Concept: $N \equiv \text{sum of digits} \pmod 3 \implies N \equiv 23 \equiv 2 \pmod 3$.

Given $N \equiv 7 \pmod{11}$.

Let $N = 11k + 7$. Testing values of $k$:

• $k = 0 \implies N = 7 \implies 7 \equiv 1 \pmod 3$

• $k = 1 \implies N = 18 \implies 18 \equiv 0 \pmod 3$

• $k = 2 \implies N = 29 \implies 29 \equiv 2 \pmod 3$

So, $N \equiv 29 \pmod{33}$.

Correct Answer: (b) 29

Concept: $13^{100} + 17^{100} \pmod{25}$.

Note $13 \equiv -12 \pmod{25}$ and $17 \equiv -8 \pmod{25}$.

Using Euler's Totient function $\phi(25) = 20$:

• $13^{100} = (13^{20})^5 \equiv 1^5 \equiv 1 \pmod{25}$

• $17^{100} = (17^{20})^5 \equiv 1^5 \equiv 1 \pmod{25}$

Sum $= 1 + 1 = 2 \pmod{25}$.

Correct Answer: (b) 2

Concept: $N = 18k + 7$.

Dividing $N$ by $12$:

• For even $k$ ($k = 0, 2, 4\dots$): $n = 7 \pmod{12}$

• For odd $k$ ($k = 1, 3, 5\dots$): $n = 13 \equiv 1 \pmod{12}$

$n$ can take $2$ possible values ($1$ and $7$).

Correct Answer: (a) 2

Concept: $N = 33k + 4$.

$N \pmod{55} = (33k + 4) \pmod{55}$.

Testing $k = 0, 1, 2, 3, 4$:

• $k = 0 \implies 4 \pmod{55}$

• $k = 1 \implies 37 \pmod{55}$

• $k = 2 \implies 70 \equiv 15 \pmod{55}$

• $k = 3 \implies 103 \equiv 48 \pmod{55}$

• $k = 4 \implies 136 \equiv 26 \pmod{55}$

There are $5$ possible remainders.

Correct Answer: (b) 5

Concept: $3^{90} + 5^{90} \pmod{34}$.

$3^{90} = (3^2)^{45} = 9^{45}$ and $5^{90} = (5^2)^{45} = 25^{45}$.

For odd $n$, $(a^n + b^n)$ is divisible by $(a + b)$.

Here $a = 9$, $b = 25 \implies a + b = 34$.

Thus, $(9^{45} + 25^{45})$ is fully divisible by $34 \implies \text{Remainder} = 0$.

Correct Answer: (a) 0

Concept: $N^2 \equiv 1 \pmod{24} \implies N^2 \equiv 1 \pmod{12}$.

Testing numbers mod $12$:

• $1^2 \equiv 1$, $5^2 = 25 \equiv 1$, $7^2 = 49 \equiv 1$, $11^2 = 121 \equiv 1 \pmod{12}$.

All odd numbers coprime to $12$ satisfy this condition ($1, 5, 7, 11$).

Correct Answer: (a) 1, 5, 7 and 11

Concept: $p = 28k + q$ where $p > 100$ is prime.

Since $p$ is prime:

• $q$ cannot be even (otherwise $28k + q$ is divisible by $2$).

• $q$ cannot be a multiple of $7$ (otherwise $28k + q$ is divisible by $7$).

Odd numbers less than $28$ not divisible by $7$:

Total values $= 12$.

Concept: $N \equiv 3 \pmod 7$ and $N \equiv 2 \pmod 4$.

Form: $N = 28k + r$.

• $28k + r \equiv 3 \pmod 7 \implies r \equiv 3 \pmod 7$ ($r \in \{3, 10, 17, 24\}$)

• $r \equiv 2 \pmod 4 \implies r = 10$.

So $N = 28k + 10$ for $0 \le N \le 1000$.

$0 \le 28k + 10 \le 1000 \implies 0 \le k \le 35$ ($36$ values).

Correct Answer: (b) 36

Concept: $43 + 47 + 49 = 139$.

$139 \equiv 9 \pmod N \implies 139 - 9 = 130$ is divisible by $N$.

Also, $N$ must be strictly greater than the individual remainders $\implies N > 49$.

Factors of $130$: $1, 2, 5, 10, 13, 26, 65, 130$.

Factors greater than $49$: $65$ and $130$.

Correct Answer: (d) More than one value is possible

Concept: $N = 14m + 3$.

We want $N \pmod{35} = (14m + 3) \pmod{35}$.

Period of $14m \pmod{35}$ repeats every $5$ steps ($m = 0, 1, 2, 3, 4$):

• $m = 0 \implies 3$

• $m = 1 \implies 17$

• $m = 2 \implies 31$

• $m = 3 \implies 45 \equiv 10$

• $m = 4 \implies 59 \equiv 24$

There are $5$ distinct values for $k$.

Correct Answer: (c) 5