Find out the smallest number, if it is divided by 15,18,27 and 35 we get 9 as remainder.

Correct option is D

Given:
A number when divided by 15, 18, 27, and 35 leaves a remainder of 9 in each case.

Formula Used:
LCM of numbers = Product of highest powers of all prime factors involved.

Solution:

Let the required number be N.

If a number N is divided by 15, 18, 27, and 35, the remainder is always 9.

This means N - 9 is divisible by all four numbers: 15, 18, 27, and 35.
So, N - 9 is the least common multiple (LCM) of 15, 18, 27, and 35.

N = LCM (15, 18, 27, 35) + 9

Prime factorizations:

15 = 3 × 5
18 = 2 × 3²
27 = 3³
35 = 5 × 7

LCM = $$2^1 \times 3^3 \times 5^1 \times 7^1 = 1890$$​​
​N = LCM + 9 = 1890+9 = 1899

So, the smallest number is 1899.

Thus, the correct answer is (d).