Find the remainder when 9 20 + 2 is divided by 4.

Correct — C. The stem prints the power flat, as "9 20 + 2". It is 9²⁰ + 2, and the key's 3 confirms that reading.

9 = 4 × 2 + 1, so 9 leaves remainder 1 on division by 4
Reduce the base first: 9²⁰ ≡ 1²⁰ (mod 4) = 1
Add the 2: 1 + 2 = 3
3 is smaller than 4, so it is already the remainder: 3 → option (c)

Why the others are wrong

• (a)
  1 — 1 is the remainder of 9²⁰ on its own, since 9 ≡ 1 (mod 4). This drops the "+ 2" the stem attaches before the division.
• (b)
  2 — 2 is what the two misreadings give: 9 × 20 + 2 = 182 and 920 + 2 = 922 both leave remainder 2. The key's 3 rules both out and fixes the power reading.
• (d)
  0 — 0 would need 9²⁰ + 2 to be a multiple of 4. 9²⁰ is odd, so 9²⁰ + 2 is odd, and no odd number is divisible by 4.

Remainder questions on large powers are solved by reducing the base before taking the power, never by computing the power.

If a ≡ b (mod m), then aⁿ ≡ bⁿ (mod m). Here 9 ≡ 1 (mod 4), so every power of 9 is ≡ 1, whatever the exponent — 9², 9²⁰ and 9²⁰⁰ all leave remainder 1 on division by 4.

Once the power is reduced, the rest is arithmetic: add the constant and, if the total reaches the divisor, reduce once more.

Key facts

• 9 ≡ 1 (mod 4), so every power of 9 leaves remainder 1 when divided by 4.
• If a ≡ b (mod m) then aⁿ ≡ bⁿ (mod m) — reduce the base, then raise it.
• An odd number increased by 2 stays odd, so it can never be divisible by 4.
• A number is divisible by 4 when the number formed by its last two digits is.

Common traps

• Trying to evaluate 9²⁰ instead of reducing 9 to 1 modulo 4.
• Reading the flattened "9 20" as 9 × 20 or as 920, both of which give 2.
• Reducing the power correctly and then forgetting to add the constant 2.

No directly related past PYQ was found.