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Questions such as "find the remainder when 2¹⁰⁰ is divided by 7" look like they need a calculator. They don't. The binomial theorem turns them into two lines of algebra, as long as you make one good rewrite at the start. This post shows that rewrite, works through remainders, last two digits and a divisibility proof, and lists the slips that cost marks.
Where does this appear in the syllabus?
The JEE (Main) 2026 syllabus lists "Binomial theorem for a positive integral index, general term and middle term and simple applications". Remainder and divisibility questions are among those simple applications. NCERT Class 11 Chapter 7, Binomial Theorem works one in full (Example 4) and sets another as Exercise 7.1, Question 13. Check your own exam year's syllabus. For the general term and coefficients, see our post on the binomial theorem for JEE.
What is the key idea?
Write the base as a multiple of the divisor plus or minus 1. Then expand. Every term except the last has the divisor as a factor, so only the last term decides the remainder.
Worked example 1: remainder when 2¹⁰⁰ is divided by 7
Find a power of 2 close to a multiple of 7: 2³ = 8 = 7 + 1.
100 = 3 × 33 + 1, so 2¹⁰⁰ = 2 × (2³)³³ = 2 × (7 + 1)³³.
Expand (7 + 1)³³: every term except 1³³ = 1 contains 7. So (7 + 1)³³ = 7k + 1 for some integer k.
2¹⁰⁰ = 2(7k + 1) = 14k + 2. The remainder is 2.
Worked example 2: when the base is "minus one"
Find the remainder when 5⁹⁹ is divided by 13.
5² = 25 = 26 − 1, and 26 = 2 × 13.
5⁹⁹ = 5 × (5²)⁴⁹ = 5 × (26 − 1)⁴⁹.
Every term of (26 − 1)⁴⁹ except (−1)⁴⁹ = −1 contains 26. So 5⁹⁹ = 13m − 5 for some integer m.
A remainder cannot be negative. Write 13m − 5 = 13(m − 1) + 8. The remainder is 8.
Worked example 3: last two digits
The last two digits of a number are its remainder on division by 100. Find the last two digits of 11²⁵.
Write 11 = 10 + 1. Then (1 + 10)²⁵ = 1 + 25 × 10 + ²⁵C₂ × 10² + ... Every term from the third on contains 100, so 11²⁵ leaves the same remainder as 1 + 250 = 251. The last two digits are 51.
Here you must keep two terms, not one, because the second term, 250, is not a multiple of 100.
Worked example 4: a divisibility proof
NCERT Example 4 proves that 6ⁿ − 5n always leaves remainder 1 when divided by 25, by expanding (1 + 5)ⁿ. Try the same method on Exercise 7.1, Question 13: show that 9ⁿ⁺¹ − 8n − 9 is divisible by 64 for every positive integer n.
9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹ = 1 + (n + 1) × 8 + ⁿ⁺¹C₂ × 8² + ... + 8ⁿ⁺¹.
The first two terms are 1 + 8n + 8 = 8n + 9.
Every remaining term contains 8² = 64. So 9ⁿ⁺¹ = 8n + 9 + 64k.
Hence 9ⁿ⁺¹ − 8n − 9 = 64k, which is divisible by 64. ∎
Check with n = 1: 81 − 8 − 9 = 64. Correct.
Common mistakes and how to avoid them
Leaving a negative remainder. Add the divisor until the answer is between 0 and divisor − 1.
Dropping too many terms. When working mod 100 with a base of 10 + 1, the second term matters. Keep every term until the divisor clearly divides it.
Getting the exponent split wrong. Write 100 = 3 × 33 + 1 explicitly before rewriting 2¹⁰⁰.
Forgetting the sign of (−1)ⁿ. An odd power of −1 is −1; an even power is +1.
Two methods, one answer
You can also find 2¹⁰⁰ mod 7 by listing powers: 2, 4, 1, 2, 4, 1, ... The cycle has length 3, and 100 leaves remainder 1 on division by 3, so the answer is the first value, 2. Doing both methods is a strong check. Rittle-Johnson and Star (2007) found that 70 seventh-graders who compared alternative solution methods side by side made greater gains in procedural knowledge and flexibility than those who studied the methods one at a time.
A practice routine
Do five problems a week: two remainders with "+1" bases, two with "−1" bases, one last-two-digits problem.
Solve each one by the binomial method and by listing a cycle, and confirm they agree.
Mix these into sets with general-term and coefficient questions from the same chapter, so you have to recognise the type. Rohrer, Dedrick and Stershic found that 126 seventh-graders who practised mixed problem types scored higher on tests one day and 30 days later.
Prove one divisibility statement a week in full sentences, like NCERT's Example 4.
If you get a different answer from the two methods and cannot find the slip, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: find the remainder when 7¹⁰³ is divided by 25, using 7² = 49 = 50 − 1, then check that your answer is 18.
