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

Hint:

Notice remainder = divisor minus 2 in every case, so x + 2 is a common multiple of 6, 7, 8, 9, 10.

Answer 1

The Correct Option is A

Step 1: Set up the relation between divisor and remainder.
Check the gap between each divisor and its remainder: \(6-4=2\), \(7-5=2\), \(8-6=2\), \(9-7=2\), \(10-8=2\). Every divisor leaves a remainder exactly 2 less than itself. So if such a number is increased by 2, it becomes exactly divisible by 6, 7, 8, 9 and 10 all at once.

Step 2: Find the LCM of the divisors.
\(6 = 2 \times 3\), \(7 = 7\), \(8 = 2^3\), \(9 = 3^2\), \(10 = 2 \times 5\).
\(\text{LCM}(6,7,8,9,10) = 2^3 \times 3^2 \times 5 \times 7 = 2520\).
So \(x + 2\) must be a multiple of 2520, meaning \(x = 2520k - 2\) for some positive integer \(k\).

Step 3: Find the greatest 4-digit value.
For \(k = 3\): \(x = 2520 \times 3 - 2 = 7560 - 2 = 7558\), a 4-digit number.
For \(k = 4\): \(x = 2520 \times 4 - 2 = 10078\), a 5-digit number, too large.
So the greatest 4-digit value is \(x = 7558\).

Final Answer:
Sum of digits \(= 7 + 5 + 5 + 8 = 25\). \[ \boxed{25} \]