Find the smallest number to be added to 999, so that 99 divides the sum exactly.

Correct — A. Divide first, then top up whatever is missing.

999 ÷ 99: 99 × 10 = 990, so 999 leaves a remainder of 9.
The next multiple of 99 is 99 × 11 = 1,089.
Number to be added = 1,089 − 999 = 90.

The shortcut is the same arithmetic: add divisor − remainder = 99 − 9 = 90.
Check: 999 + 90 = 1,089 = 99 × 11 exactly → option (a)

Why the others are wrong

• (b)
  0 — 0 claims 999 is already a multiple of 99. It is not: 99 × 10 = 990 and 99 × 11 = 1,089 straddle it, leaving a remainder of 9.
• (c)
  9 — 9 is the remainder, which is what you subtract, not what you add. Adding it gives 1,008, and 99 does not divide 1,008.
• (d)
  99 — 99 adds a whole divisor to a number that was never a multiple, so 999 + 99 = 1,098 is still 9 past a multiple of 99.

Write the number as N = dq + r, with r the remainder on division by d.

To reach the next multiple upwards you must supply what the remainder is short of a full divisor: add d − r.

To reach the nearest multiple downwards you strip the remainder: subtract r. Both moves start from the same division, which is why the wrong one is such a tempting option.

Here d = 99, N = 999 and r = 9, so the addition is 99 − 9 = 90 and the subtraction would be 9.

Key facts

• To make N divisible by d you add d − (N mod d), the complement of the remainder.
• 999 leaves remainder 9 on division by 99, so 90 is the smallest number that can be added.
• 999 + 90 = 1,089 = 99 × 11.
• Because 100 ≡ 1 (mod 99), the remainder on 99 equals the remainder of the sum of the two-digit groups taken from the right.

Common traps

• Answering the remainder 9, which is the number to subtract rather than to add
• Adding the full divisor 99 on the assumption that a multiple of the divisor always preserves divisibility

No directly related past PYQ was found.