Number Play — NCERT Solutions

A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

(i) 123
Sum of digits $= 1 + 2 + 3 = 6$
6 is not divisible by 9.
∴ 123 is not divisible by 9.

(ii) 405
Sum of digits $= 4 + 0 + 5 = 9$
9 is divisible by 9.
∴ 405 is divisible by 9.

(iii) 8888
Sum of digits $= 8 + 8 + 8 + 8 = 32$
32 is not divisible by 9.
∴ 8888 is not divisible by 9.

(iv) 93547
Sum of digits $= 9 + 3 + 5 + 4 + 7 = 28$
28 is not divisible by 9.
∴ 93547 is not divisible by 9.

(v) 358095
Sum of digits $= 3 + 5 + 8 + 0 + 9 + 5 = 30$
30 is not divisible by 9.
∴ 358095 is not divisible by 9.

Given: We need the smallest multiple of 9 that contains only even digits (0, 2, 4, 6, 8).

Method: List multiples of 9 and check whether all digits are even.

• $9 \times 1 = 9$ → digit 9 is odd ✗
• $9 \times 2 = 18$ → digit 1 is odd ✗
• $9 \times 3 = 27$ → both digits odd ✗
• $9 \times 4 = 36$ → digit 3 is odd ✗
• $9 \times 5 = 45$ → both digits odd ✗
• $9 \times 6 = 54$ → digit 5 is odd ✗
• $9 \times 7 = 63$ → both digits odd ✗
• $9 \times 8 = 72$ → digit 7 is odd ✗
• $9 \times 9 = 81$ → both digits odd ✗
• $9 \times 10 = 90$ → digit 9 is odd ✗
• Continue checking two-digit multiples... all two-digit multiples of 9 contain at least one odd digit.
• Check three-digit multiples with only even digits. The smallest even-digit number is 200. $200 ÷ 9 = 22.2...$, so next multiple is $9 × 23 = 207$ (odd digit). Keep checking.
• We need digit sum divisible by 9 using only even digits (0,2,4,6,8).
• Smallest such sum = 18 (e.g., digits 2+8+8 or 4+6+8 or 6+6+6, etc.)
• Smallest number with digit sum 18 using only even digits: try 288 → $2+8+8=18$ ✓, all even digits ✓.
• Check: $288 ÷ 9 = 32$ ✓

Answer: The smallest multiple of 9 with no odd digits is $\boxed{288}$.

Given: Find the multiple of 9 nearest to 6000.

Step 1: Divide 6000 by 9.
$$6000 ÷ 9 = 666.\overline{6}$$

Step 2: The two nearest multiples of 9 are:

• $9 \times 666 = 5994$
• $9 \times 667 = 6003$

Step 3: Find the distances:

• $6000 - 5994 = 6$
• $6003 - 6000 = 3$

Step 4: Since 6003 is closer to 6000 (distance 3 < 6),

Answer: The multiple of 9 closest to 6000 is $\boxed{6003}$.

Given: Find multiples of 9 strictly between 4300 and 4400.

Step 1: Find the smallest multiple of 9 greater than 4300.
$$4300 ÷ 9 = 477.\overline{7}$$
So the next multiple is $9 \times 478 = 4302$.

Step 2: Find the largest multiple of 9 less than 4400.
$$4400 ÷ 9 = 488.\overline{8}$$
So the largest multiple less than 4400 is $9 \times 488 = 4392$.

Step 3: Count the multiples from $9 \times 478$ to $9 \times 488$:
$$488 - 478 + 1 = 11$$

Verification: The multiples are 4302, 4311, 4320, 4329, 4338, 4347, 4356, 4365, 4374, 4383, 4392 — that is 11 multiples.

Answer: There are $\boxed{11}$ multiples of 9 between 4300 and 4400.

Given: Digital root of an 8-digit number $N$ is 5.

Concept: The digital root of a number equals the remainder when divided by 9 (with the convention that if the remainder is 0, the digital root is 9).

So $N \equiv 5 \pmod{9}$.

$N + 10 \equiv 5 + 10 = 15 \equiv 15 - 9 = 6 \pmod{9}$

Answer: The digital root of $(N + 10)$ is $\boxed{6}$.

Example: Start with 5.
Sequence: 5, 16, 27, 38, 49, 60, 71, 82, 93, 104, 115, …

Digital roots: 5, 7, 9, 2, 4, 6, 8, 1, 3, 5, 7, …

Observation: Since $11 \equiv 2 \pmod{9}$, each time we add 11, the digital root increases by 2 (mod 9). The digital roots form a repeating cycle of length 9, cycling through 9 values with a step of 2:
$$\ldots, 5, 7, 9, 2, 4, 6, 8, 1, 3, 5, 7, \ldots$$

Generalisation: If the starting number has digital root $d$, the sequence of digital roots is $d,\ d+2,\ d+4,\ \ldots$ (all mod 9, using 9 instead of 0). The pattern repeats every 9 terms and covers all digits 1–9.

Given expression: $9a + 36b + 13$

Step 1: Simplify modulo 9.
$$9a \equiv 0 \pmod{9}$$
$$36b = 4 \times 9 \times b \equiv 0 \pmod{9}$$
$$13 \equiv 4 \pmod{9}$$

Step 2: Therefore,
$$9a + 36b + 13 \equiv 0 + 0 + 4 = 4 \pmod{9}$$

Answer: The digital root of $9a + 36b + 13$ is $\boxed{4}$ (for any whole-number values of $a$ and $b$).

(i) Parity and digital root:

Examine examples:

• Even numbers: 2 (DR=2), 4 (DR=4), 12 (DR=3), 18 (DR=9), 20 (DR=2), 22 (DR=4)
• Odd numbers: 1 (DR=1), 3 (DR=3), 11 (DR=2), 19 (DR=1)

Conjecture: There is no fixed relationship between the parity of a number and its digital root. Even numbers can have any digital root (1–9), and so can odd numbers. For example, 12 is even but has digital root 3 (odd), while 11 is odd but has digital root 2 (even).

(ii) Digital root and remainder when divided by 3 or 9:

Examine examples:

Conjecture 1: The digital root of a number equals its remainder when divided by 9, except when the remainder is 0, in which case the digital root is 9.

Conjecture 2: If the digital root is divisible by 3 (i.e., DR ∈ {3, 6, 9}), then the number is divisible by 3. The remainder when divided by 3 equals (digital root) mod 3. So the digital root and the remainder on division by 3 are directly related.

Given:
$$\begin{array}{r} A1 \\ +\;1B \\ \hline B0 \end{array}$$

Step 1 — Units column: $1 + B = 0$ or $1 + B = 10$.
Since digits are 0–9, $1 + B = 10 \Rightarrow B = 9$ (with carry 1 to tens column).

Step 2 — Tens column (with carry 1): $A + 1 + 1 = B = 9$
$$A + 2 = 9 \Rightarrow A = 7$$

Verification: $71 + 19 = 90$ ✓ (B = 9, units digit 0 ✓)

Answer: $A = 7,\ B = 9$, i.e., $71 + 19 = 90$.

Given:
$$\begin{array}{r} AB \\ +\;37 \\ \hline 6A \end{array}$$

Step 1 — Units column: $B + 7 = A$ or $B + 7 = 10 + A$ (with carry).

Step 2 — Tens column (no carry): $A + 3 = 6 \Rightarrow A = 3$.

Step 3: Back to units: $B + 7 = A = 3$ is impossible (would give negative B).
So there must be a carry: $B + 7 = 10 + A = 13 \Rightarrow B = 6$.
But then tens column with carry: $A + 3 + 1 = 6 \Rightarrow A = 2$.
Check units: $B + 7 = 13 \Rightarrow B = 6$. ✓

Verification: $26 + 37 = 63$ ✓ (tens digit 6 ✓, units digit 3 = A ✓)

Answer: $A = 2,\ B = 6$, i.e., $26 + 37 = 63$.

Given:
$$\begin{array}{r} ON \\ ON \\ +\;ON \\ \hline PO \end{array}$$

This means $3 \times \overline{ON} = \overline{PO}$.

Step 1 — Units column: $3 \times N$ ends in O.

Step 2 — Tens column: $3 \times O + \text{carry} = 10P + O$ (since the tens digit of the result is P and units digit is O).
Wait — the result is $\overline{PO}$, a 2-digit number, so $3 \times O + \text{carry}$ gives tens digit P and we need the units digit of the result to be O.

Actually the result $\overline{PO}$ has tens digit P and units digit O.

Step 3: Units: $3N \equiv O \pmod{10}$, carry $c_1 = \lfloor 3N/10 \rfloor$.
Tens: $3O + c_1 = 10P + O$ (no further carry since result is 2-digit, so P is a single digit and $3 \times \overline{ON} < 100$, meaning $\overline{ON} \leq 33$, so $O \leq 3$).

From tens: $2O + c_1 = 10P$. Since $O \leq 3$ and $c_1 \leq 2$, $2O + c_1 \leq 8$, so $P = 0$? But P is the leading digit — P cannot be 0.

Let me reconsider: $3O + c_1$ gives tens digit P and units digit O.
So $3O + c_1 = 10P + O \Rightarrow 2O + c_1 = 10P$.
For $P \geq 1$: $2O + c_1 \geq 10$. With $c_1 \leq 2$, we need $2O \geq 8$, so $O \geq 4$.
But if $O \geq 4$, then $\overline{ON} \geq 40$, and $3 \times 40 = 120$ — a 3-digit number, contradiction.

So $P = 0$ is forced, but P can't be 0 as leading digit. Let me re-read: result is $\overline{PO}$, a 2-digit number. Actually if $P=0$ the result would be a 1-digit number O, which contradicts the layout.

Re-examine: perhaps the result is meant to be a 2-digit number with tens digit P ≠ 0. Try small values of O:

• $O=1$: $3 \times 1N$. For result to have tens digit P and units digit 1: $3N$ ends in 1 → $N=7$ (since $3×7=21$, carry 2). Tens: $3×1+2=5$. Result tens digit = 5 = P, units = 1 = O ✓. So $\overline{ON}=17$, $3×17=51=\overline{PO}$ with P=5, O=1 ✓. All digits: O=1, N=7, P=5 — all different ✓.

Verification: $17 + 17 + 17 = 51$ ✓

Answer: $O = 1,\ N = 7,\ P = 5$, i.e., $17 + 17 + 17 = 51$.

Given:
$$\begin{array}{r} QR \\ QR \\ +\;QR \\ \hline PRR \end{array}$$

This means $3 \times \overline{QR} = \overline{PRR}$, a 3-digit number.

Step 1: $\overline{PRR}$ has the same digit R in both tens and units places.

Step 2 — Units column: $3R$ ends in R, so $3R \equiv R \pmod{10} \Rightarrow 2R \equiv 0 \pmod{10} \Rightarrow R = 0$ or $R = 5$.

Case R = 0: $3 \times \overline{Q0} = \overline{P00}$. Units: $3×0=0$ ✓, carry=0. Tens: $3Q$ ends in 0 → $Q=0$ (but then QR=00, not a proper 2-digit number) or no valid Q. Actually $3Q$ ends in 0 means $Q=0$ or $Q=10$ — impossible for a digit. So R=0 gives no valid solution.

Case R = 5: $3 \times \overline{Q5} = \overline{P55}$.
Units: $3×5=15$, units digit 5 ✓, carry = 1.
Tens: $3Q + 1$ ends in 5 → $3Q + 1 \equiv 5 \pmod{10} \Rightarrow 3Q \equiv 4 \pmod{10}$.
Trying digits: $Q=8$: $3×8=24$, $24+1=25$, tens digit 5 ✓, carry 2.
Hundreds: carry = 2 = P. So P=2.

Check: $\overline{QR}=85$, $3×85=255=\overline{PRR}$ with P=2, R=5 ✓. All letters: Q=8, R=5, P=2 — all different ✓.

Verification: $85 + 85 + 85 = 255$ ✓

Answer: $Q = 8,\ R = 5,\ P = 2$, i.e., $85 + 85 + 85 = 255$.

Given: $\overline{PQ} \times 8 = \overline{RS}$, where P, Q, R, S are distinct digits and the result is a 2-digit number.

Constraints:

• $\overline{PQ}$ is a 2-digit number, so $10 \leq \overline{PQ} \leq 99$.
• $\overline{PQ} \times 8$ must also be a 2-digit number, so $\overline{PQ} \leq 12$ (since $13 \times 8 = 104$, a 3-digit number).
• $\overline{PQ} \geq 10$, so $\overline{PQ} \in \{10, 11, 12\}$.
• All four digits P, Q, R, S must be distinct.

Check each:

• $10 \times 8 = 80$: P=1, Q=0, R=8, S=0. But Q=S=0, not all distinct. ✗
• $11 \times 8 = 88$: P=Q=1, not distinct. ✗
• $12 \times 8 = 96$: P=1, Q=2, R=9, S=6. All distinct ✓.

Answer: $PQ = 12,\ RS = 96$, i.e., $12 \times 8 = 96$.

Given: $\overline{GH} \times H = \overline{9K}$

Condition: The units digit of $\overline{GH}$ is H, and H is also the multiplier. The result is a 2-digit number in the 90s (tens digit = 9).

Check each option for the condition $GH \times H$ (units digit of the 2-digit number equals the multiplier):

• $11 \times 9 = 99$: G=1, H=1 — but G and H must be different digits. ✗
• $12 \times 8 = 96$: Units digit of 12 is 2, but multiplier is 8. H should be units digit = 2, but multiplier = 8 ≠ 2. ✗
• $46 \times 2 = 92$: Units digit of 46 is 6, multiplier is 2. 6 ≠ 2. ✗
• $24 \times 4 = 96$: Units digit of 24 is 4, multiplier is 4. H = 4 ✓. G=2, H=4, K=6 — all different ✓. Result is 96 (in the 90s) ✓.
• $47 imes 2 = 94$: Units digit of 47 is 7, multiplier is 2. 7 ≠ 2. ✗
• $31 \times 3 = 93$: Units digit of 31 is 1, multiplier is 3. 1 ≠ 3. ✗
• $16 \times 6 = 96$: Units digit of 16 is 6, multiplier is 6. H = 6 ✓. G=1, H=6, K=6 — but H=K=6, not all distinct. ✗

Answer: $GH \times H = 24 \times 4 = 96$, so $G=2,\ H=4,\ K=6$.

Given: $\overline{BYE} \times 6 = \overline{RAY}$, with B=1, Y is even, $Y < 7$.

So $\overline{1YE} \times 6 = \overline{RAY}$.

Possible even values of Y: 0, 2, 4, 6 — but Y < 7, so Y ∈ {0, 2, 4, 6}.

Units digit condition: $6 \times E$ must end in Y.

Try Y = 2: $6E$ ends in 2 → $E = 2$ (gives 12) or $E = 7$ (gives 42). But E ≠ Y = 2, so E = 7.
$\overline{1YE} = 127$. $127 \times 6 = 762$. Result = $\overline{RAY} = 762$: R=7, A=6, Y=2 ✓. Check Y=2 in result ✓. All digits B=1,Y=2,E=7,R=7,A=6 — but E=R=7, not all distinct. ✗

Try Y = 4: $6E$ ends in 4 → $E = 4$ (gives 24, but E≠Y) or $E = 9$ (gives 54).
E = 9: $\overline{1YE} = 149$. $149 \times 6 = 894$. Result $\overline{RAY}$: R=8, A=9, Y=4 ✓. Check: A=9=E — not all distinct. ✗
E = 4: E=Y=4, not distinct. ✗

Try Y = 6: $6E$ ends in 6 → $E = 1$ (gives 6, no — $6×1=6$ ✓) or $E=6$ (E=Y ✗).
E=1: but B=1=E, not distinct. ✗
Also $E=6$: E=Y ✗.

Try Y = 0: $6E$ ends in 0 → $E = 0$ (E=Y ✗) or $E=5$.
E=5: $\overline{1YE}=105$. $105 \times 6=630$. Result $\overline{RAY}=630$: R=6, A=3, Y=0 ✓. All digits: B=1,Y=0,E=5,R=6,A=3 — all distinct ✓.

Verification: $105 \times 6 = 630$ ✓

Answer: $B=1,\ Y=0,\ E=5,\ R=6,\ A=3$, i.e., $105 \times 6 = 630$.

Given: $\overline{UT} \times 3 = \overline{PUT}$

$\overline{PUT}$ is a 3-digit number and $\overline{UT}$ is a 2-digit number.

$\overline{PUT} = 100P + \overline{UT}$

So: $3 \times \overline{UT} = 100P + \overline{UT}$
$$2 \times \overline{UT} = 100P$$
$$\overline{UT} = 50P$$

Since $\overline{UT}$ is a 2-digit number: $10 \leq 50P \leq 99$.
Only $P=1$ works: $\overline{UT} = 50$.

Check: $U=5, T=0, P=1$. All distinct ✓.
$50 \times 3 = 150 = \overline{PUT}$ with P=1, U=5, T=0 ✓.

Answer: $U=5,\ T=0,\ P=1$, i.e., $50 \times 3 = 150$.

Given: $\overline{AB} \times 5 = \overline{BC}$

Both are 2-digit numbers, so $\overline{AB} \leq 19$ (since $20 \times 5 = 100$, a 3-digit number).
Also $\overline{AB} \geq 10$, so $\overline{AB} \in \{10, 11, ..., 19\}$, meaning $A=1$.

$\overline{BC}$ has tens digit B (same as units digit of $\overline{AB}$).

So $\overline{1B} \times 5 = \overline{BC}$.

Units digit of $5 \times B$ must equal C, and tens digit of result must equal B.

Try values of B (0–9):

• $B=0$: $10 \times 5=50$. Result $\overline{BC}=50$: B=5≠0. ✗
• $B=1$: $11 \times 5=55$. Result: tens=5≠1. ✗
• $B=2$: $12 \times 5=60$. Result: tens=6≠2. ✗
• $B=3$: $13 \times 5=65$. Result: tens=6≠3. ✗
• $B=4$: $14 \times 5=70$. Result: tens=7≠4. ✗
• $B=5$: $15 \times 5=75$. Result: tens=7≠5. ✗
• $B=6$: $16 \times 5=80$. Result: tens=8≠6. ✗
• $B=7$: $17 \times 5=85$. Result: tens=8≠7. ✗
• $B=8$: $18 \times 5=90$. Result: tens=9≠8. ✗
• $B=9$: $19 \times 5=95$. Result: tens=9=B ✓, units=5=C. A=1,B=9,C=5 — all distinct ✓.

Answer: $A=1,\ B=9,\ C=5$, i.e., $19 \times 5 = 95$.

Given: $\overline{L2N} \times 2 = \overline{2NP}$

The result $\overline{2NP}$ starts with 2, so $200 \leq \overline{2NP} \leq 299$, meaning $100 \leq \overline{L2N} \leq 149$, so $L=1$.

$\overline{12N} \times 2 = \overline{2NP}$

Units: $2N$ ends in P, carry $c = \lfloor 2N/10 \rfloor$.
Tens: $2 \times 2 + c = 4 + c$ must end in N (tens digit of result is N).
Hundreds: carry from tens gives hundreds digit = 2 ✓ (already fixed).

From tens: $4 + c \equiv N \pmod{10}$, and carry to hundreds = $\lfloor(4+c)/10\rfloor$.
Since result hundreds digit is 2 and $\overline{12N} \times 2 < 300$, the hundreds carry must be 0 or 1.
Actually $\overline{12N} \times 2$: max is $129 \times 2 = 258$, min is $120 \times 2 = 240$. So hundreds digit is always 2 ✓.

From tens column: $4 + c$ gives tens digit N (no carry to hundreds since result is in 200s and $4+c \leq 5$, so no carry). Thus $N = 4 + c$.

$c = 0$: $N=4$, then units: $2×4=8$, P=8. Check: $\overline{L2N}=124$, $124×2=248=\overline{2NP}$: 2,4,8 ✓. L=1,N=4,P=8 — all distinct ✓.
$c=1$: $N=5$, then $2N$ has carry 1 means $2×5=10$, P=0. Check: $125×2=250=\overline{2NP}$: 2,5,0 ✓. L=1,N=5,P=0 — all distinct ✓.

Both are valid solutions.

Answer: $L=1,\ N=4,\ P=8$ (i.e., $124 \times 2 = 248$) or $L=1,\ N=5,\ P=0$ (i.e., $125 \times 2 = 250$).

Given: $\overline{XY} \times 4 = \overline{ZX}$

Both are 2-digit numbers, so $\overline{XY} \leq 24$ (since $25 \times 4 = 100$).
Also $\overline{XY} \geq 10$.

The units digit of the result $\overline{ZX}$ is X (same as tens digit of $\overline{XY}$).

Try X = 1 (so $\overline{XY} \in \{10,...,19\}$):
Result $\overline{ZX}$ ends in 1. $4Y$ ends in 1 — impossible (4×any digit is even or ends in 0,4,8,2,6). ✗

Try X = 2 (so $\overline{XY} \in \{20,21,22,23,24\}$):
Result $\overline{ZX}$ ends in 2. $4Y$ ends in 2 → $Y=3$ (4×3=12) or $Y=8$ (4×8=32).

• $Y=3$: $23 \times 4=92$. Result $\overline{ZX}=92$: Z=9, X=2 ✓. X=2,Y=3,Z=9 — all distinct ✓.
• $Y=8$: $28 \times 4=112$, 3-digit. ✗

Answer: $X=2,\ Y=3,\ Z=9$, i.e., $23 \times 4 = 92$.