Exams › JEE Main › Maths

When 2³⁵ * 3¹⁶ is divided by 11, what is the remainder?

1. 1
2. 3
3. 5
4. 8

Correct answer: 1

Solution

2¹⁰ ≡ 1 (mod 11) by Fermat's little theorem. 2³⁵ = (2¹⁰)³ * 2⁵ ≡ 1 * 32 = 32 ≡ 10 (mod 11). For 3: 3¹=3, 3²=9, 3³=27≡5, 3⁴≡15≡4, 3⁵≡12≡1 (mod 11). So order of 3 is 5. 3¹⁶ = 3^(15+1) = (3⁵)³ * 3 ≡ 1*3 = 3 (mod 11). Product: 2³⁵ * 3¹⁶ ≡ 10 * 3 = 30 ≡ 8 (mod 11). Remainder = 8.

Related JEE Main Maths questions

• Let A be the set {(n, 2n): n ∈ N} and let B be the set {(2n, 3n): n ∈ N}. What is the intersection A ∩ B?
• Let set A contain 3 elements and set B contain 6 elements. Then the cardinality of their union must satisfy
• Given the sets A = {1, 2, 5} and B = {3, 4, 5, 9}, what is A ∩ B?
• At a conference with 100 attendees, 29 are Indian women and 23 are Indian men. Among the Indian attendees, 4 are doctors, and 24 are either men or doctors. If there are no foreign doctors, how many foreigners and how many women doctors are present at the conference?
• Let X and Y be two non-empty sets, and let A be a non-empty set such that X ∩ A = Y ∩ A = A and X ∪ A = Y ∪ A. Which of the following must be true?
• If A and B are non-empty sets with A containing B, then which of the following is true?

⚔️ Practice JEE Main Maths free + battle 1v1 →