Question

Let x denote the greatest 4-digit number which when divided by 6, 7, 8, 9 and 10 leaves a remainder of 4, 5, 6 7 and 8 respectively. Then, the sum of the four-digits of x is

• A. 25
• B. 18
• C. 20
• D. 22

Answer 1

The Correct Option is A

To solve this problem, we need to find the greatest 4-digit number \(x\) such that:

• \(x \equiv 4 \pmod{6}\)
• \(x \equiv 5 \pmod{7}\)
• \(x \equiv 6 \pmod{8}\)
• \(x \equiv 7 \pmod{9}\)
• \(x \equiv 8 \pmod{10}\)

By Number Theory, when the remainders add up to give a certain constant when each divisor minus that remainder is calculated and is the same, this can be expressed by the equivalence \(x = k \cdot N + (N-r)\). Here, the given conditions can be reframed as \(x + 2\) is divisible by 6, 7, 8, 9, and 10. Therefore,:

\(x + 2 \equiv 0 \pmod{6, 7, 8, 9, 10}\)

Find \(LCM(6, 7, 8, 9, 10)\).

Prime factorizations are:

• 6 = \(2 \times 3\)
• 7 = \(7\)
• 8 = \(2^3\)
• 9 = \(3^2\)
• 10 = \(2 \times 5\)

The LCM is \(2^3 \times 3^2 \times 5 \times 7 = 2520\).

We seek the largest 4-digit number, hence:

\(x + 2 = 2520k\) where \(k\) is an integer.

The largest 4-digit number is 9999.

\(2520k \leq 10001\)

Solving for \(k\): \(k \leq \frac{10001}{2520} \approx 3.968\)

Max integer \(k\) is 3 giving:

\(x + 2 = 2520 \times 3 = 7560\)

Thus, \(x = 7560 - 2 = 7558\).

Sum of digits in 7558:

7 + 5 + 5 + 8 = 25

Thus, the sum of the digits of \(x\) is 25, and the correct answer is 25.