Property: A 6-digit number formed by repeating a 3-digit number takes the algebraic form $abcabc = abc \times 1001$.

Factorization: $1001 = 7 \times 11 \times 13$.

The number $1001$ always divides any such number.

Correct Option: B) 1001

Property: $101$ divides any repunit of even length formed by pairs, because $10^2 \equiv -1 \pmod{101}$.

Analysis: A 100-digit number of all 7s consists of 50 pairs of "77". Since 100 is even, it forms 50 complete blocks of $7777$, each divisible by 101.

Remainder $= 0$.

Property: Any 6-digit repeating sequence of a digit (e.g., $777777$) is divisible by $1001 = 7 \times 11 \times 13$.

Both $N$ (96 sevens) and $M$ (96 threes) are formed by 96 digits. Since 96 is a multiple of 6 ($96 = 6 \times 16$), both $N$ and $M$ are completely divisible by 1001.

Thus, $N \times M$ leaves a remainder of $0$ when divided by 1001.

Correct Option: A) 0

Modulo Analysis: $10^3 = 1000 \equiv 1 \pmod{999}$.

Therefore, $10^3 \equiv 1$, $10^6 = (10^3)^2 \equiv 1^2 = 1$, $\dots$, up to $10^{99} = (10^3)^{33} \equiv 1 \pmod{999}$.

The sequence has 33 terms ($10^3, 10^6, \dots, 10^{99}$).

Sum $\equiv 1 + 1 + \dots + 1 \text{ (33 times)} = 33 \pmod{999}$.

Correct Option: B) 33

Rule for 99: A number is divisible by 99 if the sum of its 2-digit blocks from right to left is divisible by 99.

Blocks: $23 + 56 + 34 + 0A + 0B \equiv 23 + 56 + 34 + A + B = 113 + A + B$.

For $113 + A + B$ to be a multiple of 99 (next multiple is 198):

Alternatively, using standard divisibility by 9 and 11:

• Divisibility by 9: $3 + 4 + A + 5 + 6 + B + 2 + 3 = 23 + A + B$ must be a multiple of 9.

• Possible values for $A + B$ are $4$ or $13$.

• Divisibility by 11: $(3 + B + 5 + 3) - (2 + 6 + A + 4) = (11 + B) - (12 + A) = B - A - 1$ must be $0$ or a multiple of 11.

• If $B - A = 1$ and $A + B = 13$, then $2B = 14 \implies B = 7, A = 6$ (Valid).

Correct Option: B) 13

Expanded Form: $ABBA = 1000A + 100B + 10B + A = 1001A + 110B$.

Modulo 101: $1001 = 101 \times 9 + 92 \equiv 92 \equiv -9 \pmod{101}$, and $110 \equiv 9 \pmod{101}$.

Thus, $1001A + 110B \equiv -9A + 9B = 9(B - A) \pmod{101}$.

For this to be divisible by 101, $9(B - A)$ must be a multiple of 101, which requires $B - A = 0 \implies A = B$.

Since $A$ is the leading digit of a 4-digit number, $A \in \{1, 2, \dots, 9\}$ (9 choices).

Correct Option: A) 9

Property: The block-sum test of 3-digit groups from right to left corresponds to divisibility by $10^3 - 1 = 999$.

Since $999 = 27 \times 37$, any number whose 3-digit block sum is a multiple of 37 must be divisible by 37.

Correct Option: D) 37

Factorization: $10^{12} - 1 = (10^6 - 1)(10^6 + 1) = (10^3 - 1)(10^3 + 1)(10^6 + 1)$.

Note that $10^3 + 1 = 1001$.

Thus, $10^{12} - 1$ contains $1001$ as a factor, so the remainder when divided by 1001 is $0$.

Correct Option: A) 0

Definitions:

• $Y = 10^{60} - 1$.

• $X = \frac{5}{9}(10^{60} - 1)$.

$X \times Y = \frac{5}{9}(10^{60} - 1)^2$.

Fermat's Little Theorem: $10^{p-1} \equiv 1 \pmod p$ for prime $p$ coprime to 10.

• For $p = 7$: $10^6 \equiv 1 \implies 10^{60} \equiv 1 \pmod 7$.

• For $p = 11$: $10^{10} \equiv 1 \implies 10^{60} \equiv 1 \pmod{11}$.

• For $p = 13$: $10^{12} \equiv 1 \implies 10^{60} \equiv 1 \pmod{13}$.

• For $p = 17$: $p - 1 = 16$. Since 60 is not a multiple of 16, $10^{60} \not\equiv 1 \pmod{17}$.

Correct Option: D) 17

Given: $A = \frac{10^{50} - 1}{9}$.

Expression: $C = A(10^{50} + 1) + 10^{100} = \frac{(10^{50} - 1)(10^{50} + 1)}{9} + 10^{100} = \frac{10^{100} - 1}{9} + 10^{100}$.

Modulo 101: $10^2 = 100 \equiv -1 \pmod{101} \implies 10^{100} = (10^2)^{50} \equiv (-1)^{50} = 1 \pmod{101}$.

Substitute $10^{100} \equiv 1 \pmod{101}$:

Correct Option: A) 1