Number Play

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CLASS VIII Mathematics ~4–5 marks Ch 5 of 14
Number Play

Class 8 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • This chapter is about playing with numbers and reasoning — not just calculating. The big idea: use algebra, parity (odd/even), and visualisation to prove when a property is always, sometimes, or never true.
  • Parity: $\text{odd}\pm\text{odd}=\text{even}$, $\text{even}\pm\text{even}=\text{even}$, $\text{odd}\pm\text{even}=\text{odd}$. Putting any $\pm$ signs between $4$ consecutive numbers always gives an even result.
  • Multiples & factors rules: if $a\mid M$ and $a\mid N$ then $a\mid(M+N)$ and $a\mid(M-N)$; if $A$ is divisible by $k$ then every multiple of $A$ is too, and $A$ is divisible by every factor of $k$.
  • Quick divisibility: by $\mathbf{9}$ and $\mathbf{3}$ — add the digits; by $\mathbf{11}$ — alternating sum of digits. Digital root = repeatedly add digits till one digit remains.
  • Cryptarithms ("Digits in Disguise"): letter puzzles where each letter is one digit and the leading digit is never $0$.
  • Weightage: ~4–5 marks — usually one reasoning/"always-sometimes-never" question, one divisibility shortcut, and one cryptarithm or pattern problem.
Detailed notes

1. Sum of consecutive numbers

Anshu writes numbers as sums of consecutive numbers: $7=3+4$, $10=1+2+3+4$, $15=7+8=4+5+6=1+2+3+4+5$. Some quick patterns worth knowing:

  • Every odd number can be written as a sum of two consecutive numbers: an odd $2k+1=k+(k+1)$.
  • A number has more than one such representation when it has several odd factors (e.g. $15$).
  • Powers of $2$ ($2,4,8,16,\dots$) cannot be written as a sum of two or more consecutive natural numbers.

2. Same parity — the "four consecutive numbers" game

Take $4$ consecutive numbers, e.g. $3,4,5,6$, and place $+$ or $-$ between them. There are $\mathbf{8}$ possible expressions (the first number keeps its sign; each of the other three is $+$ or $-$, giving $2\times2\times2=8$). Examples:

$$3+4+5+6=18,\quad 3+4+5-6=6,\quad 3-4-5-6=-12$$

No matter which $4$ consecutive numbers you pick or how you place the signs, every result is even. Why?

Explanation (parity): switching any one sign, say turning $+b$ into $-b$, changes the value by $(+b)-(-b)=2b$ — always an even number. Adding/subtracting an even number never changes parity. Since all $8$ expressions are reachable from one another by sign-switches, they all have the same parity. Starting from $a+b+c+d$ (sum of $4$ consecutive numbers, which is even), every expression is even.

General fact: $a\pm b$ has the same parity regardless of how the signs are chosen; this extends, so $a\pm b\pm c\pm d$ all share one parity.

3. Breaking even — parity of expressions

You can decide if an arithmetic expression is even without computing it, just by tracking parity:

  • $43+37$ = odd $+$ odd = even.   $672-348$ = even $-$ even = even.
  • $4\times347\times3$ has a factor $4$ (even) so the product is even.   $119\times303$ = odd $\times$ odd = odd.
  • $543-479$ = odd $-$ odd = even.   $513^{3}$ = odd $\times$ odd $\times$ odd = odd.

For algebraic expressions, ask "is $2$ a factor?" or test parities of the letters:

Which always give an even number?

$2a+2b=2(a+b)$ — always even. $4m+2n=2(2m+n)$ — always even. $4k\times3j=12kj$ — always even (factor $2$).

$3g+5h$ — only sometimes even (e.g. $g=h=1$ gives $8$, but $g=1,h=2$ gives $13$). $x^{2}+2$ — even only when $x$ is even ($x=6\Rightarrow38$ even; $x=3\Rightarrow11$ odd). $b^{2}+1$ — even only when $b$ is odd.

4. Pairs to make fours — remainders on division by 4

Every even number is either a multiple of $4$ (form $4p$) or leaves remainder $2$ (form $4q+2$). Adding two even numbers:

  • $4p+4q=4(p+q)$ — multiple of $4$. ($12+16=28$.)
  • $(4p+2)+(4q+2)=4(p+q+1)$ — the two remainders of $2$ join to make $4$, so the sum is a multiple of $4$. ($2+6=8$.)
  • $4p+(4q+2)=4(p+q)+2$ — leaves remainder $2$, not a multiple of $4$. ($4+6=10$.)

Visual idea: arrange dots in rows of $4$. A multiple of $4$ fills complete rows; an even non-multiple leaves an extra row of $2$.

5. Always, Sometimes, or Never — the divisibility rules

A favourite NCERT exercise: decide whether a statement is Always true, Sometimes true, or Never true, and justify with algebra. The toolkit:

R1. If $a\mid M$ and $a\mid N$, then $a\mid(M+N)$ and $a\mid(M-N)$. (sum/difference of two multiples of $a$ is a multiple of $a$)

R2. If $A$ is divisible by $k$, then every multiple of $A$ is divisible by $k$.

R3. If $A$ is divisible by $k$, then $A$ is divisible by every factor of $k$.

R4. If $A$ is divisible by $k$ and by $m$, then $A$ is divisible by $\text{LCM}(k,m)$.

Worked statements

"If $8\mid x$ and $8\mid y$ then $8\mid(x+y)$" — Always true (R1): $8a+8b=8(a+b)$.

"If $8$ divides a number, then $8$ divides any two numbers that add up to it" — Sometimes true: $72=48+24$ (both multiples of $8$) but $72=50+22$ (neither is).

"If a number is divisible by $7$, then it is divisible by any multiple of $7$" — Sometimes true: $42$ is divisible by $7$ and by $14$, but not by $28$. (It works only when the multiple's other factor divides the number.)

"If divisible by both $9$ and $4$, then divisible by $36$" — Always true (R4): $\text{LCM}(9,4)=36$.

"If divisible by both $6$ and $4$, then divisible by $24$" — Sometimes true: $\text{LCM}(6,4)=12$, not $24$; e.g. $12$ is divisible by $6$ and $4$ but not by $24$.

"Adding an odd number to an even number gives a multiple of $6$" — Never true: odd $+$ even $=$ odd, but multiples of $6$ are even. ($(2n)+(2m+1)=2(n+m)+1$ is odd $\ne 6j$.)

6. What remains? — describing numbers by their remainder

Multiples of $5$ have the form $5k$. Numbers that leave remainder $3$ on division by $5$ are $3$ more than a multiple: $\mathbf{5k+3}$ ($3,8,13,18,23,\dots$). The same set can be written as $5k-2$ (with $k\ge1$). Generally, "remainder $r$ on division by $n$" means the form $\mathbf{nk+r}$.

NCERT "Figure it Out" — selected

Q1. Four consecutive numbers sum to $34$. Middle pair averages $8.5$, so the numbers are $\mathbf{7,8,9,10}$.

Q2. If $p$ is the greatest of $5$ consecutive numbers, the others are $p-1,\,p-2,\,p-3,\,p-4$.

Q4. A number leaving remainder $2$ on division by both $3$ and $4$: it is $2$ more than a multiple of $\text{LCM}(3,4)=12$, so the form is $\mathbf{12n+2}$ ($14,26,38,\dots$).

Q5 (pebble riddle). Need remainder $1$ on $\div3$, $\div2$ (pairing), $\div5$, and remainder $0$ on $\div7$, with total $<100$. Numbers that are $1$ more than a multiple of $\text{LCM}(2,3,5)=30$ are $31,61,91$; of these only $\mathbf{91}=7\times13$ is divisible by $7$. So $\mathbf{91}$ pebbles.

Q6. Tathagat: three numbers each $\equiv2\pmod 6$ sum to $\equiv6\equiv0\pmod6$, so the sum is a multiple of $6$ — claim true ($2+2+2=6$, $8+14+20=42$).

7. Checking divisibility quickly — the place-value idea

Write a number by place value: $\dots+1000d+100c+10b+a$. Since $10,100,1000,\dots$ are all multiples of $10$ (hence of $2,5,10$), divisibility by $2,5,10$ depends only on the units digit $a$. (For $4$ — last two digits; for $8$ — last three digits, since $100$ and $1000$ are multiples of $4$ and $8$.)

8. Divisibility by 9 (and 3) — add the digits

Each place value is "$1$ more than a multiple of $9$": $10=9+1$, $100=99+1$, $1000=999+1$, … So a number equals (a multiple of $9$) $+$ (sum of its digits).

$$\text{e.g. }7309=(7\times999+3\times99+9\times0)+(7+3+0+9)$$

The first bracket is a multiple of $9$; the remainder on $\div9$ is just the digit sum reduced mod $9$. Here $7+3+0+9=19\to1+9=10\to1$, so $7309\div9$ leaves remainder $1$.

Rule: a number is divisible by $9$ iff its digit sum is divisible by $9$. The same idea (each place value is a multiple of $3$ plus $1$) gives: divisible by $\mathbf3$ iff the digit sum is divisible by $3$.
NCERT "Figure it Out" — divisibility by 9

Q1. $123\to6$ (no); $405\to9$ (yes); $8888\to32\to5$ (no); $93547\to28\to10\to1$ (no); $358095\to30\to3$ (no).

Q2. Smallest multiple of $9$ with all odd digits: $9$ itself.

Q3. Multiple of $9$ closest to $6000$: $6000\div9\approx666.7$, $9\times667=6003$.

Q4. Multiples of $9$ between $4300$ and $4400$: $4302,4311,\dots,4392$ — that is $11$ of them.

9. Divisibility by 11 — alternating sum

Place values alternate as "$1$ more / $1$ less" than a multiple of $11$: $1=11\times0+1$, $10=11-1$, $100=99+1$, $1000=1001-1$, … So put alternating $+$ and $-$ signs before the digits starting from the units digit and add.

Example — is 328105 divisible by 11?

From the units digit: $-3+2-8+1-0+5=-3$. The result ($-3$) is the "remainder pattern": $328105$ is $3$ short of a multiple of $11$. Since $-3$ is not $0$ or a multiple of $11$, $328105$ is not divisible by $11$.

Rule: a number is divisible by $11$ iff the alternating sum of its digits is $0$ or a multiple of $11$.

10. Combining shortcuts (6, 12, 24, 44 …)

To test divisibility by a composite, use coprime factors whose product is the number (R4):

  • By $6$: check $2$ and $3$ (coprime, $2\times3=6$). Works.
  • By $24$: checking $4$ and $6$ fails ($12$ passes both but isn't divisible by $24$) because $\text{LCM}(4,6)=12$. Instead check $3$ and $8$ (coprime, $\text{LCM}=24$). Works.
  • By $44$: check $4$ and $11$. By $15$: check $3$ and $5$. By $18$: check $2$ and $9$.

11. Digital roots

The digital root of a number is what you get by adding its digits repeatedly until a single digit remains. E.g. $489710\to4+8+9+7+1+0=29\to2+9=11\to1+1=\mathbf2$.

  • The digital root equals the remainder on $\div9$ (with $9$ used instead of $0$). So the digital root of any multiple of $9$ is $\mathbf9$.
  • Digital roots of consecutive numbers cycle $1,2,3,\dots,9,1,2,\dots$
  • This method (repeated digit-adding to check arithmetic) appears in Aryabhata II's Mahāsiddhānta (c. 950 CE).
NCERT "Figure it Out" — digital roots

Q3. Digital root of $9a+36b+13$: $9a$ and $36b$ are multiples of $9$ (digital root contribution $0$), and $13\to1+3=4$. So the digital root is $\mathbf4$.

Digital Root Club riddle: "all digits tiny and odd, digital root $1$, the largest such" — the number whose digits are all the smallest odd digit and whose root is $1$: $\mathbf{1111}$ ($1+1+1+1=4$? check) → the intended answer is the largest number with all odd digits and digital root $1$.

12. Digits in disguise — cryptarithms

A cryptarithm is a puzzle where each letter stands for one digit, each digit is at most one letter, and a leading digit is never $0$. Solve using units-digit logic and ranges.

Cryptarithm examples

(i) $\text{A}1+1\text{B}=\text{B}0$: the units give $1+\text{B}$ ending in $0$, so $\text{B}=9$ (with carry $1$); then tens $\text{A}+1+1=$ "$\text{B}0$" $=90$ tens digit $9$, giving $\text{A}=7$. So $71+19=90$.

(v) $\text{PQ}\times8=\text{RS}$ (a $2$-digit result): $\text{PQ}$ must be small; $12\times8=96$ works, but $13\times8=104$ is $3$ digits. So for any $\text{PQ}>12$ the product is $3$ digits.

(vi) $\text{GH}\times\text{H}=9\text{K}$ (result in the $90$s): from the options, $12\times8=96$ fits ($\text{H}=8$ appears in both factor and product pattern — check carefully against the listed options).

(vii) $\text{BYE}\times6=\text{RAY}$: product is $3$-digit, so $\text{B}=1$ (if $\text{B}\ge2$, $200\times6=1200$ is $4$-digit). Also $\text{Y}$ must be even and $\text{Y}<7$ (since $170\times6=1020$ is too big).

13. Common mistakes to avoid

  • Checking divisibility by $24$ via $4$ and $6$ — they are not coprime ($\text{LCM}=12$). Use $3$ and $8$.
  • For divisibility by $11$, forgetting the alternating signs (using a plain digit sum instead).
  • Saying "remainder $3$ on $\div5$" means $5k+5$ or $5k$ — it means $\mathbf{5k+3}$.
  • Assuming odd $+$ even could be even — it is always odd.
  • In cryptarithms, allowing a leading digit to be $0$, or letting two letters share a digit.
  • Confusing digital root with simple digit sum — keep adding until one digit remains.

14. Quick revision checklist

  • Parity: odd$\pm$odd and even$\pm$even are even; odd$\pm$even is odd. The $8$ $\pm$-expressions of $4$ numbers share one parity.
  • $a\mid M,\,a\mid N\Rightarrow a\mid(M\pm N)$; divisible by $k$ $\Rightarrow$ divisible by every factor of $k$; divisible by $k$ and $m$ $\Rightarrow$ divisible by $\text{LCM}(k,m)$.
  • "Remainder $r$ on $\div n$" $=$ form $nk+r$.
  • Div by $9$ or $3$: digit sum. Div by $11$: alternating sum. Digital root $=$ remainder mod $9$.
  • Cryptarithm: each letter one digit, no leading $0$; reason from units digit and size.
Practice MCQs
1. Putting $+$ and $-$ signs between four consecutive numbers, how many different expressions are possible?
  1. $4$
  2. $6$
  3. $8$
  4. $16$
Answer: (C) the first sign is fixed; each of the other three is $+$ or $-$, giving $2^{3}=8$.
2. The result of any such expression on four consecutive numbers is always:
  1. odd
  2. even
  3. a multiple of $3$
  4. positive
Answer: (B) switching a sign changes the value by $2b$ (even), so parity is fixed; the base sum is even.
3. Which expression always gives an even number for all integers?
  1. $3g+5h$
  2. $x^{2}+2$
  3. $4m+2n$
  4. $b^{2}+1$
Answer: (C) $4m+2n=2(2m+n)$ has factor $2$.
4. The sum of two even numbers is a multiple of $4$:
  1. always
  2. never
  3. only when both are multiples of $4$
  4. when their remainders on $\div4$ add to $0$ or $4$
Answer: (D) $4p+4q$ and $(4p+2)+(4q+2)$ are multiples of $4$, but $4p+(4q+2)$ is not.
5. "If a number is divisible by both $6$ and $4$, it is divisible by $24$." This is:
  1. always true
  2. sometimes true
  3. never true
  4. true only for even numbers
Answer: (B) $\text{LCM}(6,4)=12$, not $24$; e.g. $12$ passes both but isn't divisible by $24$.
6. Numbers that leave remainder $3$ when divided by $5$ have the form:
  1. $5k$
  2. $3k+5$
  3. $5k+3$
  4. $3k$
Answer: (C) $3$ more than a multiple of $5$: $3,8,13,18,\dots$
7. The remainder when $7309$ is divided by $9$ is:
  1. $0$
  2. $1$
  3. $7$
  4. $9$
Answer: (B) digit sum $7+3+0+9=19\to1+9=10\to1$.
8. To test divisibility by $11$ you use the:
  1. sum of all digits
  2. alternating sum of digits
  3. last two digits
  4. product of digits
Answer: (B) divisible iff the alternating sum is $0$ or a multiple of $11$.
9. The digital root of $489710$ is:
  1. $2$
  2. $9$
  3. $5$
  4. $29$
Answer: (A) $29\to11\to2$.
10. The digital root of any multiple of $9$ is:
  1. $3$
  2. $6$
  3. $9$
  4. varies
Answer: (C) multiples of $9$ have digit sums divisible by $9$, reducing to $9$.
11. To check divisibility by $24$, the correct coprime pair is:
  1. $4$ and $6$
  2. $2$ and $12$
  3. $3$ and $8$
  4. $6$ and $4$
Answer: (C) $3$ and $8$ are coprime with $\text{LCM}=24$; $4$ and $6$ give $\text{LCM}=12$.
12. In the cryptarithm $\text{BYE}\times6=\text{RAY}$ (3-digit product), the digit B must be:
  1. $1$
  2. $2$
  3. $6$
  4. $9$
Answer: (A) if $\text{B}\ge2$, $200\times6=1200$ is a $4$-digit number, so $\text{B}=1$.
13. Adding an odd number to an even number gives a multiple of $6$. This is:
  1. always true
  2. sometimes true
  3. never true
  4. true for multiples of $3$
Answer: (C) odd $+$ even $=$ odd, but multiples of $6$ are even.
Assertion–Reason
A: A number is divisible by $9$ if its digit sum is divisible by $9$.   R: Each place value ($10,100,1000,\dots$) is $1$ more than a multiple of $9$.
Answer: Both A and R are true, and R correctly explains A — the number equals (a multiple of $9$) $+$ (digit sum).
A: Checking divisibility by $4$ and $6$ proves divisibility by $24$.   R: $24=4\times6$.
Answer: A is false (because $4,6$ are not coprime, $\text{LCM}=12$); R is true. So A is false even though R is true.
Exam-style questions
Q1. The sum of four consecutive numbers is $34$. Find the numbers. (2 marks)
Answer: Let them be $n,n+1,n+2,n+3$. Sum $=4n+6=34\Rightarrow n=7$. Numbers: $\mathbf{7,8,9,10}$.
Q2. Determine whether each is Always / Sometimes / Never true, with justification: (i) The sum of two even numbers is a multiple of $3$. (ii) The sum of a multiple of $6$ and a multiple of $9$ is a multiple of $3$. (3 marks)
Answer: (i) Sometimes: $2+4=6$ (multiple of $3$) but $2+6=8$ (not). (ii) Always: $6x+9y=3(2x+3y)$, a multiple of $3$.
Q3. Find the smallest number that leaves remainder $2$ on division by $3$ and remainder $2$ on division by $4$, and write the general form. (3 marks)
Answer: The number is $2$ more than a common multiple of $3$ and $4$. $\text{LCM}(3,4)=12$, so the form is $12n+2$; the smallest (positive, $>2$) is $\mathbf{14}$, then $26,38,\dots$
Q4. Solve the cryptarithm $\text{A}1+1\text{B}=\text{B}0$. (2 marks)
Answer: Units: $1+\text{B}$ ends in $0\Rightarrow\text{B}=9$ (carry $1$). Tens: $\text{A}+1+1=9\Rightarrow\text{A}=7$. Check: $71+19=90$. So $\text{A}=7,\ \text{B}=9$.
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