Which of the following is the largest 5-digit number divisible by 47?

Correct — C. Start at the top of the 5-digit range and strip off the remainder.

Largest 5-digit number = 99999
47 × 2000 = 94000, leaving 5999
47 × 100 = 4700, leaving 1299 (quotient so far 2100)
47 × 27 = 1269, leaving 30 (quotient 2127)

So 99999 = 47 × 2127 + 30.
99999 − 30 = 99969 → option (c)

Check: 99969 ÷ 47 = 2127 exactly, and the next multiple 47 × 2128 = 100016 has six digits.

Why the others are wrong

• (a)
  99999 — 99999 leaves a remainder of 30 on division by 47, since 99999 = 47 × 2127 + 30. It is the largest 5-digit number, not the largest multiple of 47.
• (b)
  10000 — 10000 is the smallest 5-digit number, and it is not a multiple of 47 either — it leaves a remainder of 36.
• (d)
  98888 — 98888 genuinely is a multiple of 47 (47 × 2104), which is what makes it tempting. It falls 1,081 short of 99969, so it is not the largest.

To find the largest n-digit multiple of d, divide the largest n-digit number by d and subtract the remainder. For the smallest, divide the smallest n-digit number and add (d − remainder).

Here 99999 = 47 × 2127 + 30, so 99999 − 30 = 99969 is the largest 5-digit multiple of 47.

47 is prime, so no divisibility rule shortens the work. One careful long division beats testing four options one at a time, especially when more than one option is a genuine multiple.

Key facts

• 99999 = 47 × 2127 + 30, so the largest 5-digit multiple of 47 is 99969.
• 47 × 2128 = 100016, which is a 6-digit number.
• Largest n-digit multiple of d = largest n-digit number minus its remainder on division by d.
• 98888 = 47 × 2104 is also a multiple of 47, but a smaller one.

Common traps

• Choosing 99999 because it is the largest 5-digit number, without dividing at all.
• Stopping at 98888 once it is confirmed to be a multiple of 47.

No directly related past PYQ was found.