When 12, 15, 18, 20 and 25 divide the least number x, the remainder in each case was 3 but x was divisible by 7. What is the value of x?

Correct option is C

Given:

When x is divided by 12, 15, 18, 20, and 25, the remainder in each case is 3.

x is divisible by 7.

Formula Used:

If a number x leaves the same remainder r when divided by several numbers, then:

x = LCM(a, b, c,… )×k + r

where k is a positive integer.

Solution:

LCM of 12, 15, 18, 20, and 25

Prime factorization of the numbers:

$$12 = 2^2 \times 3^1, \quad 15 = 3^1 \times 5^1, \quad 18 = 2^1 \times 3^2, \quad 20 = 2^2 \times 5^1, \quad 25 = 5^2$$​​

LCM(12, 15, 18, 20, 25) = $$2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25$$​ = 900

So, x = 900k + 3

Checking divisibility by 7

x = 900k + 3 to be divisible by 7.

Putting k = 1 , 2, 3, .. 

For k = 1 

= 900(1) + 3 = 903 divisible by 7 

So, the least number which satisfy the given condition is 903.