Without doing the actual division, find the remainder when 28735429 is divided by 9.

Correct — A. A number and its digit sum leave the same remainder on division by 9, because every power of 10 is one more than a multiple of 9.

Add the digits of 28735429:
2 + 8 + 7 + 3 + 5 + 4 + 2 + 9 = 40

Divide that digit sum by 9:
40 = 4 × 9 + 4, remainder 4 → option (a)

Faster still, cast out nines as you add: drop the 9, drop the 5 + 4, and 2 + 8 + 7 + 3 + 2 = 22, whose digits give 4.

Why the others are wrong

• (b)
  8 — 8 is this number's remainder on division by 11, not by 9: it is the alternating digit sum 9 − 2 + 4 − 5 + 3 − 7 + 8 − 2 = 8.
• (c)
  2 — 2 is what 28735429 leaves on division by 7, since 7 × 4105061 = 28735427, and no digit-sum shortcut can produce it — 7 has no digit-sum rule.
• (d)
  9 — A remainder on division by 9 must lie between 0 and 8, so 9 is impossible. If the digit sum lands on 9, or on any multiple of 9, the remainder is 0.

10 ≡ 1 (mod 9), and so is 10², 10³ and every higher power.

Write a number as digits times powers of ten, and modulo 9 it therefore collapses to the plain sum of its digits — the basis of both the divisibility test for 9 and the old 'casting out nines' check on long multiplication.

The same argument with 10 ≡ 1 (mod 3) gives the digit-sum test for 3.

For 11 the sign changes: 10 ≡ −1 (mod 11), so the digit sum alternates, which is why 11 has a rule of its own.

Key facts

• A number's remainder mod 9 equals its digit sum's remainder mod 9.
• The digit sum of 28735429 is 40, and 40 = 4 × 9 + 4, so the remainder is 4.
• Mod 11 uses the alternating digit sum, because 10 ≡ −1 (mod 11).
• A remainder on division by n always lies between 0 and n − 1.

Common traps

• Reading a digital root of 9 as a remainder of 9 — it means divisible by 9, remainder 0.
• Applying the alternating sum, which belongs to 11, not to 9.
• Mis-adding eight digits under time pressure instead of dropping 9s and pairs summing to 9 as you go.

No directly related past PYQ was found.