What is the greatest four digit number which when divided by 6, 20, 33 and 66, leaves 2,16, 29 and 62 as remainders, respectively?

Correct option is A

Given:

We are asked to find the greatest four-digit number which, when divided by 6, 20, 33, and 66, leaves remainders 2, 16, 29, and 62, respectively.

Concept Used:

The problem is based on a relationship where the difference between each divisor and its corresponding remainder is constant.

The solution can be expressed in the form LCM(a, b, c, d) × k − common difference,

where a, b, c, d are the divisors

Solution: 

Let the divisors be a, b, c, d and the remainders be x, y, z, w.

We have:  6 - 2 = 4 ,  20 - 16 = 4 ,  33 - 29 = 4 ,  66 - 62 = 4

Common difference = 4

Prime Factorization of the divisors:

6 = $$2 \times 3$$​​

20 = $$2^2 \times 5$$​​

33 = $$3 \times 11$$​​

66 = $$2 \times 3 \times 11$$​​

LCM of 6, 20, 33, and 66 $$= 2^2 \times 3 \times 5 \times 11 = 660$$​​

The number N can be expressed as:

N = 660k - 4

Now, 

​$$660k - 4 \leq 9999$$​​

​$$k \leq \frac{10003}{660} \approx 15.156$$​​

Since k must be an integer, the greatest possible value for k is 15

Substitute k = 15 into the expression for N:

N = 660 × 15 − 4 = 9900 − 4 = 9896

Therefore, the greatest four-digit number that satisfies the conditions is 9896.