Class 8 › Mathematics › Chapter 5
Number Play
Parity, multiples and remainders explained with algebra and pictures; divisibility shortcuts for 3, 9 and 11 and why they work; digital roots; and cryptarithms.
1. Sums, Signs and Parity
- Consecutive sums: $7 = 3+4$, $10 = 1+2+3+4$, $12 = 3+4+5$, $15 = 7+8 = 4+5+6 = 1+2+3+4+5$. Some numbers can be written in more than one way. Every odd number is a sum of two consecutive numbers: $2n+1 = n + (n+1)$.
- Sign puzzle: take 4 consecutive numbers (e.g. $3,4,5,6$) and put $+$ or $-$ between them. With the first sign fixed there are $2 \times 2 \times 2 = 8$ expressions: $3+4+5+6$, $3+4+5-6$, $3+4-5+6$, $3+4-5-6$, $3-4+5+6$, $3-4+5-6$, $3-4-5+6$, $3-4-5-6$. Their values ($18, 6, 8, -4, 10, -2, 0, -12$) are all even.
- This holds for any 4 numbers $a, b, c, d$ (not just consecutive): all 8 expressions $a \pm b \pm c \pm d$ have the same parity.
Why all the expressions have the same parity
Explanation 1 (switch one sign): $(a+b-c-d) - (a-b-c-d) = 2b$, which is even. So switching a sign changes the value by an even number, and two numbers that differ by an even number have the same parity. Switching a $-$ to $+$ changes the value by $2c$ or $2d$ (also even). Any expression can be reached from any other by switching signs one at a time, so all 8 share the parity.
Explanation 2 (parity rules): odd $\pm$ odd $=$ even, even $\pm$ even $=$ even, odd $\pm$ even $=$ odd. So $a+b$ and $a-b$ always have the same parity; extending this, $a \pm b \pm c \pm d$ all have the same parity as $a+b+c+d$.
Explanation 3: use the positive/negative token model from Integers.
Generalisation: this is not limited to 4 numbers; it works for any number of terms. Asking "What if…?" and "Will it always happen?" is part of mathematics.
2. Breaking Even
Parity rules: even $\pm$ even $=$ even; odd $\pm$ odd $=$ even; even $\pm$ odd $=$ odd; even $\times$ anything $=$ even; odd $\times$ odd $=$ odd. An even number has factor $2$ (negative even numbers like $-2, -4$ count too).
- Arithmetic check without computing: $43+37$ (even), $672-348$ (even), $4 \times 347 \times 3$ (even), $543-479$ (even); $708-477$, $809+214$ and $119 \times 303$ are odd.
- Algebraic expressions that are always even (for any integers): $2a+2b$ (factor 2), $4m+2n$, $2u-4v$, $13k-5k = 8k$, $4k \times 3j = 12kj$.
- Not always even: $3g+5h$ (depends on $g, h$), $6m - 3n$ (parity of $3n$), $x^2 + 2$ ($x=6$ gives $38$, even; $x=3$ gives $11$, odd), $b^2 + 1$.
- Two ways to justify: (i) each term is even, so the sum is even; (ii) rewrite with a factor 2: $4m + 2q = 2(2m+q)$. Example: $m=4$, $q=-9$ gives $-2$.
3. Pairs to Make Fours
Even numbers are of two types by remainder on division by $4$: multiples of $4$ ($4p$, remainder $0$) and the others ($4p+2$, remainder $2$).
| Adding | Algebra | Result | Example |
|---|---|---|---|
| Two multiples of 4 | $4p + 4q = 4(p+q)$ | always a multiple of 4 | $12 + 16 = 28$ |
| Two non-multiples of 4 (both remainder 2) | $(4p+2)+(4q+2) = 4(p+q+1)$ | always a multiple of 4 (2 + 2 = 4) | $6 + 10 = 16$ |
| One of each | $4p + (4q+2) = 4(p+q) + 2$ | never a multiple of 4; remainder 2 | $4 + 6 = 10$ |
- Visual proof: a multiple of $4$ is rows of $4$ dots. Joining $p$ rows and $q$ rows gives $p+q$ rows. A number $4p+2$ is $p$ full rows plus a row of $2$; two such leftover rows of $2$ combine into one more full row.
- This is just like even + odd: the "remainder 2" part behaves like the odd part.
4. Always, Sometimes or Never True
Divisibility properties
1. If $a$ divides $M$ and $a$ divides $N$, then $a$ divides $M+N$ and $M-N$.
2. If $A$ is divisible by $k$, then all multiples of $A$ are divisible by $k$.
3. If $A$ is divisible by $k$, then $A$ is divisible by all the factors of $k$.
4. If $A$ is divisible by $k$ and by $m$, then $A$ is divisible by the LCM of $k$ and $m$ (the prime factorisation of $A$ contains that of the LCM).
| Statement | Verdict | Reason |
|---|---|---|
| 1. If 8 divides two numbers, it divides their sum. | Always | $8a + 8b = 8(a+b)$ (also true for subtraction) |
| 2. If a number is divisible by 8, then 8 divides any two numbers that add up to it. | Sometimes | $72 = 48 + 24$ works; $72 = 50 + 22$ does not |
| 3. If a number is divisible by 7, all its multiples are divisible by 7. | Always | $(7j) \times m$ has factor 7 |
| 4. If a number is divisible by 12, it is divisible by all factors of 12. | Always | $12m = 2 \times 6 \times m = 3 \times 4 \times m$, and so on |
| 5. If a number is divisible by 7, it is divisible by any multiple of 7. | Sometimes | $7k$ is divisible by $7m$ only if $m$ divides $k$: $42$ is divisible by $14$ but not by $28$ |
| 6. If a number is divisible by both 9 and 4, it is divisible by 36. | Always | LCM$(9,4) = 36$ |
| 7. If a number is divisible by both 6 and 4, it is divisible by 24. | Sometimes | LCM$(6,4) = 12$, not 24: $12$ qualifies for 6 and 4 but is not divisible by 24; $24$ is |
| 8. Odd number + even number is a multiple of 6. | Never | Multiples of 6 are even, odd + even is odd. Algebra: $2n + (2m+1) = 6j$ gives $2(n+m) = 6j - 1$, even $=$ odd, impossible |
5. What Remains? (Remainders)
Numbers with a given remainder
$$\text{Numbers leaving remainder } r \text{ on division by } n:\quad nk + r \quad (0 \leq r < n)$$
Numbers leaving remainder $3$ on division by $5$: $5k + 3 = 3, 8, 13, 18, 23, \dots$ Also written $5k - 2$ ($k \geq 1$), since "3 more than a multiple of 5" is "2 less than the next multiple".
Blue: multiples of 5 ($5k$). Red: numbers that leave remainder 3 ($5k + 3$).
- Combining remainders: $661 = 7q + 3$ and $4779 = 7p + 5$ (remainders on division by $7$). Then $4779 + 661 = 7(p+q+1) + 1$, remainder $1$; and $4779 - 661 = 7(p-q) + 2$, remainder $2$. Add or subtract the remainders (and reduce if needed).
- Two remainder conditions: numbers leaving remainder $2$ on division by $3$ and by $4$: $x - 2$ is a multiple of both, so a multiple of LCM$=12$: $x = 12n + 2 = 14, 26, 38, \dots$
- Remainder is 1 less than each divisor: add $1$ to get a common multiple. Remainders $2, 3, 4$ on division by $3, 4, 5$: the number is (a multiple of $60$) $- 1$, smallest $= 59$.
- Pebble riddle: remainder $1$ when grouped by $3$, by $2$ (one odd pebble left) and by $5$; divisible by $7$; less than $100$. The number is $1$ more than a multiple of LCM$(3,2,5) = 30$: $30k + 1 = 31, 61, 91$; only $91 = 7 \times 13$ works, so there are 91 pebbles.
- Tathagat's claim: numbers leaving remainder $2$ on division by $6$: $6a+2$, $6b+2$, $6c+2$ have sum $6(a+b+c+1)$, always a multiple of $6$. True.
- Counter-example (Snehal): $(12n+8) + (12m-4) = 12(n+m) + 4$, which is not always a multiple of $8$ (e.g. $k = 2$ gives $28$). False.
6. Divisibility Shortcuts for 2, 4, 5, 8 and 10 (Why They Work)
Place-value form
$$\dots dcba = \dots + 1000d + 100c + 10b + a$$
Here $a, b, c, d, \dots$ are the units, tens, hundreds, thousands digits. Example: in $4075$, $d=4$, $c=0$, $b=7$, $a=5$.
| Divisor | Test | Reason |
|---|---|---|
| 10 | Units digit is $0$ | $10b, 100c, \dots$ are all multiples of 10, so only $a$ matters |
| 5, 2 | Units digit is $0$ or $5$ / even | Every other place value is a multiple of 10, hence of 5 and 2 |
| 4 | Last two digits form a multiple of 4 | $100$ and higher place values are multiples of 4 |
| 8 | Last three digits form a multiple of 8 | $1000$ and higher place values are multiples of 8 |
7. Divisibility by 9 and 3
Why the digit-sum test works
Every power of 10 is one more than a multiple of 9: $1 = 0 + 1$, $10 = 9 + 1$, $100 = 99 + 1$, $1000 = 999 + 1$, … So each place value leaves remainder $1$ on division by $9$.
$7309 = 7(999+1) + 3(99+1) + 0(9+1) + 9 = (7 \times 999 + 3 \times 99 + 0 \times 9) + (7 + 3 + 0 + 9)$. The first bracket is a multiple of 9, so $7309$ leaves the same remainder as its digit sum $19$, i.e. as $1 + 9 = 10$, i.e. as $1 + 0 = 1$.
Check: $427$ has digit sum $4 + 2 + 7 = 13$, then $1 + 3 = 4$, so the remainder on division by $9$ is $4$.
Test for 9: a number is divisible by $9$ if and only if the sum of its digits is divisible by $9$. Repeatedly adding digits until one digit remains gives the remainder on division by $9$ (a final $9$ means remainder $0$).
Test for 3: same method, since each power of 10 also leaves remainder $1$ on division by $3$: divisible by $3$ if and only if the digit sum is divisible by $3$. ($15$, $33$, $87$ are multiples of 3 but not 9.)
Test for 3: same method, since each power of 10 also leaves remainder $1$ on division by $3$: divisible by $3$ if and only if the digit sum is divisible by $3$. ($15$, $33$, $87$ are multiples of 3 but not 9.)
- Both directions are true: if the number is divisible by 9, its digit sum is; if the digit sum is divisible by 9, the number is; and the contrapositives (not divisible $\leftrightarrow$ digit sum not divisible) hold too.
- Units digit alone is not enough: $99$ is a multiple of 9, $109$ is not, though both end in 9. Any number made only of digits $0$ and $9$ is divisible by $9$.
- Examples: $405$ is divisible by 9 (sum $9$); $123$ (sum $6$), $8888$ (sum $32$), $93547$ (sum $28$), $358095$ (sum $30$) are not.
- Smallest multiple of 9 with no odd digit: digit sum must be even, so at least $18$: digits $2, 8, 8$ give $288$. Closest multiple of 9 to $6000$: $6003$. Multiples of 9 between $4300$ and $4400$: $4302, \dots, 4392$, which is $11$ numbers.
- Rearranging digits keeps the digit sum, so any shuffle (including reversal) of a multiple of 9 is again a multiple of 9. Example: $31z5$ is a multiple of 9 when $9 + z$ is a multiple of $9$, so $z = 0$ or $9$ (two answers, since $0$ and $9$ are both digits).
8. Divisibility by 11
Place values alternate between "1 more" and "1 less" than a multiple of $11$: $1 = 0 + 1$, $10 = 11 - 1$, $100 = 99 + 1$, $1000 = 1001 - 1$, $10000 = 9999 + 1$, …
For 328105: each place value is 1 more (+) or 1 less (−) than a multiple of 11, alternating from the units place.
Test for 11 (two equivalent ways)
Method 1: add the digits in the "1 more" places (units, hundreds, ten-thousands, …) to get the excess; add the digits in the "1 less" places (tens, thousands, …) to get the shortage. Compute (excess) $-$ (shortage). For $328105$: excess $5+1+2 = 8$, shortage $0+8+3 = 11$, difference $= -3$.
Method 2: put alternating signs on the digits starting with $+$ at the units digit: $-3+2-8+1-0+5 = -3$. The result is the remainder: $328105$ is $3$ less than a multiple of 11, which is also $8$ more than a multiple of 11 ($-3 + 11 = 8$).
Result: if the difference is $0$, $11$, or any multiple of $11$, the number is divisible by 11 (remainder zero). If it is negative, add 11 to get the remainder.
- Examples: $462$: $2 - 6 + 4 = 0$, divisible. $158$ remainder $4$; $841$ remainder $5$; $481$ remainder $8$ ($1 - 8 + 4 = -3$); $5529$ remainder $7$; $90904$ divisible ($4 - 0 + 9 - 0 + 9 = 22$); $857076$ divisible ($6-7+0-7+5-8 = -11$).
- Doubling a multiple of 11: $2 \times 11k = 11(2k)$, so doubling always gives a multiple of 11. (Deepak's claim that some do not is false.)
9. Combined Divisibility Tests
Rule: to test divisibility by a composite number, split it into factors that are coprime (no common factor) and test each. Using non-coprime factors can fail.
| Divisible by | Test both | Notes |
|---|---|---|
| 6 | 2 and 3 | $38$ (no), $225$ (no), $186$ (yes), $64$ (no) |
| 12 | 3 and 4 | |
| 15 | 3 and 5 | |
| 18 | 2 and 9 | |
| 24 | 3 and 8 | Not 4 and 6: $12$ is divisible by 4 and 6 but not by 24 (prime factors $2^3 \times 3$ vs $2^2 \times 3$) |
| 36 | 4 and 9 | |
| 44 | 4 and 11 |
| Number | 2 | 3 | 4 | 5 | 6 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|---|
| 1586 | Y | N | N | N | N | N | N | N | N |
| 6686 | Y | N | N | N | N | N | N | N | N |
| 639210 | Y | Y | N | Y | Y | N | N | Y | Y |
| 429714 | Y | Y | N | N | Y | N | Y | N | N |
| 2856 | Y | Y | Y | N | Y | Y | N | N | N |
| 3060 | Y | Y | Y | Y | Y | N | Y | Y | N |
| 406839 | N | Y | N | N | N | N | N | N | N |
- Worked puzzles: $48a23b$ is a multiple of $18$ (divisible by 2 and 9): $b$ even and $17 + a + b$ a multiple of 9 gives $(a, b) = (1,0), (8,2), (6,4), (4,6), (2,8)$. $3p7q8$ divisible by $44$ (4 and 11): $q8$ divisible by 4 and $18 - (p+q) = 11$ give $(p, q) = (7,0), (5,2), (3,4), (1,6)$.
- Three consecutive numbers that are multiples of $2, 3, 4$ respectively: $2, 3, 4$, and the pattern repeats every $12$ ($14, 15, 16$, …).
- Product of consecutive integers: 2 consecutive $\to$ multiple of $2$; 3 $\to$ multiple of $6$; 4 $\to$ multiple of $24$; 5 $\to$ multiple of $120$.
- Sums of multiples of 3: $3m + 3n = 3(m+n)$ is a multiple of $6$ exactly when $m + n$ is even. Sum of any three numbers is divisible by 3 when their remainders on division by 3 are all equal, or are $0, 1, 2$ in some order.
- Nesting of multiples: multiples of $32 \subset$ multiples of $8 \subset$ multiples of $4$ (a 3-level nested Venn diagram).
10. Digital Roots
Digital root: add the digits of a number repeatedly until a single digit remains. $489710 \to 29 \to 11 \to 2$, so its digital root is $2$.
- Link with 9: the digital root equals the remainder on division by 9, except that remainder $0$ is written as $9$. So multiples of $9$ have digital root $9$.
- Link with 3: digital root $1, 4, 7 \Rightarrow$ remainder $1$ on division by $3$; $2, 5, 8 \Rightarrow$ remainder $2$; $3, 6, 9 \Rightarrow$ remainder $0$.
- Adding $10$ raises the digital root by $1$ (e.g. root $5 \to 6$). Adding $11$ repeatedly gives roots rising by $2$: $1, 3, 5, 7, 9, 2, 4, 6, 8, 1, \dots$
- Consecutive numbers cycle through $1, 2, \dots, 9$. Consecutive multiples: of $3$: $3, 6, 9$ repeating; of $6$: $6, 3, 9$ repeating; of $4$: $4, 8, 3, 7, 2, 6, 1, 5, 9$ repeating. Numbers one more than a multiple of $6$ have roots $7, 4, 1$ repeating.
- 600 to 700: digital root $5$: $608, 617, \dots, 698$; root $7$: $601, 610, \dots, 691$; root $3$: $606, 615, \dots, 696$ (each differs by 9).
- Algebra: $9a + 36b + 13 = 9(a + 4b + 1) + 4$ has digital root $4$.
- No relation between the parity of a number and its digital root.
- History: Aryabhata II (c. 950 CE) described repeatedly adding digits; it was used to check arithmetic calculations.
11. Digits in Disguise (Cryptarithms)
Rules: each letter stands for one digit; different letters stand for different digits (a digit is shown by at most one letter); the first digit of a number is never $0$.
Strategies
Use the size of numbers (a 2-digit number times 8 that is still 2-digit must be at most $12$), the units digit pattern, carries, parity, and divisibility facts. Example: $BYE \times 6 = RAY$. Since the product has 3 digits, $B = 1$ (if $B=2$ the product exceeds $1200$); $Y$ must be even, and $Y \le 6$ because $Y=7$ gives $1020$. Trying $Y = 0$, $E = 5$: $105 \times 6 = 630$, so $B=1, Y=0, E=5, R=6, A=3$.
| Cryptarithm | Solution |
|---|---|
| $A1 + 1B = B0$ | $71 + 19 = 90$ ($A=7$, $B=9$) |
| $AB + 37 = 6A$ | $25 + 37 = 62$ ($A=2$, $B=5$) |
| $ON + ON + ON = PO$ | $31 \times 3 = 93$ ($O=3, N=1, P=9$). Also fits: $17 \times 3 = 51$ and $24 \times 3 = 72$ |
| $QR + QR + QR = PRR$ | $85 \times 3 = 255$ ($Q=8, R=5, P=2$) |
| $PQ \times 8 = RS$ | $12 \times 8 = 96$ |
| $GH \times H = 9K$ | $24 \times 4 = 96$ ($G=2, H=4, K=6$) |
| $UT \times 3 = PUT$ | $50 \times 3 = 150$ |
| $AB \times 5 = BC$ | $19 \times 5 = 95$ |
| $L2N \times 2 = 2NP$ | $125 \times 2 = 250$ |
| $XY \times 4 = ZX$ | $23 \times 4 = 92$ |
| $PP \times QQ = PRP$ | $22 \times 11 = 242$ |
| $JK \times 6 = KKK$ | $74 \times 6 = 444$ |
| $EF \times E = GGG$ | $37 \times 3 = 111$ |
| $WOW \times 5 = MEOW$ | $575 \times 5 = 2875$ |
12. Summary and Quick Revision
| Topic | Key result |
|---|---|
| Parity of expressions | $a \pm b \pm c \pm \dots$ has the same parity as $a + b + c + \dots$; switching a sign changes the value by an even number |
| Sum of two multiples of 4 / two numbers with remainder 2 / mixed | Multiple of 4 / multiple of 4 / remainder 2 |
| Divisibility laws | $a \mid M, a \mid N \Rightarrow a \mid (M \pm N)$; multiples keep divisibility; factors of a divisor divide it; divisible by $k$ and $m \Rightarrow$ divisible by LCM$(k, m)$ |
| Remainder form | $nk + r$, $0 \leq r < n$ |
| Test for 3 / 9 | Digit sum divisible by 3 / 9 (digital root gives remainder mod 9) |
| Test for 11 | Alternating digit sum from the units digit is $0$ or a multiple of $11$ |
| Test for 4 / 8 | Last two / three digits divisible by 4 / 8 |
| Composite divisors | Test coprime factors (24: use 3 and 8, not 4 and 6) |
Common mistakes:
• Testing divisibility by 24 with 4 and 6 (not coprime): $12$ is a counter-example.
• Using the units digit to test divisibility by 9 or 11 (use digit sums / alternating sums).
• Forgetting that the 11-test starts with a $+$ at the units digit and alternates, and not adding 11 to a negative result to read the remainder.
• Writing $5k - 3$ for "remainder 3 on division by 5": that gives remainder $2$. Use $5k + 3$ or $5k - 2$.
• Assuming "divisible by 6 and 4" means "divisible by 24": use the LCM, which is $12$.
• Giving a cryptarithm a leading zero or reusing a digit for two letters.
• Concluding a property is always true from a few examples; use algebra for a general proof and a single counter-example to disprove.
• Testing divisibility by 24 with 4 and 6 (not coprime): $12$ is a counter-example.
• Using the units digit to test divisibility by 9 or 11 (use digit sums / alternating sums).
• Forgetting that the 11-test starts with a $+$ at the units digit and alternates, and not adding 11 to a negative result to read the remainder.
• Writing $5k - 3$ for "remainder 3 on division by 5": that gives remainder $2$. Use $5k + 3$ or $5k - 2$.
• Assuming "divisible by 6 and 4" means "divisible by 24": use the LCM, which is $12$.
• Giving a cryptarithm a leading zero or reusing a digit for two letters.
• Concluding a property is always true from a few examples; use algebra for a general proof and a single counter-example to disprove.