Find the remainder, if 19 200 is divided by 20.

Correct — B. Work in remainders, not in powers. The stem asks for 19²⁰⁰ ÷ 20.

19 = 20 − 1, so 19 ≡ −1 (mod 20)
19²⁰⁰ ≡ (−1)²⁰⁰ (mod 20)
200 is even, so (−1)²⁰⁰ = +1

The remainder is 1 → option (b).

Sanity check by hand: 19² = 361 = 18 × 20 + 1, so every even power of 19 already leaves 1.

Why the others are wrong

• (a)
  3 — 3 is not in the cycle. Powers of 19 leave only 19 when the exponent is odd and 1 when it is even, so 3 never appears.
• (c)
  −1 — −1 is a residue, not a remainder. It is true that 19 ≡ −1 (mod 20), but the even exponent turns that into +1, and a remainder must lie between 0 and 19 in any case.
• (d)
  2 — 2 would require a third value in a cycle that has only two. Odd powers of 19 leave 19 and even powers leave 1.

When the base sits one below the divisor, the problem collapses. Write the base as (divisor − 1) and it becomes −1, and −1 raised to a power is +1 for an even exponent and −1 for an odd one.

Here 19 = 20 − 1, so 19²⁰⁰ ≡ (−1)²⁰⁰ ≡ 1 (mod 20), and the even exponent settles it in a single line.

Convert back before you answer. A remainder is by definition the least non-negative value in its class, so a −1 would be reported as 19, never as −1.

Key facts

• 19 ≡ −1 (mod 20), so 19 raised to an even power leaves remainder 1 and to an odd power leaves 19.
• A remainder on division by n lies in the range 0 to n − 1 by definition, so it is never negative.
• 19² = 361 = 18 × 20 + 1, confirming the remainder 1 by direct arithmetic.

Common traps

• Choosing −1 because 19 ≡ −1, and missing that the exponent 200 is even.
• Treating a remainder as a quantity that can be negative.
• Settling it by the unit digit: 19²⁰⁰ ends in 1, which only narrows the remainder modulo 20 to 1 or 11.

No directly related past PYQ was found.