Key Concepts & Formulae at a Glance
- Parity rules: odd ± odd = even; even ± even = even; odd ± even = odd. So \(a \pm b\) always has the same parity, no matter which sign is used.
- Combining multiples: if \(a\) divides \(M\) and \(a\) divides \(N\), then \(a\) divides \(M+N\) and \(M-N\).
- If \(A\) is divisible by \(k\), then every multiple of \(A\) is also divisible by \(k\); and \(A\) itself is divisible by every factor of \(k\).
- If \(A\) is divisible by both \(k\) and \(m\), then \(A\) is divisible by \(\text{LCM}(k,m)\) — not simply by \(k\times m\) unless \(k,m\) are coprime.
- Divisibility by 10, 5, 2: check only the units digit (0 for 10; 0 or 5 for 5; even for 2).
- Divisibility by 4: last two digits divisible by 4. Divisibility by 8: last three digits divisible by 8.
- Divisibility by 9 and 3: sum of all digits divisible by 9 (or 3).
- Divisibility by 11: the alternating sum of digits (from the units digit) is divisible by 11.
- Divisibility by 6: divisible by 2 and 3. Divisibility by 24: divisible by 3 and 8 (checking 4 and 6 is not enough, since \(\text{LCM}(4,6)=12\ne24\)).
- Digital root: add the digits repeatedly until one digit remains. It equals the remainder on dividing by 9 (or 9 itself, if the number is a multiple of 9).
- Cryptarithm rules: each letter stands for one digit, each digit maps to at most one letter, and the leading digit of a number is never 0.
5.1 Is This a Multiple Of? — Sum of Consecutive Numbers
Anshu notices patterns like \(7=3+4\), \(10=1+2+3+4\), \(15=7+8=4+5+6=1+2+3+4+5\), and wonders which numbers can be written as a sum of consecutive numbers, in how many ways, and whether 0 can be written this way using negative numbers too. (Exploring these is open-ended — try a few small numbers and look for a pattern in which ones work in more than one way.)
MTTake 4 consecutive numbers (e.g. 3, 4, 5, 6). Place + and − signs between them. How many expressions are possible? Evaluate each — what do you notice about the parity of the results?
8 expressions are possible (2 choices of sign, 3 times over — \(2^3=8\)). Evaluating all of them: 18, 6, 8, −4, 10, −2, 0, −12 — every result is even.
Why: switching any one sign in an expression changes its value by twice that number (e.g. changing \(+b\) to \(-b\) changes the total by \(2b\), always even), so every expression built from the same 4 numbers has the same parity. Since \(a+(a{+}1)+(a{+}2)+(a{+}3)=4a+6\) is always even, all 8 sign-combinations of any 4 consecutive numbers are even.
Breaking Even & Pairs to Make Fours
Breaking Even — checking which of these are even, without computing: \(43+37\) (even, odd+odd), \(672-348\) (even, even−even), \(4\times347\times3\) (even, has factor 2), \(708-477\) (odd, even−odd), \(809+214\) (odd, odd+even), \(119\times303\) (odd, odd×odd), \(543-479\) (even, odd−odd), \(513^3\) (odd, odd×odd×odd).
For algebraic expressions, an expression is always even only if every term is guaranteed to carry a factor of 2:
| Expression | Always even? | Why |
|---|---|---|
| \(2a+2b\) | Yes | \(=2(a+b)\) |
| \(3g+5h\) | No | odd coefficients — parity depends on \(g,h\) |
| \(4m+2n\) | Yes | \(=2(2m+n)\) |
| \(2u-4v\) | Yes | \(=2(u-2v)\) |
| \(13k-5k\) | Yes | \(=8k=2(4k)\) |
| \(6m-3n\) | No | \(=3(2m-n)\) — even only if \(n\) is even |
| \(x^2+2\) | No | \(x^2\) has the same parity as \(x\); e.g. \(x=3\Rightarrow11\) (odd) |
| \(b^2+1\) | No | even only when \(b\) is odd |
| \(4k\times3j\) | Yes | \(=12kj\), always a multiple of 2 |
Pairs to Make Fours — even numbers are of two kinds: multiples of 4 (remainder 0), or 2 more than a multiple of 4 (remainder 2). Adding two even numbers gives a multiple of 4 only when both are the same kind:
Always, Sometimes, or Never
| # | Statement | Verdict | Reason |
|---|---|---|---|
| 1 | 8 divides two numbers separately ⟹ 8 divides their sum | Always | \(8a+8b=8(a+b)\) |
| 2 | 8 divides a number ⟹ 8 divides any two numbers that add up to it | Sometimes | e.g. \(16=8+8\) works, \(16=1+15\) doesn't |
| 3 | 7 divides a number ⟹ 7 divides all its multiples | Always | \((7j)\times m=7jm\) |
| 4 | 12 divides a number ⟹ it's divisible by all factors of 12 | Always | a factor of 12 divides every multiple of 12 |
| 5 | 7 divides a number ⟹ it's divisible by any multiple of 7 | Sometimes | \(7k\) is divisible by \(7m\) only if \(m\) divides \(k\) |
| 6 | divisible by both 9 and 4 ⟹ divisible by 36 | Always | \(\text{LCM}(9,4)=36\) (9, 4 coprime) |
| 7 | divisible by both 6 and 4 ⟹ divisible by 24 | Sometimes | \(\text{LCM}(6,4)=12\ne24\); e.g. 12 is divisible by 6 and 4, not 24 |
| 8 | odd number + even number = multiple of 6 | Never | odd + even is always odd; multiples of 6 are even |
In general: if \(A\) is divisible by both \(k\) and \(m\), \(A\) is divisible by \(\text{LCM}(k,m)\) — this is why Statement 6 is always true but Statement 7 is only sometimes true.
What Remains?
Numbers leaving remainder 3 on division by 5 are 3 more than a multiple of 5, i.e. of the form \(5k+3\) (equivalently \(5k-2\) for \(k\ge1\)). Checking each option against this: \(3k+5\), \(3k-5\), and \(\frac{3k}{5}\) do not always give remainder 3 mod 5 (their values drift as \(k\) changes); only \(5k+3\) and \(5k-2\) do.
Note: it's often handy to write an even number as \(2n\) and an odd number as \(2n+1\), for integer \(n\), rather than just "multiple of 2" / "not a multiple of 2".
Figure it Out — Multiples & Remainders
Eight questions from page 122 of the textbook.
1The sum of four consecutive numbers is 34. What are these numbers?
Let the numbers be \(n, n{+}1, n{+}2, n{+}3\). Sum \(=4n+6=34 \Rightarrow n=7\).
2Suppose p is the greatest of five consecutive numbers. Describe the other four in terms of p.
Counting down from the greatest: \(p-1, p-2, p-3, p-4\).
3Determine Always/Sometimes/Never for: (i) sum of two even numbers is a multiple of 3; (ii) not divisible by 18 ⟹ not divisible by 9; (iii) two numbers not divisible by 6 ⟹ their sum not divisible by 6; (iv) multiple of 6 + multiple of 9 is a multiple of 3; (v) multiple of 6 + multiple of 3 is a multiple of 9.
(i) Sometimes — \(2+4=6\) (mult. of 3) but \(2+6=8\) (not).
(ii) Sometimes — 27 is not divisible by 18 but is divisible by 9; 30 is divisible by neither.
(iii) Sometimes — 8 and 10 aren't divisible by 6, but their sum 18 is.
(iv) Always — \(6x+9y=3(2x+3y)\).
(v) Sometimes — \(18+9=27\) (mult. of 9) but \(12+9=21\) (not).
4Find numbers that leave remainder 2 when divided by 3, and remainder 2 when divided by 4. Write an algebraic expression for all such numbers.
The number is 2 more than a common multiple of 3 and 4, i.e. 2 more than a multiple of \(\text{LCM}(3,4)=12\).
5Pebble riddle: remainder 1 when grouped by 3, 1 when paired (by 2), 1 when grouped by 5, but exactly divisible by 7; fewer than 100. How many pebbles?
The number is 1 more than a multiple of \(\text{LCM}(2,3,5)=30\), so of the form \(30k+1\): 31, 61, 91, … . Among these under 100, only 91 is divisible by 7 (\(91=7\times13\)).
6Tathagat's claim: numbers leaving remainder 2 mod 6 — the sum of any three of them is always a multiple of 6. True?
Let the numbers be \(6a{+}2, 6b{+}2, 6c{+}2\). Sum \(=6(a+b+c)+6=6(a+b+c+1)\), always a multiple of 6.
7661 ÷ 7 leaves remainder 3, and 4779 ÷ 7 leaves remainder 5. Without calculating, find the remainder for (i) 4779+661 (ii) 4779−661, on division by 7.
\(4779=7p+5\), \(661=7q+3\).
(i) \(4779+661=7(p+q)+8=7(p+q+1)+1\) → remainder 1.
(ii) \(4779-661=7(p-q)+2\) → remainder 2.
8Find a number leaving remainder 2 mod 3, remainder 3 mod 4, and remainder 4 mod 5. What is the smallest such number, and why?
Each remainder is exactly 1 less than its divisor, so the number is 1 less than a common multiple of 3, 4, 5. The smallest common multiple is \(\text{LCM}(3,4,5)=60\).
5.2 Checking Divisibility Quickly
Writing a number in expanded place-value form, e.g. \(\ldots+1000d+100c+10b+a\): every place value except the units place is a multiple of 10, so divisibility by 10 depends only on the units digit being 0. The same idea extends: divisibility by 5 needs the units digit to be 0 or 5; divisibility by 2 needs the units digit even; divisibility by 4 needs the last two digits (the "tens+units" part, since 100 is a multiple of 4) divisible by 4; divisibility by 8 needs the last three digits divisible by 8 (since 1000 is a multiple of 8).
A Shortcut for Divisibility by 9: every place value is 1 more than a multiple of 9 (\(10=9{+}1\), \(100=99{+}1\), \(1000=999{+}1\), …), so a number equals (sum of its digits) + a multiple of 9. Hence a number is divisible by 9 exactly when the sum of its digits is. For example, \(7309\): digit sum \(=7+3+0+9=19\), and \(1+9=10\), \(1+0=1\) — so 7309 leaves remainder 1 on division by 9.
Of the four statements "(i) divisible by 9 ⟹ digit sum divisible by 9", "(ii) digit sum divisible by 9 ⟹ divisible by 9", "(iii) not divisible by 9 ⟹ digit sum not divisible by 9", "(iv) digit sum not divisible by 9 ⟹ not divisible by 9" — all four are true, since they're just different phrasings of the same if-and-only-if rule.
Figure it Out — Divisibility by 9
Four questions from page 126 of the textbook.
1Without dividing, check divisibility by 9 for: (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095.
Digit sums: (i) 6 (ii) 9 (iii) 32 (iv) 28 (v) 30. Only 9 is a multiple of 9.
2Find the smallest multiple of 9 with no odd digits.
Even digits are 0, 2, 4, 6, 8. Their smallest combination summing to a multiple of 9 is \(2+8+8=18\); arranging smallest-first gives 288.
3Find the multiple of 9 closest to 6000.
\(6000\div9=666.67\); \(9\times667=6003\), which is closer to 6000 than \(9\times666=5994\).
4How many multiples of 9 are there between 4300 and 4400?
First multiple ≥4300: \(9\times478=4302\). Last multiple ≤4400: \(9\times488=4392\). Count \(=488-478+1=11\).
Shortcuts for Divisibility by 3 and 11
Divisibility by 3 works the same way as 9, since every power of 10 is also 1 more than a multiple of 3 — a number is divisible by 3 exactly when its digit sum is.
Divisibility by 11 is different: place values alternate between 1 more and 1 less than a multiple of 11 (\(1=11(0){+}1\), \(10=11(1){-}1\), \(100=11(9){+}1\), \(1000=11(91){-}1\), …).
| Place value | Relation to multiple of 11 |
|---|---|
| Units (1) | 1 more |
| Tens (10) | 1 less |
| Hundreds (100) | 1 more |
| Thousands (1000) | 1 less |
So the alternating sum of digits (units \(-\) tens \(+\) hundreds \(-\) thousands \(+\ldots\)) gives the remainder on division by 11. For 328105: \(-3+2-8+1-0+5=-3\), so 328105 is 3 short of (or 8 more than) a multiple of 11.
Checking a few numbers this way: (i) 158 → remainder 4 (ii) 841 → remainder 5 (iii) 481 → remainder 8 (iv) 5529 → remainder 7 (v) 90904 → divisible by 11 (vi) 857076 → divisible by 11.
MTFill in the table: for each number, mark Yes/No for divisibility by 2, 3, 4, 5, 6, 8, 9, 10, 11.
| 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 |
More Divisibility Shortcuts & Digital Roots
Divisibility by 6 = divisible by 2 and 3 (its coprime prime-power factors). But divisibility by 24 is not captured by checking 4 and 6, since \(\text{LCM}(4,6)=12\ne24\) — e.g. 12 is divisible by both 4 and 6 but not 24. Instead, check divisibility by 3 and 8 (since \(24=3\times8\) and \(\gcd(3,8)=1\)).
Digital Roots: add the digits of a number repeatedly until a single digit remains — e.g. digital root of 489710 is \(4{+}8{+}9{+}7{+}1{+}0=29\to2{+}9=11\to1{+}1=2\). A number's digital root is always 9 for multiples of 9, and otherwise equals the remainder on dividing by 9.
MTRiddle: my digits are each the tiniest odd digit; my digit count, digit sum, and digital root all point to the largest odd single digit. What's my number?
The tiniest odd digit is 1, and the largest odd single digit is 9. So the number has nine digits, all 1's, with digit sum 9 and digital root 9.
Figure it Out — Digital Roots
Four questions from page 131 of the textbook.
1The digital root of an 8-digit number is 5. What is the digital root of 10 more than that number?
Adding 10 typically adds 1 to the tens digit without affecting the digit sum's structure much — the digital root increases by 1 (mod 9). E.g. for 40000001 (digital root 5), adding 10 gives 40000011 (digit sum 6).
2Generate a sequence by repeatedly adding 11 to a number. What happens to the digital roots?
Starting at 10: 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, … with digital roots 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, 3, … — each step the digital root increases by 2 (mod 9, wrapping after 9), since 11 itself has digital root 2.
3What is the digital root of 9a + 36b + 13?
\(9a+36b+13=9(a+4b+1)+4\). The first part is a multiple of 9 (digital root 9, contributing nothing extra), leaving digital root 4.
4Are there patterns between (i) a number's parity and its digital root, and (ii) its digital root and the remainder on dividing by 3 or 9?
(i) No consistent pattern — a number's digital root doesn't determine whether it's odd or even.
(ii) Digital root 9 (or a multiple of 9) ⟺ remainder 0 on dividing by 9; otherwise, the remainder on dividing by 9 equals the digital root itself. For division by 3: digital roots 3, 6, 9 give remainder 0; digital roots 1, 4, 7 give remainder 1; digital roots 2, 5, 8 give remainder 2.
5.3 Digits in Disguise
In a cryptarithm, each letter stands for one digit, each digit maps to at most one letter, and a number's leading digit is never 0.
1Solve: (i) A1 + 1B = B0 (ii) AB + 37 = 6A (iii) ON + ON + ON = PO (iv) QR + QR + QR = PRR
(i) \(A1+1B=B0\): trying \(A=7,B=9\): \(71+19=90\) ✓.
(ii) \(AB+37=6A\): trying \(A=2,B=5\): \(25+37=62\) ✓.
(iii) \(ON+ON+ON=PO\): trying \(N=1,O=3,P=9\): \(31+31+31=93\) ✓.
(iv) \(QR+QR+QR=PRR\): trying \(Q=8,R=5,P=2\): \(85+85+85=255\) ✓.
Cryptarithms with multiplication — for \(PQ\times8=RS\) (a 2-digit number × 8 gives another 2-digit number): \(10\times8=80\) shares its units digit with 10, ruled out; 11 is ruled out since \(P\ne Q\) required; \(12\times8=96\) works; \(13\times8=104\) is a 3-digit number, and every 2-digit number above 12 overshoots — so PQ = 12 is the only solution.
MTGH × H = 9K. This is a 2-digit number times a 1-digit number giving a 2-digit number in the 90s. Pick the answer from: 11×9=99, 12×8=96, 46×2=92, 24×4=96, 47×2=94, 31×3=93, 16×6=96.
The multiplier must equal the units digit of GH (both are "H"). Checking each option's units digit against its multiplier: only \(24\times4=96\) (units digit 4 = multiplier 4) and \(16\times6=96\) (units digit 6 = multiplier 6) satisfy this.
For \(BYE\times6=RAY\) (a 3-digit product): since \(2\times6=1200\) already exceeds 3 digits, \(B\) must be 1. For \(Y\): if \(Y=7\), \(170\times6=1020\) (4 digits) — too big — so \(Y \lt 7\); and since \(Y\times6\) must end in \(Y\) itself is not required (only the units digit of the whole product cycles), \(Y\) turns out to be even.
2Solve: (i) UT×3=PUT (ii) AB×5=BC (iii) L2N×2=2NP (iv) XY×4=ZX (v) PP×QQ=PRP (vi) JK×6=KKK
(i) \(U{=}5,T{=}0,P{=}1\): \(50\times3=150\) ✓. (ii) \(A{=}1,B{=}9,C{=}5\): \(19\times5=95\) ✓. (iii) \(L{=}1,N{=}5,P{=}0\): \(125\times2=250\) ✓. (iv) \(X{=}2,Y{=}3,Z{=}9\): \(23\times4=92\) ✓. (v) \(P{=}2,Q{=}1,R{=}4\): \(22\times11=242\) ✓. (vi) \(J{=}7,K{=}4\): \(74\times6=444\) ✓.
Figure it Out — Chapter Review
Sixteen questions from pages 132–133 of the textbook.
1If 31z5 is a multiple of 9, where z is a digit, what is z? Why are there two answers?
\(3+1+z+5=9+z\) must be a multiple of 9. Since \(z\) is a single digit (0–9), \(9+z\) ranges from 9 to 18 — both 9 (\(z=0\)) and 18 (\(z=9\)) are multiples of 9.
2Snehal claims: a number leaving remainder 8 mod 12, plus another that's 4 short of a multiple of 12 — their sum is always a multiple of 8. True?
Let \(a=12n+8\), \(b=12m-4\). Sum \(=12(n+m)+4=12k+4\) where \(k=n+m\). For \(k=0\): sum \(=4\) (not a multiple of 8); for \(k=1\): sum \(=16\) (is); for \(k=2\): sum \(=28\) (isn't).
3When is the sum of two multiples of 3 also a multiple of 6?
\(3m+3n=3(m+n)\), which is a multiple of 6 exactly when \(m+n\) is even — i.e. \(m\) and \(n\) have the same parity.
4Sreelatha: "I have a number divisible by 9 — reversing its digits still gives a multiple of 9." (i) Always true? (ii) Any other digit shuffles that work?
(i) True for any multiple of 9 — reversing digits doesn't change their sum.
(ii) Yes — any rearrangement (shuffle) of the digits keeps the same digit sum, so the result is still a multiple of 9.
5If 48a23b is a multiple of 18, list all possible pairs (a, b).
Multiple of 18 = multiple of 2 and 9. For divisibility by 2: \(b\in\{0,2,4,6,8\}\). For divisibility by 9: \(4+8+a+2+3+b=17+a+b\) must be a multiple of 9.
Checking each even \(b\): \(b{=}0\Rightarrow a{=}1\); \(b{=}2\Rightarrow a{=}8\); \(b{=}4\Rightarrow a{=}6\); \(b{=}6\Rightarrow a{=}4\); \(b{=}8\Rightarrow a{=}2\).
6If 3p7q8 is divisible by 44, list all possible pairs (p, q).
Divisible by 44 = divisible by 4 and 11. For divisibility by 4, "\(q8\)" must be a multiple of 4 — true for every even \(q\). For divisibility by 11: (sum of digits in odd positions) − (sum in even positions) \(=(8+7+3)-(p+q)=18-(p+q)\) must be 0 or a multiple of 11. Since \(p+q=18\) isn't possible for digits, we need \(18-(p+q)=11\Rightarrow p+q=7\).
7Try This — Find three consecutive numbers where the 1st is a multiple of 2, 2nd of 3, 3rd of 4. Are there more? How often?
2, 3, 4 is one such set. Since the pattern depends on remainders mod \(\text{LCM}(2,3,4)=12\), it repeats every 12 numbers: the next set is 14, 15, 16.
8Math Talk — Write five multiples of 36 between 45,000 and 47,000.
A multiple of 36 must be a multiple of both 4 and 9. Starting from the first one past 45,000: \(36\times1251=45036\).
9The middle number of 5 consecutive even numbers is 5p. Express the other four in terms of p.
Consecutive even numbers differ by 2, so counting out from the middle: \(5p-4, 5p-2, 5p+2, 5p+4\).
10Write a 6-digit number divisible by 15 such that, when reversed, it's divisible by 6.
Try 200025: digit sum \(=9\) (divisible by 3) and it ends in 5 (divisible by 5), so it's divisible by 15. Reversed: 520002 — ends in 2 (even) and digit sum \(=9\) (divisible by 3), so it's divisible by 6.
11Deepak claims: some multiples of 11 stay multiples of 11 when doubled, but others don't. True?
If \(n=11k\), then \(2n=22k=11\times(2k)\) — always still a multiple of 11, for every multiple of 11.
12Always/Sometimes/Never: (i) product of a multiple of 6 and a multiple of 3 is a multiple of 9; (ii) sum of 3 consecutive even numbers is divisible by 6; (iii) if abcdef is a multiple of 6, so is badcef; (iv) 8(7b−3) − 4(11b+1) is a multiple of 12.
(i) Always — \(6a\times3b=18ab\), and 18 is a multiple of 9.
(ii) Always — \(n+(n{+}2)+(n{+}4)=3(n+2)\); since \(n\) is even, \(n+2\) is even, so this is \(3\times\)even \(=\) a multiple of 6.
(iii) Always — swapping \(a\leftrightarrow b\) and \(c\leftrightarrow d\) doesn't change the last digit (still determines divisibility by 2) or the digit sum (still determines divisibility by 3).
(iv) Never — \(8(7b-3)-4(11b+1)=4(3b-7)\); since \(3b-7\) is never a multiple of 3 (as \(3b\equiv0\), so \(3b-7\equiv-7\equiv2\pmod3\)), the expression is only ever guaranteed a factor of 4, never the extra factor of 3 needed for 12.
13Choose any 3 numbers. When is their sum divisible by 3?
Each number leaves remainder 0, 1, or 2 mod 3. The sum of the three remainders is a multiple of 3 exactly when all three remainders are the same, or all three are different (i.e. 0, 1, 2 in some order) — since \(0{+}1{+}2=3\).
14Is the product of 2 consecutive integers always a multiple of 2? Of 3 consecutive integers, always a multiple of 6? What about 4 and 5 consecutive integers?
2 consecutive: always a multiple of 2 (one of the two is even).
3 consecutive: always a multiple of \(6=2\times3\) (contains at least one multiple of 2 and one of 3).
4 consecutive: always a multiple of \(24=2\times3\times4\).
5 consecutive: always a multiple of \(120=2\times3\times4\times5\).
15Solve: (i) EF×E=GGG (ii) WOW×5=MEOW
(i) \(E{=}3,F{=}7,G{=}1\): \(37\times3=111\) ✓.
(ii) \(W{=}5,O{=}7,M{=}2,E{=}8\): \(575\times5=2875\) ✓.
16Which Venn diagram correctly shows the relationship between the multiples of 4, 8, and 32?
Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4 — so the correct picture is three nested circles, smallest (32) inside middle (8) inside largest (4).
Puzzle Time — Navakankari
A Traditional Strategy Game
Navakankari (also Sālu Mane Āṭa, Chār-Pār, or Navkakri) is the Indian version of "Nine Men's Morris" — a two-player strategy game where the goal is to line up three pawns in a row to capture the opponent's pieces.
Gameplay: each player starts with 9 pawns and takes turns placing them on the board's marked intersections (one pawn per intersection). Once all pawns are placed, players take turns sliding a pawn to an adjacent empty intersection, trying to form a horizontal or vertical line of three. Completing such a line lets a player remove one opposing pawn (as long as it isn't part of the opponent's own completed line). A player wins once their opponent has fewer than 3 pawns left, or can no longer move.
Frequently Asked Questions
Continue with Chapter 6
Move on to We Distribute, Yet Things Multiply next.