We use the parity rules: odd ± odd = even, even ± even = even, odd ± even = odd, odd × odd = odd, even × any = even.

Always Even: \(2a + 2b\), \(4m + 2n\), \(2u – 4v\), \(13k – 5k = 8k\), \(4k \times 3j = 12kj\)

All contain factor 2. Example: \(4m + 2n = 2(2m + n)\) which is clearly even.

Not Always Even: \(3g + 5h\) (odd+odd=even, but even+even=even too — actually sum of odd coeffs), \(x^2 + 2\) (even when x even: \(6^2+2=38\); odd when x odd: \(3^2+2=11\)), \(b^2 + 1\), \(6m – 3n\)

Even numbers fall into two categories when divided by 4:

• Type A: Multiples of 4: \(4, 8, 12, 16, …\) (form: \(4p\))
• Type B: Not multiples of 4: \(2, 6, 10, 14, …\) (form: \(4p + 2\))

Rule: The sum of two even numbers is divisible by 4 iff both even numbers have the same remainder when divided by 4 (both are \(0 \pmod 4\) or both are \(2 \pmod 4\)).

Let the two numbers be \(8a\) and \(8b\) (both multiples of 8).

Then: \(8a + 8b = 8(a + b)\)

This is clearly a multiple of 8. Also true for subtraction: \(8a – 8b = 8(a-b)\)

General Rule: If \(a\) divides \(M\) and \(a\) divides \(N\), then \(a\) divides \(M + N\) and \(M – N\).

A number divisible by 8 can be written as: \(8m = 8a + 8b\) (sum of two multiples of 8) OR \(8m = p + q\) where neither \(p\) nor \(q\) is a multiple of 8.

Let the number be \(7j\) (divisible by 7). Any multiple is \((7j) \times m = 7(jm) = 7 \times (\text{integer})\).

This has 7 as a factor, so it’s divisible by 7.

General Rule: If \(A\) is divisible by \(k\), then all multiples of \(A\) are divisible by \(k\).

A number divisible by 12 can be written as \(12m\).

Since \(12 = 2 \times 6 = 3 \times 4 = 2 \times 2 \times 3\), we have:

\(12m = 2 \times 6 \times m = 3 \times 4 \times m = 2 \times 2 \times 3 \times m\)

Factors of 12: 1, 2, 3, 4, 6, 12. Each factor divides \(12m\).

General Rule: If \(A\) is divisible by \(k\), then \(A\) is divisible by all factors of \(k\).

Let the number be \(7k\). For it to be divisible by \(7m\) (a multiple of 7), we need \(m\) to be a factor of \(k\).

\(7k \div 7m = \frac{k}{m}\) — this is an integer only when \(m\) divides \(k\).

If a number is divisible by both 9 and 4, it must be divisible by \(\text{LCM}(9, 4) = 36\).

This is because 9 and 4 are co-prime (\(\gcd(9,4) = 1\)).

\(\text{LCM}(6, 4) = 12\), not 24. So the number must be divisible by 12, not necessarily 24.

odd + even = odd. But all multiples of 6 are even.

An odd number can never equal an even number.

Numbers with remainder 3 when divided by 5 are: 3, 8, 13, 18, 23, …

These are 3 more than multiples of 5. Multiples of 5 are \(5k\), so numbers are \(5k + 3\).

\(5k – 2\) also works for \(k \geq 1\):

So \(5k + 3\) and \(5k – 2\) (for \(k \geq 1\)) both capture all such numbers.

Rule: A number is divisible by 9 iff the sum of its digits is divisible by 9.

Only 405 is divisible by 9.

Step 1: Even digits are: 2, 4, 6, 8. We need to form a number using only these digits whose digit sum is divisible by 9.

Step 2: Sum of 9 (smallest multiple of 9) is odd, but sum of even digits is always even. So we need the next multiple: 18.

Step 3: Find even digits that sum to 18: \(2 + 8 + 8 = 18\)

Step 4: Arrange 2, 8, 8 to form the smallest number: 288

Verification: \(2 + 8 + 8 = 18 = 9 \times 2\) ✓

Step 1: Divide 6000 by 9: \(6000 \div 9 = 666.67\)

Step 2: So the nearest multiples are: \(9 \times 666 = 5994\) and \(9 \times 667 = 6003\)

Step 3: Distance from 6000:

Answer: 6003 (only 3 away from 6000)

Step 1: Find the first multiple of 9 ≥ 4300:

Step 2: Find the last multiple of 9 ≤ 4400:

Step 3: Count: from 478 to 488 inclusive:

Answer: 11 multiples of 9

Rule: Place alternating + and – signs starting from the unit’s digit. Evaluate. If result is 0 or multiple of 11, the number is divisible by 11.

For numbers 600-700, digit sum = \(6 + 0 + 0 + \text{other digits}\) = \(6 + \text{sum of last 2 digits}\).

Observation: For multiples of 3 and 6, the digital roots cycle through {3, 6, 9}. Numbers that are 1 more than a multiple of 6 have digital roots that cycle through {1, 4, 7} or {2, 5, 8}.

Clues decoded:

• “Each tiniest and odd” → all digits are odd (1 is the tiniest odd digit)
• “No shared ground with root #1” → digital root is not 1
• “Digits count, their sum, my root” → number of digits = sum of digits = digital root
• “Largest odd single-digit” → 9

So: number of digits = 9, digital root = 9, all digits are 1 (odd).

Answer: 111111111 (Nine 1s)

Written in words: Eleven crore eleven lakh eleven thousand one hundred eleven

\(= 11,11,11,111\)

Adding 10 increases the digit sum by 1 (since \(10 \rightarrow 1 + 0 = 1\)).

Digital root of new number = \(5 + 1 = \mathbf{6}\)

Example: If number is 40000001 (digit sum = 5), then 40000011 has digit sum = 6.

Let the starting number be 10. Sequence: 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, …

Observation: The digital roots follow a repeating cycle: 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, …

This happens because adding 11 (whose digital root is 2) shifts the digital root by 2 each time (mod 9).

Step 1: Rearrange the expression:

Step 2: \(9(a + 4b + 1)\) is a multiple of 9, so its digital root is 9.

Step 3: Digital root of the entire expression = digital root of \(9 + 4 = 13 \rightarrow 1 + 3 = \mathbf{4}\)

Step 1: Sum of digits = \(3 + 1 + z + 5 = 9 + z\)

Step 2: For divisibility by 9, \(9 + z\) must be divisible by 9.

Step 3: Since \(z\) is a single digit (0-9):

Two answers: \(z = 0\) or \(z = 9\)

Numbers: 3105 (sum=9) and 3195 (sum=18) — both divisible by 9.

Step 1: Let the numbers be:

\(a = 12n + 8\) and \(b = 12m – 4\)

Step 2: Their sum:

Step 3: \(12k + 4 = 4(3k + 1)\). This is always divisible by 4, but not always by 8.

Counter-example: Let \(k = 2\): \(12(2) + 4 = 28\), which is NOT divisible by 8.

Snehal’s claim is FALSE

Let the two multiples of 3 be \(3m\) and \(3n\).

Sum: \(3m + 3n = 3(m + n)\)

This is a multiple of 6 when \((m + n)\) is even.

Generalisation: The sum of two multiples of 3 is a multiple of 6 iff the sum of their quotients (when divided by 3) is even.

Step 1: For divisibility by 18, the number must be divisible by both 2 and 9.

Step 2 (Divisibility by 2): \(b\) must be even. So \(b \in \{0, 2, 4, 6, 8\}\)

Step 3 (Divisibility by 9): Sum of digits must be divisible by 9.

Sum = \(4 + 8 + a + 2 + 3 + b = 17 + a + b\)

\(17 + a + b = 18\) or \(27\) (next multiple of 9, since max \(a + b = 17\))

Possible pairs \((a, b)\): (1, 0), (8, 2), (6, 4), (4, 6), (2, 8)

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

Step 2 (Divisibility by 4): Last two digits \(q8\) must be divisible by 4.

Possible: \(08, 28, 48, 68, 88\) → \(q \in \{0, 2, 4, 6, 8\}\)

Step 3 (Divisibility by 11): Alternating sum must be 0 or multiple of 11.

For \(3p7q8\): \((8 + 7 + 3) – (q + p) = 18 – (p + q)\)

Solving \(p + q = 7\) with valid \(q\):

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

Step 1: Let the numbers be \(n, n+1, n+2\).

\(n \equiv 0 \pmod 2\), \(n + 1 \equiv 0 \pmod 3\), \(n + 2 \equiv 0 \pmod 4\)

Step 2: From \(n + 2 \equiv 0 \pmod 4\): \(n \equiv 2 \pmod 4\), so \(n\) is even. ✓

From \(n + 1 \equiv 0 \pmod 3\): \(n \equiv 2 \pmod 3\)

Step 3: Need \(n \equiv 2 \pmod 4\) and \(n \equiv 2 \pmod 3\)

So \(n – 2\) is divisible by both 3 and 4, i.e., by LCM(3,4) = 12.

\(n = 12k + 2\) for integer \(k \geq 0\)

Answer: Such triples occur infinitely often, every 12 numbers (the LCM of 2, 3, 4).

Step 1: For a multiple of 36, the number must be divisible by both 9 (sum of digits) and 4 (last two digits).

Step 2: Find multiples of 36 near 45,000:

Five multiples: 45036, 45072, 45108, 45144, 45180

Each is 36 more than the previous one.

Five consecutive even numbers with middle term \(5p\):

Since consecutive even numbers differ by 2:

Answer: The other four numbers are \(5p – 4\), \(5p – 2\), \(5p + 2\), \(5p + 4\)

Step 1: Divisible by 15 means divisible by 3 and 5 → last digit is 0 or 5.

Step 2: Reversed number divisible by 6 means divisible by 2 and 3.

For reversed to be divisible by 2, the first digit of original must be even (this becomes the last digit of reversed).

Step 3: If original ends in 0, reversed starts with 0 → not a 6-digit number.

So original must end in 5, and first digit must be even (2, 4, 6, 8).

Step 4: Sum of digits must be divisible by 3.

Other examples: 200055, 200085, 202005, …

Step 1: Let \(n\) be a multiple of 11. Then \(n = 11k\) for some integer \(k\).

Step 2: When doubled: \(2n = 2 \times 11k = 11(2k)\)

Step 3: \(2n = 11 \times (\text{integer})\), which is clearly a multiple of 11.

Deepak’s conjecture is FALSE

Reason: If \(n = 11k\), then \(2n = 11(2k)\). Since \(2k\) is always an integer, \(2n\) is always divisible by 11.

Let the three numbers be \(n_1, n_2, n_3\).

When divided by 3, each leaves remainder 0, 1, or 2.

Generalisation: The sum of three numbers is divisible by 3 iff the sum of their remainders (when divided by 3) is divisible by 3 (i.e., equals 0, 3, or 6).

Key Relationships:

• Every multiple of 32 is a multiple of 8 (since \(32 = 8 \times 4\))
• Every multiple of 8 is a multiple of 4 (since \(8 = 4 \times 2\))
• So: Multiples of 32 ⊂ Multiples of 8 ⊂ Multiples of 4

This means 32 is the innermost set, 8 is in the middle, and 4 is the outermost set.

Correct Answer: Option (iv) — Three concentric circles with 32 innermost, 8 in middle, 4 outermost.