Question

Find the largest four-digit number which when divided by 4, 7 and 13 leaves a remainder of 3 in each case.

Answer 1

Given: Find the largest four-digit number which, when divided by 4, 7 and 13, leaves remainder 3 in each case.

Step-wise calculation:

1. If n leaves remainder 3 on division by 4, 7 and 13, then n ≡ 3 (mod 4), n ≡ 3 (mod 7), n ≡ 3 (mod 13).

2. Hence, n – 3 is divisible by 4, 7 and 13, so n – 3 is a multiple of lcm(4, 7, 13).

3. lcm(4, 7, 13) = 4 × 7 × 13 = 364.

So, n = 364k + 3 for some integer k.

4. The largest four-digit number is 9999.

So, 364k + 3 ≤ 9999

⇒ 364k ≤ 9996

⇒ `k ≤ 9996/364`

5. Compute k:

364 × 27 = 9828

364 × 28 = 10192 (> 9996)

So, kmax = 27.

6. Therefore n = 364 × 27 + 3

= 9828 + 3

= 9831

7. Verification:

9831 ÷ 4 = 2457 R3,

9831 ÷ 7 = 1404 R3,

9831 ÷ 13 = 756 R3.

The largest four-digit number with the required property is 9831.