Question

Let \(x\) be the greatest number of 4 digits, which when divided by 15, 20 and 28 leaves in each case the remainder 2. The sum of digits of \(x\) is:

• A. \(19\)
• B. \(21\)
• C. \(23\)
• D. \(25\)

Hint:

When a number leaves the same remainder with multiple divisors, subtract the remainder first and then use the LCM of the divisors to find the required number.

Answer 1

The Correct Option is C

Step 1: Interpret the condition.
If \(x\) leaves remainder 2 when divided by 15, 20, and 28, it means: \[ x - 2 \ \text{is divisible by each of } 15, 20, \text{ and } 28. \] Step 2: Find the LCM of 15, 20, and 28.
Prime factorisations: \(15 = 3 \times 5\)
\(20 = 2^2 \times 5\)
\(28 = 2^2 \times 7\)
LCM \(= 2^2 \times 3 \times 5 \times 7 = 420\). So: \[ x - 2 = 420k \] where \(k\) is a positive integer. Step 3: Find the greatest 4-digit \(x\).
We need \(x \leq 9999\). Thus: \[ 420k + 2 \leq 9999 \quad \Rightarrow \quad 420k \leq 9997 \quad \Rightarrow \quad k \leq \frac{9997}{420} \approx 23.80. \] The largest integer \(k\) is \(23\). Therefore: \[ x = 420 \times 23 + 2 = 9660 + 2 = 9662. \] Step 4: Sum of digits.
Sum of digits of \(9662\) = \(9 + 6 + 6 + 2 = 23\). \[ \boxed{23} \]