For divisibility by 8, 37N must be divisible by 8.

37N for values of N from 0 to 9 to find when = 0.

For N=6: 376 is divisible by 8 = 0 (divisible).

Therefore, the value of N is 6.

Among the given options:

For 4148

Sum of digits = 4 + 1 + 4 + 8 = 17

17 is not divisible by 3.

Hence, 4148 is not divisible by 24.

Option (a) 3816

Divisibility by 4:
Last two digits: 16.
16 ÷ 4 = 4 (no remainder).
3816 is divisible by 4.
Divisibility by 9:
Sum of digits: 3 + 8 + 1 + 6 = 18.
- 18 ÷ 9 = 2 (no remainder).
- 3816 is divisible by 9.
Conclusion: 3816 is divisible by 36.

Option (b) 3400
Divisibility by 4:
Last two digits: 00.
00 ÷ 4 = 0 (no remainder).
3400 is divisible by 4.
Divisibility by 9:
Sum of digits: 3 + 4 + 0 + 0 = 7.
7 ÷ 9 = 0.777... (remainder exists).
3400 is not divisible by 9.
Conclusion: 3400 is not divisible by 36.

Option (c) 3636
Divisibility by 4:
Last two digits: 36.
36 ÷ 4 = 9 (no remainder).
3636 is divisible by 4.
Divisibility by 9:
Sum of digits: 3 + 6 + 3 + 6 = 18.
18 ÷ 9 = 2 (no remainder).
3636 is divisible by 9.
Conclusion: 3636 is divisible by 36.

Option(d) 3708
Divisibility by 4:
Last two digits: 08.
08 ÷ 4 = 2 (no remainder)
3708 is divisible by 4.
Divisibility by 9:
Sum of digits: 3 + 7 + 0 + 8 = 18.
18 ÷ 9 = 2 (no remainder).
3708 is divisible by 9.
Conclusion: 3708 is divisible by 36.

Sum of digits = 6 + 2 + 7 + x + 5 = 20 + x

For the sum to be divisible by 9, x should be 7

Since 4081 is an odd number:

​(−1)⁴⁰⁸¹ = −1

Thus, the expression becomes:

​(265)⁴⁰⁸¹ + 9 = −1 + 9 = 8

Therefore, the remainder when is divided by 266 is 8.

According to the question;

33q + 0 = 33q

If 54 is added to the dividend then,

New number = 33q + 54

= 3 × (11q + 18)

So, we can clearly say that the new number is divisible by 3.

Let the number be 'N'. According to the problem, when the number is divided by 6 and by 5, it leaves a remainder of 4. This means:

N ≡ 4 (mod 6)

N ≡ 4 (mod 5)

To find the least number that satisfies these conditions, let's solve these two congruences using the least common multiple (LCM) method.

Since N gives a remainder of 4 when divided by both 6 and 5, we know that the difference between N and 4 must be divisible by both 6 and 5. So, we can express N as:

N = 6k + 4 (for some integer k)

Now, since N must also satisfy the condition N ≡ 4 (mod 5), substitute the expression from Step 1 into the second condition:

6k + 4 ≡ 4 (mod 5)

6k ≡ 0 (mod 5)

Since 6 ≡ 1 (mod 5), the equation becomes:

k ≡ 0 (mod 5)

Therefore, k must be a multiple of 5. Let k = 5m (for some integer m).

Substituting this into the equation for N, we get:

N = 6(5m) + 4 = 30m + 4

The smallest value of N occurs when m = 0:

N = 30(0) + 4 = 4

However, N = 4 does not satisfy the conditions for divisibility by 6 and 5. Therefore, we must find the next possible value of m that gives a valid solution.

For m = 1, we get:

N = 30(1) + 4 = 34

Thus, the least number that satisfies both conditions is N = 34.