13 13 votes Which one of the following numbers is exactly divisible by $\left ( 11^{13} +1\right )$? $11^{26} +1$ $11^{33} +1$ $11^{39} -1$ $11^{52} -1$ Quantitative Aptitude gateee-2021 quantitative-aptitude number-system number-theory + – Arjun 1.2k views answer comment Share Follow 0 reply Please log in or register to add a comment.
Best answer 21 21 votes We know that, $x^{n} – y^{n}$ is divisible by $x+y,$ if $n$ is even. $x^{n} – y^{n}$ is divisible by $x-y,$ if $n$ is odd. $x^{n} + y^{n}$ is divisible by $x+y,$ if $n$ is odd. Now, we can check each option. $11^{26} + 1 = (11^{13})^{2} + 1^{2}$ is divisible by $11^{13} + 1,$ if $2$ is odd. $11^{33} + 1 = (11^{11})^{3} + 1^{3},$ here $11^{11} + 1 \neq 11^{13} + 1.$ $11^{39} - 1 = (11^{13})^{3} - 1^{3},$ here $11^{13} - 1 \neq 11^{13} + 1.$ $11^{52} - 1 = (11^{13})^{4} - 1^{4}$ is divisible by $11^{13} + 1,$ if $4$ is even. So, the correct answer is $(D).$ Lakshman Bhaiya answered Mar 27, 2021 • selected Apr 3, 2021 by Arjun Lakshman Bhaiya comment Share Follow 0 reply Please log in or register to add a comment.
17 17 votes 11⁵² - 1 ➠ (11²⁶)² - 1² a² - b² = (a + b) (a - b) ➠ (11²⁶ + 1) (11²⁶ - 1) ➠ (11²⁶ + 1) [(11¹³)² - 1²] ➠ (11²⁶ + 1) (11¹³ + 1) (11¹³ - 1) So , 11⁵² - 1 is exactly divisible by (11¹³ + 1) I hope this helps you ! Tharun_Kumar answered Apr 12, 2021 • moved May 13 by Arjun Tharun_Kumar comment Share Follow See 1 comment 1 1 comment reply anujs commented Sep 22, 2024 i moved by Arjun May 13 reply Follow flag nice approach @Tharun_Kumar 0 0 replyShare Please log in or register to add a comment.
1 1 vote 11⁵² - 1 ➠ (11²⁶)² - 1² a² - b² = (a + b) (a - b) ➠ (11²⁶ + 1) (11²⁶ - 1) ➠ (11²⁶ + 1) [(11¹³)² - 1²] ➠ (11²⁶ + 1) (11¹³ + 1) (11¹³ - 1) So , 11⁵² - 1 is exactly divisible by (11¹³ + 1) I hope this helps you ! Tharun_Kumar answered May 7, 2022 Tharun_Kumar comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Here is another method to solve this question. kp6602 answered Feb 11 kp6602 comment Share Follow 0 reply Please log in or register to add a comment.