How to Solve Remainder Theorem Problems
Solve remainder theorem aptitude problems using modular reduction, repeated squaring and Fermat’s theorem, with a worked example.
Expected Interview Answer
Remainder theorem problems are solved by working entirely in modular arithmetic — reducing each factor or term to its remainder modulo the divisor before combining — rather than computing the full expression and dividing at the end.
The core rule is that (a × b) mod m = ((a mod m) × (b mod m)) mod m, and the same holds for addition and subtraction, so any product or sum can be reduced piece by piece to keep numbers small. For powers, this becomes repeated squaring: reduce the base mod m, then square-and-reduce at each step instead of computing the full exponent. Fermat’s little theorem (a^(p-1) ≡ 1 mod p for prime p, gcd(a,p)=1) and Euler’s theorem generalize this to shortcut very large exponents. When a divisor is composite, splitting it into coprime factors and using the Chinese Remainder Theorem to recombine partial remainders is often the fastest path.
- Keeps intermediate numbers small and manageable
- Reduces huge powers to a handful of squaring steps
- Fermat’s/Euler’s theorems shortcut prime-modulus exponent problems
AI Mentor Explanation
A scorer tracking a team’s total only needs each batter’s contribution modulo the boundary needed to win, not their exact scores — if 7 runs are needed, a batter scoring 23 contributes the same 'remaining need' impact as one scoring 23 minus any multiple of 7 already accounted for. Remainder theorem problems work identically: reduce every term to its remainder modulo the divisor first, then combine those small remainders, rather than summing huge scores and dividing once at the end.
Worked example
Reduce base
- 7 mod 5 = 2
Fermat cycle
- 2^4 ≡ 1 mod 5
- 100 mod 4 = 0
Answer
- 7^100 mod 5 = 1
Step-by-Step Explanation
Step 1
Reduce each term mod the divisor
Replace every factor with its remainder before multiplying or adding.
Step 2
Combine reduced remainders
Multiply/add the small remainders, then reduce mod the divisor again.
Step 3
Apply repeated squaring for powers
Square-and-reduce at each step instead of computing the full exponent.
Step 4
Use Fermat’s/Euler’s theorem when the modulus is prime/coprime
Shortcut the exponent using a^(p-1) ≡ 1 mod p.
What Interviewer Expects
- Correct application of modular reduction before combining terms
- Recognizing when Fermat’s little theorem applies
- Correct handling of repeated squaring for large exponents
- Awareness of the Chinese Remainder Theorem for composite divisors
Common Mistakes
- Computing the full huge expression before reducing modulo the divisor
- Misapplying Fermat’s theorem when gcd(a, p) ≠ 1
- Forgetting to reduce intermediate results during repeated squaring
- Confusing remainder with quotient in the final answer
Best Answer (HR Friendly)
“The trick is to never let the numbers get large — reduce every factor to its remainder modulo the divisor first, then multiply or add those small remainders and reduce again. For large exponents I use repeated squaring, and if the divisor is prime I check whether Fermat’s little theorem shortcuts the exponent even further.”
Follow-up Questions
- How does the Chinese Remainder Theorem help when the divisor is composite?
- What changes if gcd(a, m) is not 1 when applying Euler’s theorem?
- How would you find the remainder of a factorial divided by a prime using Wilson’s theorem?
- How does repeated squaring reduce the number of multiplications needed for a^n mod m?
MCQ Practice
1. What is the remainder when 2^50 is divided by 7?
2 mod 7 cycles as 2^1=2, 2^2=4, 2^3=1 (cycle length 3). 50 mod 3 = 2, so 2^50 ≡ 2^2 = 4 mod 7.
2. What is the remainder when (17 × 23) is divided by 5?
17 mod 5 = 2, 23 mod 5 = 3, product mod 5 = 6 mod 5 = 1.
3. By Fermat’s little theorem, for prime p and gcd(a,p)=1, a^(p-1) mod p equals?
Fermat’s little theorem states a^(p-1) ≡ 1 (mod p) whenever gcd(a,p) = 1.
Flash Cards
Core modular multiplication rule? — (a×b) mod m = ((a mod m)×(b mod m)) mod m.
Fermat’s little theorem? — a^(p-1) ≡ 1 mod p for prime p, gcd(a,p)=1.
Technique for large exponents mod m? — Repeated squaring — square and reduce at each step.
When is Chinese Remainder Theorem used? — When the divisor is composite; split into coprime factors and recombine.