Question

Difficulty: HardDivisibility Rules and Remainder Theorem

If the number 11202511^{2025} is decreased by 16202516^{2025}, and the resulting difference is divided by 99, what is the remainder?

  1. 7Answer
  2. B
    2
  3. C
    4
  4. D
    0

Answer

7
To find the remainder of 112025−16202511^{2025} - 16^{2025} divided by 99, we first simplify the bases using modular arithmetic: 11≡2(mod9)11 \equiv 2 \pmod 9 and 16≡−2(mod9)16 \equiv -2 \pmod 9. Next, we evaluate the powers. We know 23=8≡−1(mod9)2^3 = 8 \equiv -1 \pmod 9. By rewriting the exponent, 22025=(23)675≡(−1)675=−1(mod9)2^{2025} = (2^3)^{675} \equiv (-1)^{675} = -1 \pmod 9. For the second term, (−2)2025=−(22025)≡−(−1)=1(mod9)(-2)^{2025} = - (2^{2025}) \equiv -(-1) = 1 \pmod 9. Subtracting the two terms yields −1−1=−2-1 - 1 = -2. Because a remainder must be positive, we add the divisor to the negative result: −2+9=7-2 + 9 = 7.

Step-by-Step Solution

1
Reduce the base numbers modulo 9.
11≡2(mod9)11 \equiv 2 \pmod 9 and 16≡7≡−2(mod9)16 \equiv 7 \equiv -2 \pmod 9.
Reducing the bases simplifies the expression to 22025−(−2)2025(mod9)2^{2025} - (-2)^{2025} \pmod 9.
2
Determine the cyclicity pattern for powers of 2 modulo 9.
21=22^1=2, 22=42^2=4, 23=8≡−1(mod9)2^3=8 \equiv -1 \pmod 9.
Finding a power that leaves a remainder of 11 or −1-1 allows us to easily compute extremely large exponents.
3
Evaluate the first term 22025(mod9)2^{2025} \pmod 9.
22025=(23)675≡(−1)675=−1(mod9)2^{2025} = (2^3)^{675} \equiv (-1)^{675} = -1 \pmod 9.
Breaking the exponent 20252025 into 3×6753 \times 675 utilizes the known value of 23(mod9)2^3 \pmod 9.
4
Evaluate the second term (−2)2025(mod9)(-2)^{2025} \pmod 9.
(−2)2025=−(22025)≡−(−1)=1(mod9)(-2)^{2025} = - (2^{2025}) \equiv -(-1) = 1 \pmod 9.
An odd power preserves the negative sign, allowing substitution of the result from the previous step.
5
Subtract the remainders and map to a positive equivalent.
(−1)−1=−2(-1) - 1 = -2. To make it positive, add 99: −2+9=7-2 + 9 = 7.
Remainders must be strictly positive and less than the divisor. Adding the divisor to a negative modular result finds its positive equivalent.

Key Concept

Modular Arithmetic Cyclicity and Negative Remainders
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