Let x be the least 4 digit number which when divided by 2 3 4 5 6 and 7 leaves a remainder of in eac...........

Quantitative Aptitude (CPO) Number System

Let x be the least 4-digit number which when divided by 2, 3, 4, 5, 6 and 7 leaves a remainder of | in each case. If x lies between 2800

and 3000, then what is the sum of the digits of x?

This questions was previously asked in

SSC CPO 9th Dec 2019 Shift-2

- The problem asks for the smallest 4-digit number \( x \) which when divided by 2, 3, 4, 5, 6, and 7, leaves remainder 1 in each case.

- This means \( x - 1 \) is divisible by all these numbers. So, \( x - 1 \) must be a common multiple of 2, 3, 4, 5, 6, and 7.

- The least common multiple (LCM) of 2, 3, 4, 5, 6, and 7 is 420.

- Thus, \( x - 1 = 420k \Rightarrow x = 420k + 1 \).

- We need \( x \) between 2800 and 3000.

Let's try values of \( k \):

- For \( k = 7 \): \( x = 420 \times 7 + 1 = 2941 \)

- 2941 is between 2800 and 3000.

- Sum of digits: \( 2 + 9 + 4 + 1 = 16 \)

- Options:

1. 15

2. 16

3. 12

4. 13

Correct Answer:

Option 2, 16

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