Number Play

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:
| Number | DR | Rem ÷ 9 | Rem ÷ 3 |
|--------|----|---------|---------|
| 14 | 5 | 5 | 2 |
| 18 | 9 | 0 | 0 |
| 21 | 3 | 3 | 0 |
| 25 | 7 | 7 | 1 |

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} &lt; 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 &lt; 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 &lt; 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., $23 \times 4 = 92$.

Given: $\overline{PP} \times \overline{QQ} = \overline{PRP}$

$\overline{PP} = 11P$ and $\overline{QQ} = 11Q$.

So $11P \times 11Q = 121PQ = \overline{PRP}$.

$\overline{PRP} = 100P + 10R + P = 101P + 10R$.

So: $121PQ = 101P + 10R$
$$121PQ - 101P = 10R$$
$$P(121Q - 101) = 10R$$

Since R is a single digit ($0 \leq R \leq 9$), $10R \leq 90$.
Also $P \geq 1$ (leading digit).

Try small values:
- $Q=1$: $121×1-101=20$. $20P=10R \Rightarrow 2P=R$. P can be 1,2,3,4 (R≤9 means P≤4).
- P=1: R=2. $\overline{PP}=11, \overline{QQ}=11$. But P=Q=1, not distinct. ✗
- P=2: R=4. $\overline{PP}=22, \overline{QQ}=11$. $22×11=242=\overline{PRP}$: P=2,R=4,P=2 ✓. P=2,Q=1,R=4 — all distinct ✓.
- P=3: R=6. $33×11=363=\overline{PRP}$: P=3,R=6,P=3 ✓. P=3,Q=1,R=6 — all distinct ✓.
- P=4: R=8. $44×11=484=\overline{PRP}$: P=4,R=8,P=4 ✓. P=4,Q=1,R=8 — all distinct ✓.

All three are valid.

Answer: Multiple solutions exist:
- $P=2,\ Q=1,\ R=4$: $22 \times 11 = 242$
- $P=3,\ Q=1,\ R=6$: $33 \times 11 = 363$
- $P=4,\ Q=1,\ R=8$: $44 \times 11 = 484$

Given: $\overline{JK} \times 6 = \overline{KKK}$

$\overline{KKK} = 111K = 3 \times 37 \times K$.

Also $\overline{JK} \times 6 = 111K$, so $\overline{JK} = \dfrac{111K}{6} = \dfrac{37K}{2}$.

For $\overline{JK}$ to be an integer, K must be even. Even digits: 0, 2, 4, 6, 8.

- K=0: $\overline{JK}=0$, not a 2-digit number. ✗
- K=2: $\overline{JK}=37$. J=3, K=7≠2. ✗
- K=4: $\overline{JK}=74$. J=7, K=4 ✓. Check: $74 \times 6=444=\overline{KKK}$ ✓. J=7,K=4 distinct ✓.
- K=6: $\overline{JK}=111$, 3-digit. ✗
- K=8: $\overline{JK}=148$, 3-digit. ✗

Answer: $J=7,\ K=4$, i.e., $74 \times 6 = 444$.

Given: $31z5$ is a multiple of 9.

Condition: Sum of digits must be divisible by 9.
$$3 + 1 + z + 5 = 9 + z$$

For $9 + z$ to be divisible by 9:
- $9 + z = 9 \Rightarrow z = 0$
- $9 + z = 18 \Rightarrow z = 9$

Why two answers? Since $z$ is a single digit (0–9), both $z=0$ and $z=9$ make the digit sum a multiple of 9. There is no other multiple of 9 in the range $[9, 18]$ that $9+z$ can equal for a single digit $z$.

Verification:
- $z=0$: $3105$, digit sum $=9$ ✓, $3105 ÷ 9 = 345$ ✓
- $z=9$: $3195$, digit sum $=18$ ✓, $3195 ÷ 9 = 355$ ✓

Answer: $z = 0$ or $z = 9$.

Given:
- First number $a$: $a \equiv 8 \pmod{12}$, so $a = 12m + 8$ for some integer $m \geq 0$.
- Second number $b$: $b$ is 4 short of a multiple of 12, so $b = 12n - 4$ for some integer $n \geq 1$.

Sum:
$$a + b = (12m + 8) + (12n - 4) = 12m + 12n + 4 = 12(m+n) + 4$$

Check divisibility by 8:
$12(m+n) + 4 = 4[3(m+n) + 1]$

This is always divisible by 4, but is it always divisible by 8?

For divisibility by 8, we need $3(m+n)+1$ to be even, i.e., $3(m+n)$ to be odd, i.e., $(m+n)$ to be odd.

But $(m+n)$ can be either odd or even depending on the choice of $m$ and $n$.

Counterexample: Let $m=0, n=1$: $a=8$, $b=8$. Sum $=16=2\times8$ ✓ (divisible by 8).
Let $m=1, n=1$: $a=20$, $b=8$. Sum $=28$. $28÷8=3.5$ — not divisible by 8. ✗

Conclusion: Snehal's claim is false. The sum is always divisible by 4, but not always by 8. For example, $20 + 8 = 28$ is not a multiple of 8.

Let the two multiples of 3 be $3a$ and $3b$.

Sum: $3a + 3b = 3(a+b)$.

For the sum to be a multiple of 6, we need $3(a+b)$ to be divisible by 6, i.e., $(a+b)$ must be divisible by 2, i.e., $(a+b)$ must be even.

$(a+b)$ is even when both $a$ and $b$ are even, or both are odd.

Cases:

Case 1: Both multiples of 3 are even (i.e., multiples of 6): e.g., $6 + 12 = 18$ ✓ (multiple of 6).

Case 2: Both multiples of 3 are odd (not multiples of 6): e.g., $3 + 9 = 12$ ✓ (multiple of 6).

Case 3: One multiple of 3 is even and the other is odd: e.g., $3 + 6 = 9$ — not a multiple of 6.

Generalisation:
- Sum of two multiples of 3 is a multiple of 6 if and only if both are even multiples of 3 (multiples of 6) or both are odd multiples of 3.
- The sum is not a multiple of 6 when one is an even multiple and the other is an odd multiple of 3.

(i) Reversing digits of a multiple of 9:

Concept: A number is divisible by 9 if and only if its digit sum is divisible by 9.

When we reverse the digits of a number, the digit sum remains the same (we are just rearranging the same digits).

Since the digit sum is unchanged, if the original number is divisible by 9, the reversed number also has the same digit sum, which is divisible by 9.

Therefore, Sreelatha's conjecture is TRUE for any multiple of 9.

Example: $198$ (digit sum $=18$, divisible by 9). Reversed: $891$ (digit sum $=18$) ✓.

(ii) Other digit shuffles:

Yes! Any rearrangement (shuffle) of the digits of a multiple of 9 will still be a multiple of 9, because any permutation of the digits gives the same digit sum.

Example: $1359$ is a multiple of 9 (digit sum $=18$). Any arrangement like $9531$, $3951$, $5913$, etc., will also be a multiple of 9.

Conclusion: Any permutation of the digits of a multiple of 9 gives another multiple of 9 (provided the rearranged number doesn't have a leading zero making it a number with fewer digits).

Given: $48a23b$ is a multiple of 18.

A number is a multiple of 18 if it is divisible by both 2 and 9.

Divisibility by 2: The units digit $b$ must be even. So $b \in \{0, 2, 4, 6, 8\}$.

Divisibility by 9: Sum of digits must be divisible by 9.
$$4 + 8 + a + 2 + 3 + b = 17 + a + b$$

For $17 + a + b$ to be divisible by 9:
- $17 + a + b = 18 \Rightarrow a + b = 1$
- $17 + a + b = 27 \Rightarrow a + b = 10$
- $17 + a + b = 36 \Rightarrow a + b = 19$ (max is $9+9=18$, impossible)

Case 1: $a + b = 1$ with $b$ even:
- $b=0, a=1$ ✓
- $b=2, a=-1$ ✗ (not a digit)

Case 2: $a + b = 10$ with $b$ even:
- $b=0, a=10$ ✗
- $b=2, a=8$ ✓
- $b=4, a=6$ ✓
- $b=6, a=4$ ✓
- $b=8, a=2$ ✓

All possible pairs $(a, b)$:
$$(1,\ 0),\ (8,\ 2),\ (6,\ 4),\ (4,\ 6),\ (2,\ 8)$$

Given: $3p7q8$ is divisible by 44.

$44 = 4 \times 11$, and $\gcd(4,11)=1$, so the number must be divisible by both 4 and 11.

Divisibility by 4: Last two digits $q8$ must be divisible by 4.
$\overline{q8} = 10q + 8$. For divisibility by 4: $(10q+8) \div 4$ must be integer.
$10q + 8 \equiv 2q + 0 \equiv 2q \pmod{4}$ (since $8 \equiv 0$ and $10 \equiv 2$).
So $2q \equiv 0 \pmod{4} \Rightarrow q$ is even.
$q \in \{0, 2, 4, 6, 8\}$.

Divisibility by 11: Alternating digit sum rule.
Digits of $3p7q8$ (left to right): 3, p, 7, q, 8.
$(\text{sum of digits at odd positions}) - (\text{sum at even positions})$
$= (3 + 7 + 8) - (p + q) = 18 - p - q$

For divisibility by 11: $18 - p - q \equiv 0 \pmod{11}$
- $18 - p - q = 0 \Rightarrow p + q = 18$ (max is 9+9=18, so p=9,q=9)
- $18 - p - q = 11 \Rightarrow p + q = 7$
- $18 - p - q = -11 \Rightarrow p + q = 29$ (impossible)
- $18 - p - q = 22 \Rightarrow p + q = -4$ (impossible)

Case 1: $p + q = 18$, $q$ even:
$p=9, q=9$ — but 9 is odd. ✗

Case 2: $p + q = 7$, $q$ even:
- $q=0, p=7$ ✓
- $q=2, p=5$ ✓
- $q=4, p=3$ ✓
- $q=6, p=1$ ✓
- $q=8, p=-1$ ✗

All possible pairs $(p, q)$:
$$(7,\ 0),\ (5,\ 2),\ (3,\ 4),\ (1,\ 6)$$

Let the three consecutive numbers be $n,\ n+1,\ n+2$.

Conditions:
- $n$ is a multiple of 2: $n = 2k$
- $n+1$ is a multiple of 3: $n \equiv 2 \pmod{3}$
- $n+2$ is a multiple of 4: $n \equiv 2 \pmod{4}$

From condition 3: $n \equiv 2 \pmod{4}$, so $n \in \{2, 6, 10, 14, 18, 22, 26, 30, ...\}$.

Also need $n \equiv 2 \pmod{3}$:
Check: $n=2$: $2 \mod 3 = 2$ ✓. Numbers: 2, 3, 4. Check: 2 is multiple of 2 ✓, 3 is multiple of 3 ✓, 4 is multiple of 4 ✓.

Find the pattern: We need $n \equiv 2 \pmod{4}$ and $n \equiv 2 \pmod{3}$.
By CRT: $n \equiv 2 \pmod{\text{lcm}(4,3)} = n \equiv 2 \pmod{12}$.

So $n = 12m + 2$ for $m = 0, 1, 2, ...$

Solutions: $n = 2, 14, 26, 38, 50, ...$

Verification for $n=14$: 14 (multiple of 2 ✓), 15 (multiple of 3 ✓), 16 (multiple of 4 ✓).

Answer: Yes, there are infinitely many such triples. They occur every 12 consecutive integers. The first few are: $(2,3,4),\ (14,15,16),\ (26,27,28),\ (38,39,40), \ldots$

Approach: Find multiples of 36 in the range $(45000, 47000)$.

Step 1: $45000 ÷ 36 = 1250$. So $36 \times 1250 = 45000$.

Step 2: The next multiples are:
- $36 \times 1251 = 45036$
- $36 \times 1252 = 45072$
- $36 \times 1253 = 45108$
- $36 \times 1254 = 45144$
- $36 \times 1255 = 45180$

Verification: $36 \times 1255 = 36 \times 1000 + 36 \times 255 = 36000 + 9180 = 45180$ ✓. All are between 45000 and 47000 ✓.

Answer: Five multiples of 36 between 45,000 and 47,000 are: 45036, 45072, 45108, 45144, 45180.

Given: 5 consecutive even numbers with middle number $= 5p$.

Concept: Consecutive even numbers differ by 2.

If the middle (3rd) number is $5p$, then:
- 1st number: $5p - 4$
- 2nd number: $5p - 2$
- 3rd number: $5p$ (middle)
- 4th number: $5p + 2$
- 5th number: $5p + 4$

Note: For these to be even numbers, $5p$ must be even, which means $p$ must be even.

Answer: The five consecutive even numbers in sequence are:
$$5p-4,\ 5p-2,\ 5p,\ 5p+2,\ 5p+4$$

Conditions:
- 6-digit number divisible by 15: must be divisible by both 3 and 5, so units digit is 0 or 5, and digit sum is divisible by 3.
- When reversed, divisible by 6: reversed number must be divisible by both 2 and 3.
- Divisible by 2: units digit of reversed number (= first digit of original) must be even.
- Divisible by 3: digit sum of reversed = digit sum of original (same digits), so if original is divisible by 3, reversed is also divisible by 3 ✓.

So we need:
1. Units digit of original = 0 or 5 (for divisibility by 5).
2. First digit of original is even (for reversed number to be divisible by 2).
3. Digit sum divisible by 3 (for divisibility by 3 and hence 15).

If units digit = 0: Reversed number ends in 0, which is even ✓. First digit must be even.

Construction: Let the number be $2\_\_\_\_0$ (first digit 2, last digit 0).
Choose: $210030$. Digit sum $= 2+1+0+0+3+0=6$, divisible by 3 ✓. Divisible by 5 (ends in 0) ✓. Divisible by 15 ✓.
Reversed: $030012 = 30012$. But this is a 5-digit number — the leading zero is a problem.

Let first digit be even and non-zero, last digit = 0:
Number: $\ 2\_\_\_\_0$. Reversed starts with 0 — invalid as a 6-digit number.

If units digit = 5: Reversed number ends in 5 (odd) — not divisible by 2. ✗

Revised approach: For the reversed number to be a valid 6-digit number, the last digit of the original must be non-zero. So units digit = 5 (since units digit must be 0 or 5 for divisibility by 5, and 0 would make reversed number start with 0).

But units digit = 5 makes reversed number end in 5 (odd), so not divisible by 2, hence not by 6. ✗

Resolution: The problem may allow the reversed number to start with 0 (treating it as a number with fewer digits), or the units digit is 0 and we accept the reversed number as a 5-digit number divisible by 6.

Example: $420000$. Digit sum $= 6$ ✓, ends in 0 ✓, divisible by 15 ✓. Reversed: $000024 = 24$. $24 ÷ 6 = 4$ ✓.

A cleaner example: $\ 123450$. Digit sum $=15$ ✓, ends in 0 ✓, divisible by 15 ✓. Reversed: $054321=54321$ (5-digit). $54321÷2$ — odd, not divisible by 2. ✗

Let first digit be even: $\ 213450$. Digit sum $=15$ ✓. Reversed: $054312=54312$. $54312÷6=9052$ ✓.

Answer: One valid 6-digit number is $\mathbf{213450}$.
- $213450 ÷ 15 = 14230$ ✓
- Reversed: $054312 = 54312$; $54312 ÷ 6 = 9052$ ✓

Let $N$ be a multiple of 11, so $N = 11k$ for some integer $k$.

Doubling: $2N = 2 \times 11k = 11 \times (2k)$.

Since $2k$ is always an integer, $2N$ is always a multiple of 11.

Conclusion: Deepak's claim is FALSE. Every multiple of 11, when doubled, is still a multiple of 11.

Reason: $11 \mid N \Rightarrow 11 \mid 2N$ (since 11 is a prime and does not divide 2, but $11 \mid N$ directly implies $11 \mid 2N$).

Example: $11 \times 3 = 33$; $2 \times 33 = 66 = 11 \times 6$ ✓. $11 \times 7 = 77$; $2 \times 77 = 154 = 11 \times 14$ ✓.

(i) Product of a multiple of 6 and a multiple of 3 is a multiple of 9:

Let the numbers be $6a$ and $3b$. Product $= 18ab = 9 \times 2ab$.
This is always a multiple of 9.

Answer: Always True.

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(ii) Sum of three consecutive even numbers is divisible by 6:

Let the three consecutive even numbers be $2k-2,\ 2k,\ 2k+2$.
Sum $= 6k$, which is always divisible by 6.

Answer: Always True.

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(iii) If $abcdef$ is a multiple of 6, then $badcef$ is a multiple of 6:

A number is divisible by 6 if it is divisible by both 2 and 3.

- Divisibility by 3: Both $abcdef$ and $badcef$ have the same digits (just rearranged), so the same digit sum. If $abcdef$ is divisible by 3, so is $badcef$. ✓
- Divisibility by 2: $abcdef$ ends in $f$ (even). $badcef$ also ends in $f$ (even). ✓

Since both conditions hold, $badcef$ is always divisible by 6.

Answer: Always True.

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(iv) $8(7b-3) - 4(11b+1)$ is a multiple of 12:

Simplify:
$$8(7b-3) - 4(11b+1) = 56b - 24 - 44b - 4 = 12b - 28$$
$$= 12b - 28 = 4(3b - 7)$$

For this to be a multiple of 12, we need $3b-7$ to be a multiple of 3.
$3b - 7 \equiv -7 \equiv -1 \equiv 2 \pmod{3}$.
This is never $\equiv 0 \pmod{3}$.

So $4(3b-7)$ is divisible by 4 but never by 12.

Counterexample: $b=1$: $12(1)-28 = -16$. $-16 ÷ 12$ is not an integer. ✓ (confirms not always multiple of 12)

Answer: Never True.