Last Digits Application
When the very large number 3^{100} is computed and its final two digits are examined, the binomial theorem reveals those last two digits to be which pair?
Select the correct option:
Solution
01
Finding the last two digits means working modulo 100, and the binomial theorem makes this tractable when the base power can be rewritten around a multiple of 10, a favourite JEE Advanced trick. Note 3^4 = 81, so 3^{100} = (3^4)^{25} = 81^{25} = (80 + 1)^{25}. Expanding, (80 + 1)^{25} = C(25,0)80^{25} + ... + C(25,24)80 + C(25,25). Every term containing 80 to a power of one or more is divisible by 100 except we must keep the term linear in 80: C(25,24)·80 = 25·80 = 2000, which is itself divisible by 100. All higher powers of 80 carry factors of 80^2 = 6400, also divisible by 100. Thus modulo 100 only the constant term C(25,25) = 1 survives, giving last two digits 01. Option 43 results from an arithmetic slip in the binomial reduction. Option 00 wrongly assumes full divisibility by 100. Option 21 mishandles the linear term. Plausibility check: since 81 ≡ 1 (mod 100) would be too strong, we verified the linear term 25·80 = 2000 ≡ 0, confirming 81^{25} ≡ 1 (mod 100) and last digits 01.
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About This Question
- Subject
- mathematics
- Chapter
- binomial theorem and its simple applications
- Topic
- last digits application
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
01
Finding the last two digits means working modulo 100, and the binomial theorem makes this tractable when the base power can be rewritten around a multiple of 10, a favourite JEE Advanced trick. Note 3^4 = 81, so 3^{100} = (3^4)^{25} = 81^{25} = (80 + 1)^{25}. Expanding, (80 + 1)^{25} = C(25,0)80^{25} + ... + C(25,24)80 + C(25,25). Every term containing 80 to a power of one or more is divisible by 100 except we must keep the term linear in 80: C(25,24)·80 = 25·80 = 2000, which is itself divisible by 100. All higher powers of 80 carry factors of 80^2 = 6400, also divisible by 100. Thus modulo 100 only the constant term C(25,25) = 1 survives, giving last two digits 01. Option 43 results from an arithmetic slip in the binomial reduction. Option 00 wrongly assumes full divisibility by 100. Option 21 mishandles the linear term. Plausibility check: since 81 ≡ 1 (mod 100) would be too strong, we verified the linear term 25·80 = 2000 ≡ 0, confirming 81^{25} ≡ 1 (mod 100) and last digits 01.
This hard difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of last digits application. It appeared in the 2025 exam.
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