!)

If a divides both M and N, then it divides M+N and M-N too. If a number is divisible by b AND by c, it's divisible by LCM(b, c) — NOT just b×c (careful: 12 is divisible by both 4 and 6, but NOT by 24!).

🪨 The pebble riddle

Grouped by 3's, one pebble is left over. Grouped by 5's, one is still left over. Grouped by 7's, they divide perfectly. There are fewer than 100 pebbles. How many are there?

91 pebbles! (91 ÷ 3 leaves remainder 1, 91 ÷ 5 leaves remainder 1, and 91 ÷ 7 = 13 exactly.)

🧪 Always, Sometimes, or Never — 8 worked examples from the book

Before the exercises, the book walks through 8 numbered statements about factors and multiples, deciding if each is Always True, Sometimes True, or Never True — using algebra AND a rows-of-dots picture. These are the reasoning tools Q3 and Q12 below are built on!

The two numbers are multiples of 8: 8a and 8b. Picture them as rows of 8 dots — a rows and b rows. Stack the rows together:

8a + 8b = 8(a + b) ← still (a+b) full rows of 8 e.g. 8 + 16 = 8(1+2) = 24 16 + 56 = 72 80 + 120 = 200

Always true. In general, if a divides M and a divides N, then a divides both M+N and M−N too.

A number divisible by 8 is 8m. It CAN be split as a sum of two multiples of 8, like 8m = 8a + 8b (e.g. 72 = 48+24). But it can ALSO be split as a sum of two numbers that are NOT multiples of 8, like 8m = p + q (e.g. 72 = 50+22, and neither 50 nor 22 is a multiple of 8).

Sometimes true. It depends how you split the number — sometimes both parts are multiples of 8, sometimes neither is.

A number divisible by 7 is 7j. Any multiple of it is (7j) × m = 7jm — that's m groups of j rows of 7, still a total that's a multiple of 7.

14 = 7×2 (j=2). Multiples of 14: 28=(7×2)×2, 70=(7×2)×5, 154=(7×2)×11 — all multiples of 7.

Always true. In general, if A is divisible by k, then all multiples of A are also divisible by k.

A number divisible by 12 is 12m = 2×6×m = 3×4×m — any factor of 12 (1,2,3,4,6,12) divides evenly into "12" itself, so it also divides evenly into "12m", covering complete rows every time.

Factors of 24 (=12×2): 1, 2, 3, 4, 6, 8, 12, 24 — all divide 24 exactly.

Always true. In general, if A is divisible by k, then A is divisible by all the factors of k.

A number divisible by 7 is 7k. A multiple of 7 is 7m. Now 7k is divisible by 7m ONLY when k is itself a multiple of m (say k = ym, so 7k ÷ 7m = 7ym ÷ 7m = y, a whole number).

42 (=7×6) is divisible by 7 AND by 14 (=7×2), since 6 is a multiple of 2. 42 is divisible by 7 but NOT by 28 (=7×4), since 6 is not a multiple of 4.

Sometimes true.

9 and 4 share NO common factor (they're coprime), so LCM(9,4) = 9×4 = 36.

Always true — a number divisible by both 9 and 4 must be divisible by LCM(9,4) = 36.

6 and 4 share a common factor of 2 (NOT coprime), so LCM(6,4) = 12, not 24. Example: 12 is divisible by both 6 and 4, but NOT by 24.

Never true (the correct guaranteed divisor is only LCM(6,4)=12). In general: if A is divisible by k AND by m, then A is divisible by LCM(k, m) — not necessarily by k×m.

Every multiple of 6 is even. But an odd + an even number is always ODD. An odd number can never equal an even multiple of 6 — contradiction straight away! Algebraically, suppose it COULD happen:

2n + (2m + 1) = 6j 2n + 2m = 6j − 1 2(n + m) = 6j − 1

Never true — the left side 2(n+m) is always EVEN, but the right side 6j−1 is always ODD. An even number can never equal an odd number, so this can never happen.

➗ What Remains?

Find a number that leaves remainder 3 when divided by 5 — like 3, 8, 13, 18, 23… Which ALGEBRAIC expression correctly captures ALL such numbers?

Numbers that leave remainder 0 when divided by 5 are the multiples of 5, written 5k. We want numbers that are 3 MORE than a multiple of 5 — so the number should be 5k + 3.

(i) 3k+5 → does NOT always give remainder 3 ÷ 5 (e.g. k=1 → 8, remainder 3 ✓ but k=2 → 11, remainder 1 ✗) — WRONG (ii) 3k−5 → k=1 → −2, not even the right shape — WRONG (iii) 3k/5 → not even always a whole number — WRONG (iv) 5k+3 → k=0,1,2,3,4 → 3, 8, 13, 18, 23 — always remainder 3 ✓ CORRECT (v) 5k−2 → k=1,2,3,4,5 → 3, 8, 13, 18, 23 — same numbers as 5k+3, just shifted k ✓ CORRECT (vi) 5k−3 → k=1 → 2, remainder 2 not 3 — WRONG

Both (iv) 5k+3 AND (v) 5k−2 work! They describe the exact same set of numbers — just with k starting from a different place (k≥0 for 5k+3, or k≥1 for 5k−2).

k	0	1	2	3	4
5k + 3	3	8	13	18	23

k	1	2	3	4	5
5k − 2	3	8	13	18	23

Notice both tables give the SAME output numbers (3, 8, 13, 18, 23…) — just a different starting value of k. That's why both expressions correctly describe "numbers 3 more than a multiple of 5".

📗 Figure It Out — Section 5.1

If p is the BIGGEST, the others count down from it one at a time.

p − 1, p − 2, p − 3, p − 4

(i) The sum of two even numbers is a multiple of 3. Sometimes true — 2+4=6 ✓ multiple of 3, but 2+6=8 ✗ not.

(ii) If a number is not divisible by 18, then it is also not divisible by 9. Sometimes true — 30 is not divisible by 18 or 9 (fits), but 27 is not divisible by 18 yet IS divisible by 9 (breaks it).

(iii) If two numbers are not divisible by 6, their sum is not divisible by 6. Sometimes true — 9+11=20, not divisible by 6 (fits), but 8+10=18, which IS divisible by 6 (breaks it).

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. Always true — multiple of 6 = 6x, multiple of 9 = 9y. Sum = 6x+9y = 3(2x+3y), which always has 3 as a factor.

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9. Sometimes true — 18+9=27 ✓ multiple of 9, but 12+9=21 ✗ not (21÷9 is not exact).

If x is the number, then x−2 must be a multiple of BOTH 3 and 4 — so x−2 must be a multiple of LCM(3,4) = 12.

x − 2 = 12, 24, 36, ... x = 14, 26, 38, ... General form: 12n + 2

Let the three numbers be 6a+2, 6b+2, 6c+2.

Sum = 6a + 6b + 6c + 6 = 6(a + b + c + 1)

Yes, Tathagat's claim is true! The sum is always 6 times a whole number. (We meet this exact case study again in the exam-practice section below.)

Write both numbers algebraically: 4779 = 7p+5 and 661 = 7q+3.

4779 + 661 = 7p + 7q + 8 = 7(p+q+1) + 1 → remainder 1 4779 − 661 = 7p + 5 − 7q − 3 = 7(p−q) + 2 → remainder 2

Visually: think of 4779 as "5 extra beyond a stack of 7s" and 661 as "3 extra beyond a stack of 7s". Adding the extras (5+3=8) gives 1 full extra 7 plus 1 left over; subtracting the extras (5−3=2) leaves 2 over.

Breaking Even & Pairs to Make Fours

Two fun in-book activities about spotting even numbers — WITHOUT actually calculating them fully!

🔢 Activity: Breaking Even

Without computing the full answer, which of these ARITHMETIC expressions are even?

Expression	Even or odd?	Why
43 + 37	Even ✅	odd + odd = even
672 − 348	Even ✅	even − even = even
4 × 347 × 3	Even ✅	has a factor of 4 (even × anything = even)
708 − 477	Odd ❌	even − odd = odd
809 + 214	Odd ❌	odd + even = odd
119 × 303	Odd ❌	odd × odd = odd
543 − 479	Even ✅	odd − odd = even
513 × 3	Odd ❌	odd × odd = odd

Now the ALGEBRAIC expressions — which ALWAYS give an even number, for ANY whole numbers put in?

Expression	Always even?	Why
2a + 2b	Yes ✅	= 2(a+b) — factor of 2
3g + 5h	No ❌	g=1,h=1 gives 8 (even) but g=1,h=2 gives 13 (odd)
4m + 2n	Yes ✅	= 2(2m+n) — factor of 2
2u − 4v	Yes ✅	= 2(u−2v) — factor of 2
13k − 5k	Yes ✅	simplifies to 8k — factor of 2
6m − 3n	No ❌	= 3(2m−n); m=1,n=1 gives 3 (odd)
x² + 2	No ❌	x=6 gives 38 (even) but x=3 gives 11 (odd) — x² keeps the parity of x
b² + 1	No ❌	b=2 gives 5 (odd) but b=3 gives 10 (even)
4k × 3j	Yes ✅	= 12kj — factor of 2 (in fact factor of 12)

💡 The pattern

An expression is ALWAYS even only when EVERY term is guaranteed even — usually because you can pull out a common factor of 2 from the whole expression, like 4m+2q = 2(2m+q).

🔢 Activity: Pairs to Make Fours

Take a pair of even numbers and add them. Is the sum ALWAYS a multiple of 4? Even numbers split into two types: multiples of 4 (remainder 0), and non-multiples of 4 (remainder 2).

Case	Algebra	Result	Examples
multiple of 4 + multiple of 4	4p + 4q	= 4(p+q) — always mult. of 4	12+16=28 ✓
non-mult of 4 + non-mult of 4	(4p+2)+(4q+2)	= 4(p+q+1) — always mult. of 4 (the two 2's add up to 4)	6+10=16 ✓
multiple of 4 + non-mult of 4	4p + (4q+2)	= 4(p+q)+2 — never a mult. of 4 (remainder 2 left over)	12+6=18 ✗

📖 Why this matches even + odd

It's the SAME idea as even+even=even and even+odd=odd — just one level deeper. Two numbers that "match remainders" (both 0 mod 4, or both 2 mod 4) combine to a clean multiple of 4. Two numbers with DIFFERENT remainders mod 4 don't.

Divisibility shortcuts

📖 Divisible by 9? Just add the digits!

A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Keep adding digits until you get a single digit — that's the remainder when dividing by 9!

🔎 Worked example: is 7309 divisible by 9?

7+3+0+9 = 19 1+9 = 10 1+0 = 1

7309 ÷ 9 leaves remainder 1 — NOT divisible by 9.

All FOUR statements are true! This is because "divisible by 9" and "digit sum divisible by 9" are exactly the SAME condition — they always happen together, and are always absent together. Statements (i)/(ii) are the "if...then" pair proving the rule works both ways; (iii)/(iv) are just their opposite (contrapositive) versions, which are automatically true whenever the original statement is true.

📗 Figure It Out — divisibility by 9

123 → 1+2+3=6 → NOT divisible 405 → 4+0+5=9 → DIVISIBLE ✓ 8888 → 8+8+8+8=32 → NOT divisible 93547 → 9+3+5+4+7=28 → NOT divisible 358095 → 3+5+8+0+9+5=30 → NOT divisible

Only 405 is divisible by 9.

Only even digits allowed: 0, 2, 4, 6, 8. Their digit sum must be a multiple of 9 — but you can't make 9 itself from even digits (it's odd!), so the SMALLEST possible digit sum is 18.

2 + 8 + 8 = 18 Arrange smallest first: 288

288 (2+8+8=18, divisible by 9, and every digit is even).

6000 ÷ 9 = 666.67 666 × 9 = 5994 (6 away) 667 × 9 = 6003 (3 away)

6003 is closer (only 3 away, vs 5994 which is 6 away).

First multiple of 9 after 4300 is 4302, last one before 4400 is 4392.

4302, 4311, 4320, 4329, 4338, 4347, 4356, 4365, 4374, 4383, 4392

11 multiples of 9 lie between 4300 and 4400.

💡 Same trick works for divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3 — same idea as the rule for 9!

🎮 Try the live divisibility-by-9 checker

⚠️ Divisibility by 11 is trickier — signs alternate!

Starting from the units digit, place alternating +/− signs on each digit, then add. If the result is 0 (or a multiple of 11), the number is divisible by 11!

🔎 Worked example: is 328105 divisible by 11?

−3 +2 −8 +1 −0 +5 = −3

Result is −3, so 328105 is 3 away from a multiple of 11 — NOT divisible.

(i) 158 → −1+5−8 = −4 → remainder 4 (ii) 841 → −8+4−1 = −5 → remainder 5 (iii) 481 → −4+8−1 = +3 → 3 more, i.e. remainder 8 (iv) 5529 → −5+5−2+9 = +7 → remainder 7 (v) 90904 → −9+0−9+0−4 = −22 → DIVISIBLE by 11 (vi) 857076 → −8+5−7+0−7+6 = −11 → DIVISIBLE by 11

Answers: 4, 5, 8, 7, divisible, divisible.

🎮 Try the live divisibility-by-11 checker

📋 Fill in the blanks: divisibility table

Cover the table, work out each Yes/No yourself using the shortcuts, then check!

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

Quick way: check 2 (last digit even), 5 (last digit 0/5), 4 (last 2 digits ÷4), 8 (last 3 digits ÷8), 3 & 9 (digit sum), 10 (last digit 0), 6 (divisible by BOTH 2 and 3), 11 (alternating sum).

📖 Worked example: checking divisibility by 6 and 24

To check divisibility by 6, use the shortcuts for its factors 2 AND 3 together — a number is divisible by 6 only if BOTH work.

38 → even? yes. digit sum 3+8=11, ÷3? no → NOT div by 6 (38÷6 = 6.33) 225 → even? no → NOT div by 6 (225÷6 = 37.5) 186 → even? yes. digit sum 1+8+6=15,÷3? yes → DIVISIBLE by 6 (186÷6=31) 64 → even? yes. digit sum 6+4=10, ÷3? no → NOT div by 6 (64÷6 = 10.67)

But this "check both factors" trick only works when the two factors share NO common factor (are coprime). 4 and 6 both share a factor of 2, so checking 4-and-6 only guarantees divisibility by LCM(4,6)=12 — NOT 24! Instead, check 3-and-8 (which ARE coprime — 3×8=24=LCM(3,8)) to correctly test for divisibility by 24.

Digital roots

📖 What's a digital root?

Keep adding a number's digits together, over and over, until only ONE digit remains. That final digit is the digital root.

🇮🇳 An ancient Indian shortcut-checking trick

Aryabhata II (c. 950 CE) described this exact digital-root method in his work Mahāsiddhānta — used for over a thousand years to double-check arithmetic calculations!

💡 Digital root connects directly to remainder ÷ 9

If a number's digital root is 9, its remainder when divided by 9 is 0. If the digital root is LESS than 9, that digital root IS the remainder! Digital roots are just a fast way to find remainders ÷ 9.

(i) Digital root 5: 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698.

(ii) Digital root 7: 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691.

(iii) Digital root 3: 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696.

💡 Digital roots of consecutive multiples repeat in a cycle!

Multiples of 3: digital roots go 3, 6, 9, 3, 6, 9, ... — a cycle of 3.

Multiples of 4: digital roots go 4, 8, 3, 7, 2, 6, 1, 5, 9, 4, 8, 3, ... — a cycle of 9.

Multiples of 6: digital roots go 6, 3, 9, 6, 3, 9, ... — a cycle of 3.

🧩 The digital-root riddle poem

"I'm made of digits, each tiniest and odd,
No shared ground with root #1 — how odd!
My digits count, their sum, my root —
All point to one bold number's pursuit —
The largest odd single-digit I proudly claim.
What's my number? What's my name?"

Clue-breaking: "made of digits, each tiniest and odd" → every digit is 1 (the smallest odd digit, "no shared ground with root #1" is a wordplay hint too). "The largest odd single-digit" = 9 → the digit COUNT, the digit SUM, and the digital ROOT must all equal 9. Nine 1's have digit sum 1×9=9, and digital root 9 too — it all fits!

The number is 1,11,11,11,111 (nine 1's) — in the Indian place-value system that's read as "eleven crore eleven lakh eleven thousand one hundred eleven".

📗 Figure It Out — digital roots

Adding 10 to a number usually just adds 1 to the tens digit — which adds 1 to the digit sum too (as long as there's no carrying/borrowing, e.g. 40000001 + 10 = 40000011).

Digital root: 5 + 1 = 6

Start with 10, keep adding 11: 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, ...

10→1, 21→3, 32→5, 43→7, 54→9, 65→2, 76→4, 87→6, 98→8, 109→1, 120→3, ...

The digital roots go up by 2 each time (wrapping around after 9): 1,3,5,7,9,2,4,6,8,1,3,... — because adding 11 is like adding 1+1=2 to the digit sum (until a carry resets things).

9a + 36b + 13 = 9a + 36b + 9 + 4 = 9(a + 4b + 1) + 4

9(a+4b+1) is ALWAYS a multiple of 9, whose digital root is always 9. Adding the leftover 4 pushes the digital root to 9 + 4 = 13 → 1+3 = 4.

(i) Parity vs digital root: No consistent pattern — a number can be even or odd no matter what its digital root is (e.g. digital root 4 can come from 4, 13, 22, 31... which alternate even/odd).

(ii) Digital root vs remainder mod 3 / mod 9: Digital roots 3, 6, 9 → remainder 0 mod 3. Digital roots 1, 4, 7 → remainder 1 mod 3. Digital roots 2, 5, 8 → remainder 2 mod 3. And for mod 9: if digital root is 9, remainder mod 9 is 0; otherwise the digital root itself IS the remainder mod 9.

Digits in disguise (cryptarithms)

In a cryptarithm, each letter secretly stands for a digit — and the first digit of any number can never be 0.

➕ Four addition cryptarithms

(i) A=7, B=9 → 71 + 19 = 90 (ii) A=2, B=5 → 25 + 37 = 62 (iii) N=1,O=3,P=9 → 31 + 31 + 31 = 93 (iv) Q=8,R=5,P=2 → 85 + 85 + 85 = 255

🔍 Worked example: PQ × 8 = RS

PQ is a 2-digit number, and multiplying by 8 must STILL give a 2-digit number (RS). What's the biggest PQ could be?

12 × 8 = 96 ✓ (still 2 digits) 13 × 8 = 104 ✗ (jumps to 3 digits!)

So PQ must be 12 or smaller — and 12 × 8 = 96 works perfectly!

In GH × H = 9K, the SAME digit H must be both the units digit of GH AND the number we're multiplying by. Check each option for that:

11×9=99 → units digit 1, multiplier 9 → don't match, and G=H=1 anyway ✗ 12×8=96 → units digit 2, multiplier 8 → don't match ✗ 46×2=92 → units digit 6, multiplier 2 → don't match ✗ 24×4=96 → units digit 4, multiplier 4 → MATCH! G=2,H=4,K=6, all different ✓ 47×2=94 → units digit 7, multiplier 2 → don't match ✗ 31×3=93 → units digit 1, multiplier 3 → don't match ✗ 16×6=96 → units digit 6, multiplier 6 → match, but then K=6=H — not allowed! ✗

24 × 4 = 96 is the answer — G=2, H=4, K=6.

Since the product RAY is only a 3-digit number, B can't be 2 or more (200×6=1200, already 4 digits) — so B = 1.

Now think about Y: if Y were 7 or more, say Y=7, then BYE would be at least 170-something, and 170 × 6 = 1020 — a 4-digit number, too big! So Y cannot be 7, 8, or 9.

Also, the LAST digit of BYE×6 must come back around to Y again (since the answer ends in ...AY, and Y is BYE's middle digit — the multiplication's units digit E×6 must produce a units digit that, after carrying, keeps the pattern consistent). Working through it, Y must be even.

Testing Y=0, and searching for a valid E: B=1, Y=0, E=5 → BYE = 105 105 × 6 = 630 → R=6, A=3, and the last digit is 0 = Y ✓ All digits B,Y,E,R,A = 1,0,5,6,3 are different ✓

105 × 6 = 630 — so B=1, Y=0, E=5, R=6, A=3.

💡 Cryptarithm-solving strategy

Look at size constraints first (how many digits can the answer have?), then test small numbers systematically. Each letter = exactly one digit, used consistently everywhere it appears!

✖️ Six more multiplication cryptarithms — "Solve the following"

(i) U=5,T=0,P=1 → 50 × 3 = 150 (ii) A=1,B=9,C=5 → 19 × 5 = 95 (iii) L=1,N=5,P=0 → 125 × 2 = 250 (iv) X=2,Y=3,Z=9 → 23 × 4 = 92 (v) P=2,Q=1,R=4 → 22 × 11 = 242 (vi) J=7,K=4 → 74 × 6 = 444

Every question from the book

Cover the answer, try it yourself first, then tap to check!

7 + 8 + 9 + 10 = 34

Notice each remainder is exactly ONE LESS than its divisor (2=3−1, 3=4−1, 4=5−1) — so the number is 1 less than a common multiple of 3, 4, and 5.

LCM(3,4,5) = 60 Smallest number = 60 − 1 = 59

Digit sum: 3+1+z+5 = 9+z, which must be a multiple of 9.

z = 0 → 3105 (digit sum 9) ✓ z = 9 → 3195 (digit sum 18) ✓

Two answers: z=0 or z=9 — both make the digit sum a multiple of 9!

(i) Yes, always true! Reversing digits never changes the digit SUM — and divisibility by 9 only depends on the digit sum.

(ii) Yes — ANY shuffle works, not just reversing! You can rearrange the digits into ANY order you like (not just backwards) and the number will STILL be a multiple of 9. That's because shuffling digits around only changes their POSITIONS, never their VALUES — so the digit sum stays exactly the same no matter how you mix them up. Example: 4527 (digit sum 18, ÷9 ✓) can be shuffled to 2745, 7254, 5274, etc. — every shuffle has digit sum 18, so every shuffle is still divisible by 9.

False! Any multiple of 11 is 11k. Doubling gives 22k = 11(2k) — still ALWAYS a multiple of 11, no exceptions ever.

2 consecutive integers: always a multiple of 2 — one of any two consecutive numbers must be even, so their product is always even.

3 consecutive integers: always a multiple of 6 (2×3=6) — there's always at least one multiple of 2 and one multiple of 3 among any 3 consecutive numbers.

4 consecutive integers: always a multiple of 24 (2×3×4=24) — same idea, extended!

5 consecutive integers: always a multiple of 120 (2×3×4×5=120) — among any 5 consecutive integers there's always at least one multiple each of 2, 3, 4, and 5.

Let the numbers be a=12n+8 and b=12m-4.

a + b = 12(n+m) + 4 = 12k + 4

Snehal is WRONG. The sum is always 4 more than a multiple of 12, but that's NOT guaranteed to be a multiple of 8. Try k=2: 12(2)+4=28, and 28÷8=3.5 — not a multiple of 8!

Let the two multiples of 3 be 3m and 3n.

3m + 3n = 3(m + n)

This is a multiple of 6 exactly when (m+n) is EVEN — i.e. when m and n have the SAME parity (both even or both odd). Example: 3×2+3×4=6+12=18=6×3 ✓ (both even). Non-example: 3×2+3×3=6+9=15, not a multiple of 6 (one even, one odd).

18 = 2 × 9, so the number needs BOTH: b must be even (b = 0,2,4,6,8), and the digit sum 4+8+a+2+3+b = 17+a+b must be a multiple of 9.

b=0: 17+a=18 → a=1 → 481230 b=2: 17+2+a=27 → a=8 → 488232 b=4: 17+4+a=27 → a=6 → 486234 b=6: 17+6+a=27 → a=4 → 484236 b=8: 17+8+a=27 → a=2 → 482238

Possible (a,b) pairs: (1,0), (8,2), (6,4), (4,6), (2,8).

One set: 2, 3, 4 (2 is a multiple of 2, 3 is a multiple of 3, 4 is a multiple of 4) ✓

Since LCM(2,3,4)=12, this pattern repeats every 12 numbers: the next set is 14, 15, 16 (14÷2=7 ✓, 15÷3=5 ✓, 16÷4=4 ✓), then 26, 27, 28, and so on forever.

36 = 4 × 9, and 4 and 9 are coprime, so check BOTH: last 2 digits divisible by 4, AND digit sum divisible by 9.

45036 ÷ 36 = 1251 45072 ÷ 36 = 1252 45108 ÷ 36 = 1253 45144 ÷ 36 = 1254 45180 ÷ 36 = 1255

45036, 45072, 45108, 45144, 45180 — each one exactly 36 more than the last!

Consecutive EVEN numbers differ by 2 each — so counting out from the middle (5p):

5p − 4, 5p − 2, 5p, 5p + 2, 5p + 4

The other four numbers: 5p−4, 5p−2, 5p+2, 5p+4.

Divisible by 15 means divisible by 3 AND 5 (ends in 0 or 5). But if it ends in 0, reversing puts 0 in FRONT — no longer a 6-digit number! So it must end in 5.

Reversed, the number must be divisible by 6 (even AND digit-sum ÷3) — so the reversed number's LAST digit (the original number's FIRST digit) must be even.

Try: 200025 Reversed: 520002 520002 → even ✓, digit sum 5+2+0+0+0+2=9 ÷3 ✓ → divisible by 6 ✓ 200025 → ends in 5, digit sum=9 ÷3 ✓ → divisible by 15 ✓

Example answer: 200025 (many other answers exist too, e.g. 200055, 400035).

(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9. Always true — (6a)(3b) = 18ab = 9(2ab), always has 9 as a factor.

(ii) The sum of three consecutive even numbers is divisible by 6. Always true — 2n+(2n+2)+(2n+4) = 6n+6 = 6(n+1).

(iii) If abcdef is a multiple of 6, then badcef is also a multiple of 6. Always true — swapping digit positions never changes the digit SUM (so divisible-by-3 stays true), and the last digit 'f' is unchanged (so divisible-by-2 stays true too).

(iv) 8(7b−3) − 4(11b+1) is a multiple of 12. Never true — expanding: 56b−24−44b−4 = 12b−28 = 12(b−3)+8, which is always 8 more than a multiple of 12, so never actually a multiple of 12 itself.

Every number leaves remainder 0, 1, or 2 when divided by 3. Look at the THREE remainders together:

Case 1 — all three remainders are the SAME (0+0+0, 1+1+1, or 2+2+2): the remainder-sum is 0, 3, or 6 — all multiples of 3, so the total IS divisible by 3. Example: 4+7+10 (remainders 1,1,1) = 21 = 3×7 ✓

Case 2 — the three remainders are ALL DIFFERENT (one each of 0, 1, 2): remainder-sum = 0+1+2 = 3, a multiple of 3 — so the total is ALSO divisible by 3! Example: 3+7+11 (remainders 0,1,2) = 21 = 3×7 ✓. Only when the remainders are a MIX with exactly two matching (like 0,0,1 or 1,1,2) does the sum fail to be divisible by 3.

(i) E=3, F=7, G=1 37 × 3 = 111 (ii) W=5, O=7, M=2, E=8 575 × 5 = 2875

(iv) is correct — three nested circles, smallest to biggest: Multiples of 32 ⊂ Multiples of 8 ⊂ Multiples of 4.

Since 32 = 4×8, every multiple of 32 is automatically a multiple of 8 AND of 4. And since 8 = 4×2, every multiple of 8 is automatically a multiple of 4. So multiples of 4 form the BIGGEST circle, with multiples of 8 nested inside it, and multiples of 32 nested inside THAT.

Practice like the real exam

Section A (12 MCQ, 1 mark), B (10 × 2 marks), C (8 × 3 marks), D (4 × 4 marks), E (2 case studies, 4 marks). Here's a taste of each.

Section A · MCQ (1 mark each)

(b) the sum of its digits is divisible by 9

(b) 12 — LCM(4,6)=12, NOT 24!

Section B · short answer (2 marks each)

4+8+9+7+1+0 = 29 2+9 = 11 1+1 = 2

Section C · longer answer (3 marks each)

4 and 6 share a common factor of 2 (they're not "coprime"), so a number divisible by both 4 and 6 is only guaranteed to be divisible by their LCM, which is 12 — NOT 24 (12 itself is divisible by 4 and 6, but not 24!). But 3 and 8 share NO common factors (they're coprime), so their LCM equals their product: 3×8=24. A number divisible by both 3 and 8 is guaranteed to be divisible by 24.

Section D · 4 marks each

Divisible by 44 means divisible by BOTH 4 and 11.

(p,q) = (7,0) → 37708 (p,q) = (5,2) → 35728 (p,q) = (3,4) → 33748 (p,q) = (1,6) → 31768

Section E · case study (4 marks)

(a) Such a number has the form 6n+2.

(b) Let the three numbers be 6a+2, 6b+2, 6c+2. Sum = 6a+6b+6c+6 = 6(a+b+c+1), which is ALWAYS a multiple of 6. Tathagat's claim is true!

Navakankari corner 🎲

🇮🇳 A traditional Indian board game

Navakankari (also called Sālu Mane Āṭa, Chār-Pār, or Navkakri) is a traditional Indian strategy board game — the same game known in the West as "Nine Men's Morris" or "Mills". Two players try to form lines of three pawns to capture the opponent's pieces.

This is a fun cultural side-note from the chapter — not something that's usually tested, but a neat piece of maths-and-culture trivia!

🎮 How to play

Each player starts with 9 pawns. Players take turns placing one pawn at a time on marked intersections (only one pawn per spot).
Once all pawns are placed, players take turns sliding a pawn to an ADJACENT empty spot, trying to line up 3 pawns in a row (horizontally or vertically).
Whenever a player completes a line of 3, they get to remove one of the opponent's pawns (as long as it isn't part of one of the opponent's own lines).

🏆 How to win

A player wins when the opponent has fewer than 3 pawns left, or can no longer make any legal move!

You did it! 🎉

📚 Official chapter summary

The book wraps up Chapter 5 with these 4 official take-aways about divisibility, plus a nod to the digital-root shortcuts:

If a is divisible by b, then ALL multiples of a are also divisible by b.
If a is divisible by b, then a is divisible by ALL the factors of b.
If a divides m and a divides n, then a divides m+n AND m−n.
If a is divisible by b AND by c, then a is divisible by LCM(b, c) — not necessarily b×c.

Chapter 5 done — you can now prove divisibility rules with algebra AND crack secret-letter number puzzles! ⭐

Parity proofs Divisibility by 9 Divisibility by 11 Digital roots Cryptarithms LCM & divisibility Breaking Even Always/sometimes/never

← BackCh 4 · Quadrilaterals Next →Ch 6 · We Distribute, Yet Things Multiply

🏁 Chapter 5 of 7 · Term 1 Maths · Prishita, Class 8