Find the smallest number which when diminished by 3 is divisible by 21, 24, 30 and 45.

Correct option is C

Given:

We need to find the smallest number which when diminished by 3 is divisible by 21, 24, 30, and 45.

Solution:

Prime factorizations:

21 = $$3 \times 7$$​​

24 = $$2^3 \times 3$$​​

30 = $$2 \times 3 \times 5$$​​

45 = $$3^2 \times 5$$​​

LCM will be the product of highest powers of all prime factors:

LCM = $$2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7$$​ = 2520

Now,

x = 2520 + 3 = 2523